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	<id>https://wiki.ubc.ca/api.php?action=feedcontributions&amp;feedformat=atom&amp;user=CassandraTravlos</id>
	<title>UBC Wiki - User contributions [en]</title>
	<link rel="self" type="application/atom+xml" href="https://wiki.ubc.ca/api.php?action=feedcontributions&amp;feedformat=atom&amp;user=CassandraTravlos"/>
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	<updated>2026-09-19T12:16:23Z</updated>
	<subtitle>User contributions</subtitle>
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	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:CassandraTravlos&amp;diff=190080</id>
		<title>User:CassandraTravlos</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:CassandraTravlos&amp;diff=190080"/>
		<updated>2012-09-10T04:42:12Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: Blanked the page&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:CassandraTravlos/Essay&amp;diff=190079</id>
		<title>User:CassandraTravlos/Essay</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:CassandraTravlos/Essay&amp;diff=190079"/>
		<updated>2012-09-10T04:39:53Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: /* Calculus in The Faculty of Human Kinetics */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:CassandraTravlos/About_me&amp;diff=190078</id>
		<title>User:CassandraTravlos/About me</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:CassandraTravlos/About_me&amp;diff=190078"/>
		<updated>2012-09-10T04:39:02Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: Blanked the page&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_13&amp;diff=75272</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_13&amp;diff=75272"/>
		<updated>2011-02-04T06:24:00Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;      &lt;br /&gt;
&#039;&#039;&#039;Pick one of the topic offered below and then explain in your own words what it means that these concepts work on a logarithmic scale.&#039;&#039;&#039;     &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
* Our team has decided to look at PH and how it works on a logarithmic scale.     &lt;br /&gt;
&lt;br /&gt;
From previous knowledge we know that the pH scale is a range of acids to bases. The scale ranges from 0 to 14, with 0 being the most acidic and 14 being the most basic.     &lt;br /&gt;
&lt;br /&gt;
According to: http://www.sciencebuddies.org/science-fair-projects/project_ideas/Chem_AcidsBasespHScale.shtml     &lt;br /&gt;
&lt;br /&gt;
An acid can be defined as: a substance that has a very high concentration of hydrogen ions.     &lt;br /&gt;
From this we know that when an acid is dissolved in water, the balance between hydrogen ions and hydroxyl ions is shifted in the solution.     &lt;br /&gt;
Having more hydrogen ions=acidic solution.     &lt;br /&gt;
&lt;br /&gt;
A base can be defined as: a substance that accepts hydrogen ions.     &lt;br /&gt;
From this we know that when a base is dissolved in water, the balance between hydrogen ions and hydroxyl ions shifts the opposite way.     &lt;br /&gt;
Having more hydroxyl ions=basic solution or alkaline.&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
Acidity and alkalinity are measured on a logarithmic scale called the pH scale.     &lt;br /&gt;
&lt;br /&gt;
From our research we found that one of the reasons the pH scale uses a logarithmic scale is because if we take a really strong acidic solution, it can have one hundred million million (100,000,000,000,000) times more hydrogen ions than a strongly basic solution and a strongly basic solution can have 100,000,000,000,000 times more hydroxide ions than a strongly acidic solution the hydrogen ion and hydroxide ion concentrations these solutions can vary over that entire range.     &lt;br /&gt;
&lt;br /&gt;
The logarithmic scale shows the following pattern:     &lt;br /&gt;
Each single unit change in the pH scale corresponds to a change in hydrogen ion concentration of ten times!      &lt;br /&gt;
&lt;br /&gt;
So the benefit of using a logarithmic scale is that you don&#039;t have to write all the zeros.&lt;br /&gt;
&lt;br /&gt;
From further research on http://www.elmhurst.edu/~chm/vchembook/184ph.html we found examples to prove this:      &lt;br /&gt;
 &lt;br /&gt;
* A pH reading of 4 is ten times more acidic than a pH of 5 and 100 times (10 times 10) more acidic than a pH of 6&lt;br /&gt;
* A pH reading of 10 is ten times more basic than a pH of 9 and 100 times (10 times 10) more baisc than a pH of 8&lt;br /&gt;
&lt;br /&gt;
The following equation is used to calculate the pH or hydrogen ion concentration using logarithms:(found on http://www.elmhurst.edu/~chm/vchembook/184ph.html)    &lt;br /&gt;
&lt;br /&gt;
* &#039;&#039;&#039;pH is defined as follows: pH = -log10[H+]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This means that if you have a hydrogen ion [H+] concentrations of 0.1M (where M is the concentration in moles per liter)the equation would be used as seen below:&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
pH = - log10(0.1)&lt;br /&gt;
     &lt;br /&gt;
pH = - (-1)    &lt;br /&gt;
&lt;br /&gt;
pH = 1 , [0.1 = 10^-1, so log10(0.1) = -1]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Sample Problem:&#039;&#039;&#039;    &lt;br /&gt;
* if an acid has a hydrogen ion concentration [H+] of 0.0001 M, find the pH.     &lt;br /&gt;
&lt;br /&gt;
In order to solve this we can first convert the number to exponential notation, find the log, then solve the pH equation as shown below:&lt;br /&gt;
&lt;br /&gt;
[H+] 0.0001 M = 10^-4; log 10^-4= -4     &lt;br /&gt;
&lt;br /&gt;
pH = -log[H+]&lt;br /&gt;
&lt;br /&gt;
pH = -log(10^-4)&lt;br /&gt;
&lt;br /&gt;
pH = -(-4)&lt;br /&gt;
&lt;br /&gt;
pH = +4&lt;br /&gt;
&lt;br /&gt;
The purpose of the negative sign in the log definition is to give a positive pH value.&lt;br /&gt;
&lt;br /&gt;
Example is taken from: http://www.elmhurst.edu/~chm/vchembook/184ph.html&lt;br /&gt;
&lt;br /&gt;
[[Image:184phscale.gif|500px|left|thumb|Ph Scale]]&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_13&amp;diff=75271</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_13&amp;diff=75271"/>
		<updated>2011-02-04T06:23:34Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;      &lt;br /&gt;
&#039;&#039;&#039;Pick one of the topic offered below and then explain in your own words what it means that these concepts work on a logarithmic scale.&#039;&#039;&#039;     &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
* Our team has decided to look at PH and how it works on a logarithmic scale.     &lt;br /&gt;
&lt;br /&gt;
From previous knowledge we know that the pH scale is a range of acids to bases. The scale ranges from 0 to 14, with 0 being the most acidic and 14 being the most basic.     &lt;br /&gt;
&lt;br /&gt;
According to: http://www.sciencebuddies.org/science-fair-projects/project_ideas/Chem_AcidsBasespHScale.shtml     &lt;br /&gt;
&lt;br /&gt;
An acid can be defined as: a substance that has a very high concentration of hydrogen ions.     &lt;br /&gt;
From this we know that when an acid is dissolved in water, the balance between hydrogen ions and hydroxyl ions is shifted in the solution.     &lt;br /&gt;
Having more hydrogen ions=acidic solution.     &lt;br /&gt;
&lt;br /&gt;
A base can be defined as: a substance that accepts hydrogen ions.     &lt;br /&gt;
From this we know that when a base is dissolved in water, the balance between hydrogen ions and hydroxyl ions shifts the opposite way.     &lt;br /&gt;
Having more hydroxyl ions=basic solution or alkaline.&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
Acidity and alkalinity are measured on a logarithmic scale called the pH scale.     &lt;br /&gt;
&lt;br /&gt;
From our research we found that one of the reasons the pH scale uses a logarithmic scale is because if we take a really strong acidic solution, it can have one hundred million million (100,000,000,000,000) times more hydrogen ions than a strongly basic solution and a strongly basic solution can have 100,000,000,000,000 times more hydroxide ions than a strongly acidic solution the hydrogen ion and hydroxide ion concentrations these solutions can vary over that entire range.     &lt;br /&gt;
&lt;br /&gt;
The logarithmic scale shows the following pattern:     &lt;br /&gt;
Each single unit change in the pH scale corresponds to a change in hydrogen ion concentration of ten times!      &lt;br /&gt;
&lt;br /&gt;
So the benefit of using a logarithmic scale is that you don&#039;t have to write all the zeros.&lt;br /&gt;
&lt;br /&gt;
From further research on http://www.elmhurst.edu/~chm/vchembook/184ph.html we found examples to prove this:      &lt;br /&gt;
 &lt;br /&gt;
* A pH reading of 4 is ten times more acidic than a pH of 5 and 100 times (10 times 10) more acidic than a pH of 6&lt;br /&gt;
* A pH reading of 10 is ten times more basic than a pH of 9 and 100 times (10 times 10) more baisc than a pH of 8&lt;br /&gt;
&lt;br /&gt;
The following equation is used to calculate the pH or hydrogen ion concentration using logarithms:(found on http://www.elmhurst.edu/~chm/vchembook/184ph.html)    &lt;br /&gt;
&lt;br /&gt;
* &#039;&#039;&#039;pH is defined as follows: pH = -log10[H+]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This means that if you have a hydrogen ion [H+] concentrations of 0.1M (where M is the concentration in moles per liter)the equation would be used as seen below:&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
pH = - log10(0.1)     &lt;br /&gt;
pH = - (-1)     &lt;br /&gt;
pH = 1 , [0.1 = 10^-1, so log10(0.1) = -1]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Sample Problem:&#039;&#039;&#039;    &lt;br /&gt;
* if an acid has a hydrogen ion concentration [H+] of 0.0001 M, find the pH.     &lt;br /&gt;
&lt;br /&gt;
In order to solve this we can first convert the number to exponential notation, find the log, then solve the pH equation as shown below:&lt;br /&gt;
&lt;br /&gt;
[H+] 0.0001 M = 10^-4; log 10^-4= -4     &lt;br /&gt;
&lt;br /&gt;
pH = -log[H+]&lt;br /&gt;
pH = -log(10^-4)&lt;br /&gt;
pH = -(-4)&lt;br /&gt;
pH = +4&lt;br /&gt;
&lt;br /&gt;
The purpose of the negative sign in the log definition is to give a positive pH value.&lt;br /&gt;
&lt;br /&gt;
Example is taken from: http://www.elmhurst.edu/~chm/vchembook/184ph.html&lt;br /&gt;
&lt;br /&gt;
[[Image:184phscale.gif|500px|left|thumb|Ph Scale]]&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_13&amp;diff=75267</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_13&amp;diff=75267"/>
		<updated>2011-02-04T06:20:17Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;   &lt;br /&gt;
&#039;&#039;&#039;Pick one of the topic offered below and then explain in your own words what it means that these concepts work on a logarithmic scale.&#039;&#039;&#039;  &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
* Our team has decided to look at PH and how it works on a logarithmic scale.  &lt;br /&gt;
&lt;br /&gt;
From previous knowledge we know that the pH scale is a range of acids to bases. The scale ranges from 0 to 14, with 0 being the most acidic and 14 being the most basic.  &lt;br /&gt;
&lt;br /&gt;
According to: http://www.sciencebuddies.org/science-fair-projects/project_ideas/Chem_AcidsBasespHScale.shtml  &lt;br /&gt;
&lt;br /&gt;
An acid can be defined as: a substance that has a very high concentration of hydrogen ions.  &lt;br /&gt;
From this we know that when an acid is dissolved in water, the balance between hydrogen ions and hydroxyl ions is shifted in the solution.  &lt;br /&gt;
Having more hydrogen ions=acidic solution.  &lt;br /&gt;
&lt;br /&gt;
A base can be defined as: a substance that accepts hydrogen ions.  &lt;br /&gt;
From this we know that when a base is dissolved in water, the balance between hydrogen ions and hydroxyl ions shifts the opposite way.  &lt;br /&gt;
Having more hydroxyl ions=basic solution or alkaline.&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
Acidity and alkalinity are measured on a logarithmic scale called the pH scale.  &lt;br /&gt;
&lt;br /&gt;
From our research we found that one of the reasons the pH scale uses a logarithmic scale is because if we take a really strong acidic solution, it can have one hundred million million (100,000,000,000,000) times more hydrogen ions than a strongly basic solution and a strongly basic solution can have 100,000,000,000,000 times more hydroxide ions than a strongly acidic solution the hydrogen ion and hydroxide ion concentrations these solutions can vary over that entire range.  &lt;br /&gt;
&lt;br /&gt;
The logarithmic scale shows the following pattern:  &lt;br /&gt;
Each single unit change in the pH scale corresponds to a change in hydrogen ion concentration of ten times!   &lt;br /&gt;
&lt;br /&gt;
So the benefit of using a logarithmic scale is that you don&#039;t have to write all the zeros.&lt;br /&gt;
&lt;br /&gt;
From further research on http://www.elmhurst.edu/~chm/vchembook/184ph.html we found examples to prove this:   &lt;br /&gt;
 &lt;br /&gt;
* A pH reading of 4 is ten times more acidic than a pH of 5 and 100 times (10 times 10) more acidic than a pH of 6&lt;br /&gt;
* A pH reading of 10 is ten times more basic than a pH of 9 and 100 times (10 times 10) more baisc than a pH of 8&lt;br /&gt;
&lt;br /&gt;
The following equation is used to calculate the pH or hydrogen ion concentration using logarithms:(found on  http://www.elmhurst.edu/~chm/vchembook/184ph.html)&lt;br /&gt;
&lt;br /&gt;
* &#039;&#039;&#039;pH is defined as follows: pH = -log10[H+]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This means that if you have a hydrogen ion [H+] concentrations of 0.1M (where M is the concentration in moles per liter)the equation would be used as seen below:&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
pH = - log10(0.1)  &lt;br /&gt;
   = - (-1)  &lt;br /&gt;
   = 1 , [0.1 = 10^-1, so log10(0.1) = -1]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
An Example Problem: if an acid has a hydrogen ion [H+] concentration of 0.0001 M, find the pH.  &lt;br /&gt;
&lt;br /&gt;
In order to solve this we can first convert the number to exponential notation, find the log, then solve the pH equation as shown below:&lt;br /&gt;
&lt;br /&gt;
H^+ 0.0001 M = 10^-4; log 10^-4= -4  &lt;br /&gt;
&lt;br /&gt;
pH= -log(H^+)=-log(10^-4)=-(-4)=+4=pH&lt;br /&gt;
&lt;br /&gt;
The purpose of the negative sign in the log definition is to give a positive pH value.&lt;br /&gt;
&lt;br /&gt;
examples are taken from: http://www.elmhurst.edu/~chm/vchembook/184ph.html&lt;br /&gt;
&lt;br /&gt;
[[Image:184phscale.gif|500px|left|thumb|Ph Scale]]&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Thread:Course_talk:MATH110/003/Groups/Group_08/Basic_Skills_Project/Remarks_on_your_contribution_to_the_Basic_Skills_Project/reply&amp;diff=74329</id>
		<title>Thread:Course talk:MATH110/003/Groups/Group 08/Basic Skills Project/Remarks on your contribution to the Basic Skills Project/reply</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Thread:Course_talk:MATH110/003/Groups/Group_08/Basic_Skills_Project/Remarks_on_your_contribution_to_the_Basic_Skills_Project/reply&amp;diff=74329"/>
		<updated>2011-02-01T19:32:18Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: Reply to Remarks on your contribution to the Basic Skills Project&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;So how do we know if this work is adequate enuogh to recieve the 5%?&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_12&amp;diff=73007</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework 12</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_12&amp;diff=73007"/>
		<updated>2011-01-27T02:47:32Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;===Homework 12===&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
We are given the function:    &lt;br /&gt;
&lt;br /&gt;
1/(1+e^-t)&lt;br /&gt;
&lt;br /&gt;
* In order to change the height of the horizontal asymptote on the right, we must add a variable &#039;c&#039; to the equation:&lt;br /&gt;
  &lt;br /&gt;
(1/(1+e^-t))+ c    &lt;br /&gt;
 &lt;br /&gt;
* Here &#039;c&#039; is the factor which vertically translates the graph and this therefore changes the horizontal asymptote from its original position by a value of &#039;c&#039;. The horizontal asymptote of the original function is a y=1. Therefore we can say that K = c + 1.    &lt;br /&gt;
&lt;br /&gt;
* If we substitute in a value of 3 for &#039;c&#039; the right asymptote changes to y=4, which can also be found by using the equation:&lt;br /&gt;
 &lt;br /&gt;
K = c + 1&lt;br /&gt;
 &lt;br /&gt;
K= 3+1 =4&lt;br /&gt;
 &lt;br /&gt;
* We can see the following pattern occur by changing the value that that function is vertically translated by the value of &#039;c&#039; using this formula.    &lt;br /&gt;
&lt;br /&gt;
* To change the y-intercept to any number between 0 and K (which we chose to be 4 in this case) we can do this by adding another variable &#039;b&#039; into the equation:     &lt;br /&gt;
&lt;br /&gt;
(1/(1+(&#039;&#039;&#039;b&#039;&#039;&#039;)e^-t))+3 , where &#039;b&#039; cannot be equal to zero   &lt;br /&gt;
&lt;br /&gt;
* Here &#039;b&#039; provides us with a horizontal stretch with a dilation factor of &#039;b&#039;.    &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
The values of the original function are plotted as follows:&lt;br /&gt;
&lt;br /&gt;
x=-1, y=3.26&lt;br /&gt;
&lt;br /&gt;
x=0, y=3.5&lt;br /&gt;
&lt;br /&gt;
x=1, y=3.73&lt;br /&gt;
&lt;br /&gt;
x=2, y=3.88&lt;br /&gt;
&lt;br /&gt;
x=3, y=3.95&lt;br /&gt;
&lt;br /&gt;
x=4, y=3.982&lt;br /&gt;
&lt;br /&gt;
x=5, y=3.9933&lt;br /&gt;
&lt;br /&gt;
x=6, y=3.9975&lt;br /&gt;
&lt;br /&gt;
x=7, y=3.9991&lt;br /&gt;
&lt;br /&gt;
x=8, y=3.9997&lt;br /&gt;
&lt;br /&gt;
x=9, y=3.9999&lt;br /&gt;
&lt;br /&gt;
x=10, y=4&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we let b=7, then (1/(1+(7)e^-t))+3, then the values are shown as:&lt;br /&gt;
&lt;br /&gt;
x=-1, y=3.04&lt;br /&gt;
&lt;br /&gt;
x=0, y=3.125&lt;br /&gt;
&lt;br /&gt;
x=1, y=3.2797&lt;br /&gt;
&lt;br /&gt;
x=2, y=3.51&lt;br /&gt;
&lt;br /&gt;
x=3, y=3.74&lt;br /&gt;
&lt;br /&gt;
x=4, y=3.8864&lt;br /&gt;
&lt;br /&gt;
x=5, y=3.955&lt;br /&gt;
&lt;br /&gt;
x=6, y=3.9829&lt;br /&gt;
&lt;br /&gt;
x=7, y=3.9937&lt;br /&gt;
&lt;br /&gt;
x=8, y=3.9977&lt;br /&gt;
&lt;br /&gt;
x=9, y=3.9991&lt;br /&gt;
&lt;br /&gt;
x=10, y=3.9997&lt;br /&gt;
&lt;br /&gt;
x=11, y=3.9999&lt;br /&gt;
&lt;br /&gt;
x=12, y=4&lt;br /&gt;
&lt;br /&gt;
* Here we can see that by placing &#039;b&#039; into the equation we can alter the y-intercept but still maintain the same horizontal asymptote for both funtions!&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
BONUS&lt;br /&gt;
 &lt;br /&gt;
* In order to change the slope of the curved part so that the slope can go from very close to zero to almost vertical we must add yet another variable which we can call &#039;a&#039; to the function:&lt;br /&gt;
&lt;br /&gt;
a(1/(1+e^-)), where &#039;a&#039; cannot be equal to zero.&lt;br /&gt;
&lt;br /&gt;
* If we make &#039;a&#039; a very large number such as 1,000,000 the slope becomes almost vertical. If we make &#039;a&#039; a very small number such as 1/10000000, the slope gets close to zero. This is an example of a vertical strech of the function by a dilation factor of &#039;a&#039;.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Once you&#039;ve played with the function enough, try to find an application of the graph to model something. It can be anything which starts at a value and then goes to another one (think for a population, it goes from 0 to it&#039;s carrying capacity). Explain what you are modelling and how you decide to attribute a numerical value to each of the 2 or 3 parameters that you researched just above. Then use the model to make a prediction. For example, if your model is suppose to describe a population for which you have its initial population and carrying capacity (potentially its rate of increase if you solved the bonus part), then use that data to make a prediction for the population in 20 years, or use the model to predict when will the population reach 95% of its carrying capacity). When doing this last part, explain well where you&#039;re taking your data from (real data or imagined data), what it is that you&#039;re modelling and how you are doing the math to answer a predictive question.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
*&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_12&amp;diff=72719</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework 12</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_12&amp;diff=72719"/>
		<updated>2011-01-26T07:21:30Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;===Homework 12===&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
We are given the function:    &lt;br /&gt;
&lt;br /&gt;
1/(1+e^-t)&lt;br /&gt;
&lt;br /&gt;
* In order to change the height of the horizontal asymptote on the right, we must add a variable &#039;c&#039; to the equation:&lt;br /&gt;
  &lt;br /&gt;
(1/(1+e^-t))+ c    &lt;br /&gt;
 &lt;br /&gt;
* Here &#039;c&#039; is the factor which vertically translates the graph and this therefore changes the horizontal asymptote from its original position by a value of &#039;c&#039;. The horizontal asymptote of the original function is a y=1. Therefore we can say that K = c + 1.    &lt;br /&gt;
&lt;br /&gt;
* If we substitute in a value of 3 for &#039;c&#039; the right asymptote changes to y=4, which can also be found by using the equation:&lt;br /&gt;
 &lt;br /&gt;
K = c + 1&lt;br /&gt;
 &lt;br /&gt;
K= 3+1 =4&lt;br /&gt;
 &lt;br /&gt;
* We can see the following pattern occur by changing the value that that function is vertically translated by the value of &#039;c&#039; using this formula.    &lt;br /&gt;
&lt;br /&gt;
* To change the y-intercept to any number between 0 and K (which we chose to be 4 in this case) we can do this by adding another variable &#039;b&#039; into the equation:     &lt;br /&gt;
&lt;br /&gt;
(1/(1+(&#039;&#039;&#039;b&#039;&#039;&#039;)e^-t))+3 , where &#039;b&#039; cannot be equal to zero   &lt;br /&gt;
&lt;br /&gt;
* Here &#039;b&#039; provides us with a horizontal stretch with a dilation factor of &#039;b&#039;.    &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
The values of the original function are plotted as follows:&lt;br /&gt;
&lt;br /&gt;
x=-1, y=3.26&lt;br /&gt;
&lt;br /&gt;
x=0, y=3.5&lt;br /&gt;
&lt;br /&gt;
x=1, y=3.73&lt;br /&gt;
&lt;br /&gt;
x=2, y=3.88&lt;br /&gt;
&lt;br /&gt;
x=3, y=3.95&lt;br /&gt;
&lt;br /&gt;
x=4, y=3.982&lt;br /&gt;
&lt;br /&gt;
x=5, y=3.9933&lt;br /&gt;
&lt;br /&gt;
x=6, y=3.9975&lt;br /&gt;
&lt;br /&gt;
x=7, y=3.9991&lt;br /&gt;
&lt;br /&gt;
x=8, y=3.9997&lt;br /&gt;
&lt;br /&gt;
x=9, y=3.9999&lt;br /&gt;
&lt;br /&gt;
x=10, y=4&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we let b=7, then (1/(1+(7)e^-t))+3, then the values are shown as:&lt;br /&gt;
&lt;br /&gt;
x=-1, y=3.04&lt;br /&gt;
&lt;br /&gt;
x=0, y=3.125&lt;br /&gt;
&lt;br /&gt;
x=1, y=3.2797&lt;br /&gt;
&lt;br /&gt;
x=2, y=3.51&lt;br /&gt;
&lt;br /&gt;
x=3, y=3.74&lt;br /&gt;
&lt;br /&gt;
x=4, y=3.8864&lt;br /&gt;
&lt;br /&gt;
x=5, y=3.955&lt;br /&gt;
&lt;br /&gt;
x=6, y=3.9829&lt;br /&gt;
&lt;br /&gt;
x=7, y=3.9937&lt;br /&gt;
&lt;br /&gt;
x=8, y=3.9977&lt;br /&gt;
&lt;br /&gt;
x=9, y=3.9991&lt;br /&gt;
&lt;br /&gt;
x=10, y=3.9997&lt;br /&gt;
&lt;br /&gt;
x=11, y=3.9999&lt;br /&gt;
&lt;br /&gt;
x=12, y=4&lt;br /&gt;
&lt;br /&gt;
* Here we can see that by placing &#039;b&#039; into the equation we can alter the y-intercept but still maintain the same horizontal asymptote for both funtions!&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
BONUS&lt;br /&gt;
 &lt;br /&gt;
* In order to change the slope of the curved part so that the slope can go from very close to zero to almost vertical we must add yet another variable which we can call &#039;a&#039; to the function:&lt;br /&gt;
&lt;br /&gt;
a(1/(1+e^-)), where &#039;a&#039; cannot be equal to zero.&lt;br /&gt;
&lt;br /&gt;
* If we make &#039;a&#039; a very large number such as 1,000,000 the slope becomes almost vertical. If we make &#039;a&#039; a very small number such as 1/10000000, the slope gets close to zero. This is an example of a vertical strech of the function by a dilation factor of &#039;a&#039;.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Once you&#039;ve played with the function enough, try to find an application of the graph to model something. It can be anything which starts at a value and then goes to another one (think for a population, it goes from 0 to it&#039;s carrying capacity). Explain what you are modelling and how you decide to attribute a numerical value to each of the 2 or 3 parameters that you researched just above. Then use the model to make a prediction. For example, if your model is suppose to describe a population for which you have its initial population and carrying capacity (potentially its rate of increase if you solved the bonus part), then use that data to make a prediction for the population in 20 years, or use the model to predict when will the population reach 95% of its carrying capacity). When doing this last part, explain well where you&#039;re taking your data from (real data or imagined data), what it is that you&#039;re modelling and how you are doing the math to answer a predictive question.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
CAN SOMEONE IN THE GROUP DO THIS PART!!!!!!!!!!!!! &lt;br /&gt;
&lt;br /&gt;
If we use the example of having two fish initially and if they have babies, the graph shows the increase over time of fish in the aquarium. If we only have enough food to feed four fish then, we can see the four fish only live and the amount of fish in the aquarium plateaus.&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_12&amp;diff=72718</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework 12</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_12&amp;diff=72718"/>
		<updated>2011-01-26T07:20:49Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;===Homework 12===&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
We are given the function:    &lt;br /&gt;
&lt;br /&gt;
1/(1+e^-t)&lt;br /&gt;
&lt;br /&gt;
* In order to change the height of the horizontal asymptote on the right, we must add a variable &#039;c&#039; to the equation:&lt;br /&gt;
  &lt;br /&gt;
(1/(1+e^-t))+ c    &lt;br /&gt;
 &lt;br /&gt;
* Here &#039;c&#039; is the factor which vertically translates the graph and this therefore changes the horizontal asymptote from its original position by a value of &#039;c&#039;. The horizontal asymptote of the original function is a y=1. Therefore we can say that K = c + 1.    &lt;br /&gt;
&lt;br /&gt;
* If we substitute in a value of 3 for &#039;c&#039; the right asymptote changes to y=4, which can also be found by using the equation:&lt;br /&gt;
 &lt;br /&gt;
K = c + 1&lt;br /&gt;
 &lt;br /&gt;
K= 3+1 =4&lt;br /&gt;
 &lt;br /&gt;
* We can see the following pattern occur by changing the value that that function is vertically translated by the value of &#039;c&#039; using this formula.    &lt;br /&gt;
&lt;br /&gt;
* To change the y-intercept to any number between 0 and K (which we chose to be 4 in this case) we can do this by adding another variable &#039;b&#039; into the equation:     &lt;br /&gt;
&lt;br /&gt;
(1/(1+(&#039;&#039;&#039;b&#039;&#039;&#039;)e^-t))+3 , where &#039;b&#039; cannot be equal to zero   &lt;br /&gt;
&lt;br /&gt;
* Here &#039;b&#039; provides us with a horizontal stretch with a dilation factor of &#039;b&#039;.    &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
The values of the original function are plotted as follows:&lt;br /&gt;
&lt;br /&gt;
x=-1, y=3.26&lt;br /&gt;
&lt;br /&gt;
x=0, y=3.5&lt;br /&gt;
&lt;br /&gt;
x=1, y=3.73&lt;br /&gt;
&lt;br /&gt;
x=2, y=3.88&lt;br /&gt;
&lt;br /&gt;
x=3, y=3.95&lt;br /&gt;
&lt;br /&gt;
x=4, y=3.982&lt;br /&gt;
&lt;br /&gt;
x=5, y=3.9933&lt;br /&gt;
&lt;br /&gt;
x=6, y=3.9975&lt;br /&gt;
&lt;br /&gt;
x=7, y=3.9991&lt;br /&gt;
&lt;br /&gt;
x=8, y=3.9997&lt;br /&gt;
&lt;br /&gt;
x=9, y=3.9999&lt;br /&gt;
&lt;br /&gt;
x=10, y=4&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we let b=7, then (1/(1+(7)e^-t))+3, then the values are shown as:&lt;br /&gt;
&lt;br /&gt;
x=-1, y=3.04&lt;br /&gt;
&lt;br /&gt;
x=0, y=3.125&lt;br /&gt;
&lt;br /&gt;
x=1, y=3.2797&lt;br /&gt;
&lt;br /&gt;
x=2, y=3.51&lt;br /&gt;
&lt;br /&gt;
x=3, y=3.74&lt;br /&gt;
&lt;br /&gt;
x=4, y=3.8864&lt;br /&gt;
&lt;br /&gt;
x=5, y=3.955&lt;br /&gt;
&lt;br /&gt;
x=6, y=3.9829&lt;br /&gt;
&lt;br /&gt;
x=7, y=3.9937&lt;br /&gt;
&lt;br /&gt;
x=8, y=3.9977&lt;br /&gt;
&lt;br /&gt;
x=9, y=3.9991&lt;br /&gt;
&lt;br /&gt;
x=10, y=3.9997&lt;br /&gt;
&lt;br /&gt;
x=11, y=3.9999&lt;br /&gt;
&lt;br /&gt;
x=12, y=4&lt;br /&gt;
&lt;br /&gt;
* Here we can see that by placing &#039;b&#039; into the equation we can alter the y-intercept but still maintain the same horizontal asymptote for both funtions!&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
BONUS&lt;br /&gt;
 &lt;br /&gt;
* In order to change the slope of the curved part so that the slope can go from very close to zero to almost vertical we must add yet another variable which we can call &#039;a&#039; to the function:&lt;br /&gt;
&lt;br /&gt;
a(1/(1+e^-)), where &#039;a&#039; cannot be equal to zero.&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
* If we make &#039;a&#039; a very large number such as 1,000,000 the slope becomes almost vertical. If we make &#039;a&#039; a very small number such as 1/10000000, the slope gets close to zero. This is an example of a vertical strech of the function by a dilation factor of &#039;a&#039;.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Once you&#039;ve played with the function enough, try to find an application of the graph to model something. It can be anything which starts at a value and then goes to another one (think for a population, it goes from 0 to it&#039;s carrying capacity). Explain what you are modelling and how you decide to attribute a numerical value to each of the 2 or 3 parameters that you researched just above. Then use the model to make a prediction. For example, if your model is suppose to describe a population for which you have its initial population and carrying capacity (potentially its rate of increase if you solved the bonus part), then use that data to make a prediction for the population in 20 years, or use the model to predict when will the population reach 95% of its carrying capacity). When doing this last part, explain well where you&#039;re taking your data from (real data or imagined data), what it is that you&#039;re modelling and how you are doing the math to answer a predictive question.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
CAN SOMEONE IN THE GROUP DO THIS PART!!!!!!!!!!!!! &lt;br /&gt;
&lt;br /&gt;
If we use the example of having two fish initially and if they have babies, the graph shows the increase over time of fish in the aquarium. If we only have enough food to feed four fish then, we can see the four fish only live and the amount of fish in the aquarium plateaus.&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_12&amp;diff=72717</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework 12</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_12&amp;diff=72717"/>
		<updated>2011-01-26T07:19:56Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: /* Homework 12 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;===Homework 12===&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
We are given the function:    &lt;br /&gt;
&lt;br /&gt;
1/(1+e^-t)&lt;br /&gt;
&lt;br /&gt;
* In order to change the height of the horizontal asymptote on the right, we must add a variable &#039;c&#039; to the equation:&lt;br /&gt;
  &lt;br /&gt;
(1/(1+e^-t))+ c    &lt;br /&gt;
 &lt;br /&gt;
* Here &#039;c&#039; is the factor which vertically translates the graph and this therefore changes the horizontal asymptote from its original position by a value of &#039;c&#039;. The horizontal asymptote of the original function is a y=1. Therefore we can say that K = c + 1.    &lt;br /&gt;
&lt;br /&gt;
* If we substitute in a value of 3 for &#039;c&#039; the right asymptote changes to y=4, which can also be found by using the equation:&lt;br /&gt;
 &lt;br /&gt;
K = c + 1&lt;br /&gt;
 &lt;br /&gt;
K= 3+1 =4&lt;br /&gt;
 &lt;br /&gt;
* We can see the following pattern occur by changing the value that that function is vertically translated by the value of &#039;c&#039; using this formula.    &lt;br /&gt;
&lt;br /&gt;
* To change the y-intercept to any number between 0 and K (which we chose to be 4 in this case) we can do this by adding another variable &#039;b&#039; into the equation:     &lt;br /&gt;
&lt;br /&gt;
(1/(1+(&#039;&#039;&#039;b&#039;&#039;&#039;)e^-t))+3 , where &#039;b&#039; cannot be equal to zero   &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
* Here &#039;b&#039; provides us with a horizontal stretch with a dilation factor of &#039;b&#039;.    &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
The values of the original function are plotted as follows:&lt;br /&gt;
&lt;br /&gt;
x=-1, y=3.26&lt;br /&gt;
&lt;br /&gt;
x=0, y=3.5&lt;br /&gt;
&lt;br /&gt;
x=1, y=3.73&lt;br /&gt;
&lt;br /&gt;
x=2, y=3.88&lt;br /&gt;
&lt;br /&gt;
x=3, y=3.95&lt;br /&gt;
&lt;br /&gt;
x=4, y=3.982&lt;br /&gt;
&lt;br /&gt;
x=5, y=3.9933&lt;br /&gt;
&lt;br /&gt;
x=6, y=3.9975&lt;br /&gt;
&lt;br /&gt;
x=7, y=3.9991&lt;br /&gt;
&lt;br /&gt;
x=8, y=3.9997&lt;br /&gt;
&lt;br /&gt;
x=9, y=3.9999&lt;br /&gt;
&lt;br /&gt;
x=10, y=4&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we let b=7, then (1/(1+(7)e^-t))+3, then the values are shown as:&lt;br /&gt;
&lt;br /&gt;
x=-1, y=3.04&lt;br /&gt;
&lt;br /&gt;
x=0, y=3.125&lt;br /&gt;
&lt;br /&gt;
x=1, y=3.2797&lt;br /&gt;
&lt;br /&gt;
x=2, y=3.51&lt;br /&gt;
&lt;br /&gt;
x=3, y=3.74&lt;br /&gt;
&lt;br /&gt;
x=4, y=3.8864&lt;br /&gt;
&lt;br /&gt;
x=5, y=3.955&lt;br /&gt;
&lt;br /&gt;
x=6, y=3.9829&lt;br /&gt;
&lt;br /&gt;
x=7, y=3.9937&lt;br /&gt;
&lt;br /&gt;
x=8, y=3.9977&lt;br /&gt;
&lt;br /&gt;
x=9, y=3.9991&lt;br /&gt;
&lt;br /&gt;
x=10, y=3.9997&lt;br /&gt;
&lt;br /&gt;
x=11, y=3.9999&lt;br /&gt;
&lt;br /&gt;
x=12, y=4&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
* Here we can see that by placing &#039;b&#039; into the equation we can alter the y-intercept but still maintain the same horizontal asymptote for both funtions!&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
BONUS&lt;br /&gt;
 &lt;br /&gt;
* In order to change the slope of the curved part so that the slope can go from very close to zero to almost vertical we must add yet another variable which we can call &#039;a&#039; to the function:&lt;br /&gt;
&lt;br /&gt;
a(1/(1+e^-)), where &#039;a&#039; cannot be equal to zero.&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
* If we make &#039;a&#039; a very large number such as 1,000,000 the slope becomes almost vertical. If we make &#039;a&#039; a very small number such as 1/10000000, the slope gets close to zero. This is an example of a vertical strech of the function by a dilation factor of &#039;a&#039;.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Once you&#039;ve played with the function enough, try to find an application of the graph to model something. It can be anything which starts at a value and then goes to another one (think for a population, it goes from 0 to it&#039;s carrying capacity). Explain what you are modelling and how you decide to attribute a numerical value to each of the 2 or 3 parameters that you researched just above. Then use the model to make a prediction. For example, if your model is suppose to describe a population for which you have its initial population and carrying capacity (potentially its rate of increase if you solved the bonus part), then use that data to make a prediction for the population in 20 years, or use the model to predict when will the population reach 95% of its carrying capacity). When doing this last part, explain well where you&#039;re taking your data from (real data or imagined data), what it is that you&#039;re modelling and how you are doing the math to answer a predictive question.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
CAN SOMEONE IN THE GROUP DO THIS PART!!!!!!!!!!!!! &lt;br /&gt;
&lt;br /&gt;
If we use the example of having two fish initially and if they have babies, the graph shows the increase over time of fish in the aquarium. If we only have enough food to feed four fish then, we can see the four fish only live and the amount of fish in the aquarium plateaus.&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_12&amp;diff=72716</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework 12</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_12&amp;diff=72716"/>
		<updated>2011-01-26T07:18:35Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: /* Homework 12 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;===Homework 12===&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
We are given the function:    &lt;br /&gt;
&lt;br /&gt;
1/(1+e^-t)&lt;br /&gt;
&lt;br /&gt;
* In order to change the height of the horizontal asymptote on the right, we must add a variable &#039;c&#039; to the equation:&lt;br /&gt;
  &lt;br /&gt;
(1/(1+e^-t))+ c    &lt;br /&gt;
 &lt;br /&gt;
* Here &#039;c&#039; is the factor which vertically translates the graph and this therefore changes the horizontal asymptote from its original position by a value of &#039;c&#039;. The horizontal asymptote of the original function is a y=1. Therefore we can say that K = c + 1.    &lt;br /&gt;
&lt;br /&gt;
* If we substitute in a value of 3 for &#039;c&#039; the right asymptote changes to y=4, which can also be found by using the equation:&lt;br /&gt;
 &lt;br /&gt;
K = c + 1&lt;br /&gt;
 &lt;br /&gt;
K= 3+1 =4&lt;br /&gt;
 &lt;br /&gt;
* We can see the following pattern occur by changing the value that that function is vertically translated by the value of &#039;c&#039; using this formula.    &lt;br /&gt;
&lt;br /&gt;
* To change the y-intercept to any number between 0 and K (which we chose to be 4 in this case) we can do this by adding another variable &#039;b&#039; into the equation:     &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
(1/(1+(&#039;&#039;&#039;b&#039;&#039;&#039;)e^-t))+3 , where &#039;b&#039; cannot be equal to zero   &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
* Here &#039;b&#039; provides us with a horizontal stretch with a dilation factor of &#039;b&#039;.    &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
The values of the original function are plotted as follows:&lt;br /&gt;
&lt;br /&gt;
x=-1, y=3.26&lt;br /&gt;
&lt;br /&gt;
x=0, y=3.5&lt;br /&gt;
&lt;br /&gt;
x=1, y=3.73&lt;br /&gt;
&lt;br /&gt;
x=2, y=3.88&lt;br /&gt;
&lt;br /&gt;
x=3, y=3.95&lt;br /&gt;
&lt;br /&gt;
x=4, y=3.982&lt;br /&gt;
&lt;br /&gt;
x=5, y=3.9933&lt;br /&gt;
&lt;br /&gt;
x=6, y=3.9975&lt;br /&gt;
&lt;br /&gt;
x=7, y=3.9991&lt;br /&gt;
&lt;br /&gt;
x=8, y=3.9997&lt;br /&gt;
&lt;br /&gt;
x=9, y=3.9999&lt;br /&gt;
&lt;br /&gt;
x=10, y=4&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we let b=7, then (1/(1+(7)e^-t))+3, then the values are shown as:&lt;br /&gt;
&lt;br /&gt;
x=-1, y=3.04&lt;br /&gt;
&lt;br /&gt;
x=0, y=3.125&lt;br /&gt;
&lt;br /&gt;
x=1, y=3.2797&lt;br /&gt;
&lt;br /&gt;
x=2, y=3.51&lt;br /&gt;
&lt;br /&gt;
x=3, y=3.74&lt;br /&gt;
&lt;br /&gt;
x=4, y=3.8864&lt;br /&gt;
&lt;br /&gt;
x=5, y=3.955&lt;br /&gt;
&lt;br /&gt;
x=6, y=3.9829&lt;br /&gt;
&lt;br /&gt;
x=7, y=3.9937&lt;br /&gt;
&lt;br /&gt;
x=8, y=3.9977&lt;br /&gt;
&lt;br /&gt;
x=9, y=3.9991&lt;br /&gt;
&lt;br /&gt;
x=10, y=3.9997&lt;br /&gt;
&lt;br /&gt;
x=11, y=3.9999&lt;br /&gt;
&lt;br /&gt;
x=12, y=4&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
* Here we can see that by placing &#039;b&#039; into the equation we can alter the y-intercept but still maintain the same horizontal asymptote for both funtions!&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
* In order to change the slope of the curved part so that the slope can go from very close to zero to almost vertical we must add yet another variable which we can call &#039;a&#039; to the function:&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
a(1/(1+e^-)), where &#039;a&#039; cannot be equal to zero.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
* If we make &#039;a&#039; a very large number such as 1,000,000 the slope becomes almost vertical. If we make &#039;a&#039; a very small number such as 1/10000000, the slope gets close to zero. This is an example of a vertical strech of the function by a dilation factor of &#039;a&#039;.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Once you&#039;ve played with the function enough, try to find an application of the graph to model something. It can be anything which starts at a value and then goes to another one (think for a population, it goes from 0 to it&#039;s carrying capacity). Explain what you are modelling and how you decide to attribute a numerical value to each of the 2 or 3 parameters that you researched just above. Then use the model to make a prediction. For example, if your model is suppose to describe a population for which you have its initial population and carrying capacity (potentially its rate of increase if you solved the bonus part), then use that data to make a prediction for the population in 20 years, or use the model to predict when will the population reach 95% of its carrying capacity). When doing this last part, explain well where you&#039;re taking your data from (real data or imagined data), what it is that you&#039;re modelling and how you are doing the math to answer a predictive question.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
CAN SOMEONE IN THE GROUP DO THIS PART!!!!!!!!!!!!! &lt;br /&gt;
&lt;br /&gt;
If we use the example of having two fish initially and if they have babies, the graph shows the increase over time of fish in the aquarium. If we only have enough food to feed four fish then, we can see the four fish only live and the amount of fish in the aquarium plateaus.&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_12&amp;diff=72715</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework 12</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_12&amp;diff=72715"/>
		<updated>2011-01-26T07:17:05Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;===Homework 12===&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
We are given the function:    &lt;br /&gt;
&lt;br /&gt;
1/(1+e^-t)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
* In order to change the height of the horizontal asymptote on the right, we must add a variable &#039;c&#039; to the equation:&lt;br /&gt;
  &lt;br /&gt;
(1/(1+e^-t))+ c    &lt;br /&gt;
 &lt;br /&gt;
* Here &#039;c&#039; is the factor which vertically translates the graph and this therefore changes the horizontal asymptote from its original position by a value of &#039;c&#039;. The horizontal asymptote of the original function is a y=1. Therefore we can say that K = c + 1.    &lt;br /&gt;
&lt;br /&gt;
* If we substitute in a value of 3 for &#039;c&#039; the right asymptote changes to y=4, which can also be found by using the equation:&lt;br /&gt;
 &lt;br /&gt;
K = c + 1&lt;br /&gt;
 &lt;br /&gt;
K= 3+1 =4&lt;br /&gt;
 &lt;br /&gt;
* We can see the following pattern occur by changing the value that that function is vertically translated by the value of &#039;c&#039; using this formula.    &lt;br /&gt;
&lt;br /&gt;
* To change the y-intercept to any number between 0 and K (which we chose to be 4 in this case) we can do this by adding another variable &#039;b&#039; into the equation:     &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
(1/(1+(&#039;&#039;&#039;b&#039;&#039;&#039;)e^-t))+3 , where &#039;b&#039; cannot be equal to zero   &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
* Here &#039;b&#039; provides us with a horizontal stretch with a dilation factor of &#039;b&#039;.    &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
The values of the original function are plotted as follows:&lt;br /&gt;
&lt;br /&gt;
x=-1, y=3.26&lt;br /&gt;
&lt;br /&gt;
x=0, y=3.5&lt;br /&gt;
&lt;br /&gt;
x=1, y=3.73&lt;br /&gt;
&lt;br /&gt;
x=2, y=3.88&lt;br /&gt;
&lt;br /&gt;
x=3, y=3.95&lt;br /&gt;
&lt;br /&gt;
x=4, y=3.982&lt;br /&gt;
&lt;br /&gt;
x=5, y=3.9933&lt;br /&gt;
&lt;br /&gt;
x=6, y=3.9975&lt;br /&gt;
&lt;br /&gt;
x=7, y=3.9991&lt;br /&gt;
&lt;br /&gt;
x=8, y=3.9997&lt;br /&gt;
&lt;br /&gt;
x=9, y=3.9999&lt;br /&gt;
&lt;br /&gt;
x=10, y=4&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we let b=7, then (1/(1+(7)e^-t))+3, then the values are shown as:&lt;br /&gt;
&lt;br /&gt;
x=-1, y=3.04&lt;br /&gt;
&lt;br /&gt;
x=0, y=3.125&lt;br /&gt;
&lt;br /&gt;
x=1, y=3.2797&lt;br /&gt;
&lt;br /&gt;
x=2, y=3.51&lt;br /&gt;
&lt;br /&gt;
x=3, y=3.74&lt;br /&gt;
&lt;br /&gt;
x=4, y=3.8864&lt;br /&gt;
&lt;br /&gt;
x=5, y=3.955&lt;br /&gt;
&lt;br /&gt;
x=6, y=3.9829&lt;br /&gt;
&lt;br /&gt;
x=7, y=3.9937&lt;br /&gt;
&lt;br /&gt;
x=8, y=3.9977&lt;br /&gt;
&lt;br /&gt;
x=9, y=3.9991&lt;br /&gt;
&lt;br /&gt;
x=10, y=3.9997&lt;br /&gt;
&lt;br /&gt;
x=11, y=3.9999&lt;br /&gt;
&lt;br /&gt;
x=12, y=4&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
* Here we can see that by placing &#039;b&#039; into the equation we can alter the y-intercept but still maintain the same horizontal asymptote for both funtions!&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
* In order to change the slope of the curved part so that the slope can go from very close to zero to almost vertical we must add yet another variable which we can call &#039;a&#039; to the function:&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
a(1/(1+e^-)), where &#039;a&#039; cannot be equal to zero.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
* If we make &#039;a&#039; a very large number such as 1,000,000 the slope becomes almost vertical. If we make &#039;a&#039; a very small number such as 1/10000000, the slope gets close to zero. This is an example of a vertical strech of the function by a dilation factor of &#039;a&#039;.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Once you&#039;ve played with the function enough, try to find an application of the graph to model something. It can be anything which starts at a value and then goes to another one (think for a population, it goes from 0 to it&#039;s carrying capacity). Explain what you are modelling and how you decide to attribute a numerical value to each of the 2 or 3 parameters that you researched just above. Then use the model to make a prediction. For example, if your model is suppose to describe a population for which you have its initial population and carrying capacity (potentially its rate of increase if you solved the bonus part), then use that data to make a prediction for the population in 20 years, or use the model to predict when will the population reach 95% of its carrying capacity). When doing this last part, explain well where you&#039;re taking your data from (real data or imagined data), what it is that you&#039;re modelling and how you are doing the math to answer a predictive question.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
CAN SOMEONE IN THE GROUP DO THIS PART!!!!!!!!!!!!! &lt;br /&gt;
&lt;br /&gt;
If we use the example of having two fish initially and if they have babies, the graph shows the increase over time of fish in the aquarium. If we only have enough food to feed four fish then, we can see the four fish only live and the amount of fish in the aquarium plateaus.&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_12&amp;diff=72714</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework 12</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_12&amp;diff=72714"/>
		<updated>2011-01-26T07:15:53Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;===Homework 12===&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
We are given the function:    &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
1/(1+e^-t)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
* In order to change the height of the horizontal asymptote on the right, we must add a variable &#039;c&#039; to the equation:&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
(1/(1+e^-t))+ c    &lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
* Here &#039;c&#039; is the factor which vertically translates the graph and this therefore changes the horizontal asymptote from its original position by a value of &#039;c&#039;. The horizontal asymptote of the original function is a y=1. Therefore we can say that K = c + 1.    &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
* If we substitute in a value of 3 for &#039;c&#039; the right asymptote changes to y=4, which can also be found by using the equation:&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
K = c + 1&lt;br /&gt;
 &lt;br /&gt;
K= 3+1 =4&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
* We can see the following pattern occur by changing the value that that function is vertically translated by the value of &#039;c&#039; using this formula.    &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
* To change the y-intercept to any number between 0 and K (which we chose to be 4 in this case) we can do this by adding another variable &#039;b&#039; into the equation:     &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
(1/(1+(&#039;&#039;&#039;b&#039;&#039;&#039;)e^-t))+3 , where &#039;b&#039; cannot be equal to zero   &lt;br /&gt;
&lt;br /&gt;
* Here &#039;b&#039; provides us with a horizontal stretch with a dilation factor of &#039;b&#039;.    &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
The values of the original function are plotted as follows:&lt;br /&gt;
&lt;br /&gt;
x=-1, y=3.26&lt;br /&gt;
&lt;br /&gt;
x=0, y=3.5&lt;br /&gt;
&lt;br /&gt;
x=1, y=3.73&lt;br /&gt;
&lt;br /&gt;
x=2, y=3.88&lt;br /&gt;
&lt;br /&gt;
x=3, y=3.95&lt;br /&gt;
&lt;br /&gt;
x=4, y=3.982&lt;br /&gt;
&lt;br /&gt;
x=5, y=3.9933&lt;br /&gt;
&lt;br /&gt;
x=6, y=3.9975&lt;br /&gt;
&lt;br /&gt;
x=7, y=3.9991&lt;br /&gt;
&lt;br /&gt;
x=8, y=3.9997&lt;br /&gt;
&lt;br /&gt;
x=9, y=3.9999&lt;br /&gt;
&lt;br /&gt;
x=10, y=4&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we let b=7, then (1/(1+(7)e^-t))+3, then the values are shown as:&lt;br /&gt;
&lt;br /&gt;
x=-1, y=3.04&lt;br /&gt;
&lt;br /&gt;
x=0, y=3.125&lt;br /&gt;
&lt;br /&gt;
x=1, y=3.2797&lt;br /&gt;
&lt;br /&gt;
x=2, y=3.51&lt;br /&gt;
&lt;br /&gt;
x=3, y=3.74&lt;br /&gt;
&lt;br /&gt;
x=4, y=3.8864&lt;br /&gt;
&lt;br /&gt;
x=5, y=3.955&lt;br /&gt;
&lt;br /&gt;
x=6, y=3.9829&lt;br /&gt;
&lt;br /&gt;
x=7, y=3.9937&lt;br /&gt;
&lt;br /&gt;
x=8, y=3.9977&lt;br /&gt;
&lt;br /&gt;
x=9, y=3.9991&lt;br /&gt;
&lt;br /&gt;
x=10, y=3.9997&lt;br /&gt;
&lt;br /&gt;
x=11, y=3.9999&lt;br /&gt;
&lt;br /&gt;
x=12, y=4&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
* Here we can see that by placing &#039;b&#039; into the equation we can alter the y-intercept but still maintain the same horizontal asymptote for both funtions!&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
* In order to change the slope of the curved part so that the slope can go from very close to zero to almost vertical we must add yet another variable which we can call &#039;a&#039; to the function:&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
a(1/(1+e^-)), where &#039;a&#039; cannot be equal to zero.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
* If we make &#039;a&#039; a very large number such as 1,000,000 the slope becomes almost vertical. If we make &#039;a&#039; a very small number such as 1/10000000, the slope gets close to zero. This is an example of a vertical strech of the function by a dilation factor of &#039;a&#039;.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Once you&#039;ve played with the function enough, try to find an application of the graph to model something. It can be anything which starts at a value and then goes to another one (think for a population, it goes from 0 to it&#039;s carrying capacity). Explain what you are modelling and how you decide to attribute a numerical value to each of the 2 or 3 parameters that you researched just above. Then use the model to make a prediction. For example, if your model is suppose to describe a population for which you have its initial population and carrying capacity (potentially its rate of increase if you solved the bonus part), then use that data to make a prediction for the population in 20 years, or use the model to predict when will the population reach 95% of its carrying capacity). When doing this last part, explain well where you&#039;re taking your data from (real data or imagined data), what it is that you&#039;re modelling and how you are doing the math to answer a predictive question.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
CAN SOMEONE IN THE GROUP DO THIS PART!!!!!!!!!!!!! &lt;br /&gt;
&lt;br /&gt;
If we use the example of having two fish initially and if they have babies, the graph shows the increase over time of fish in the aquarium. If we only have enough food to feed four fish then, we can see the four fish only live and the amount of fish in the aquarium plateaus.&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_12&amp;diff=72713</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework 12</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_12&amp;diff=72713"/>
		<updated>2011-01-26T07:14:55Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;===Homework 12===&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
We are given the function:    &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
1/(1+e^-t)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
* In order to change the height of the horizontal asymptote on the right, we must add a variable &#039;c&#039; to the equation:&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
(1/(1+e^-t))+ c    &lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
* Here &#039;c&#039; is the factor which vertically translates the graph and this therefore changes the horizontal asymptote from its original position by a value of &#039;c&#039;. The horizontal asymptote of the original function is a y=1. Therefore we can say that K = c + 1.    &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
* If we substitute in a value of 3 for &#039;c&#039; the right asymptote changes to y=4, which can also be found by using the equation:&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
K = c + 1&lt;br /&gt;
 &lt;br /&gt;
K= 3+1 =4&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
* We can see the following pattern occur by changing the value that that function is vertically translated by the value of &#039;c&#039; using this formula.    &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
* To change the y-intercept to any number between 0 and K (which we chose to be 4 in this case) we can do this by adding another variable &#039;b&#039; into the equation:     &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
(1/(1+(&#039;&#039;&#039;b&#039;&#039;&#039;)e^-t))+3 , where &#039;b&#039; cannot be equal to zero   &lt;br /&gt;
&lt;br /&gt;
* Here &#039;b&#039; provides us with a horizontal stretch with a dilation factor of &#039;b&#039;.    &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
The values of the original function are plotted as follows:&lt;br /&gt;
&lt;br /&gt;
x=-1, y=3.26&lt;br /&gt;
&lt;br /&gt;
x=0, y=3.5&lt;br /&gt;
&lt;br /&gt;
x=1, y=3.73&lt;br /&gt;
&lt;br /&gt;
x=2, y=3.88&lt;br /&gt;
&lt;br /&gt;
x=3, y=3.95&lt;br /&gt;
&lt;br /&gt;
x=4, y=3.982&lt;br /&gt;
&lt;br /&gt;
x=5, y=3.9933&lt;br /&gt;
&lt;br /&gt;
x=6, y=3.9975&lt;br /&gt;
&lt;br /&gt;
x=7, y=3.9991&lt;br /&gt;
&lt;br /&gt;
x=8, y=3.9997&lt;br /&gt;
&lt;br /&gt;
x=9, y=3.9999&lt;br /&gt;
&lt;br /&gt;
x=10, y=4&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we let b=7, then (1/(1+(7)e^-t))+3, then the values are shown as:&lt;br /&gt;
&lt;br /&gt;
x=-1, y=3.04&lt;br /&gt;
&lt;br /&gt;
x=0, y=3.125&lt;br /&gt;
&lt;br /&gt;
x=1, y=3.2797&lt;br /&gt;
&lt;br /&gt;
x=2, y=3.51&lt;br /&gt;
&lt;br /&gt;
x=3, y=3.74&lt;br /&gt;
&lt;br /&gt;
x=4, y=3.8864&lt;br /&gt;
&lt;br /&gt;
x=5, y=3.955&lt;br /&gt;
&lt;br /&gt;
x=6, y=3.9829&lt;br /&gt;
&lt;br /&gt;
x=7, y=3.9937&lt;br /&gt;
&lt;br /&gt;
x=8, y=3.9977&lt;br /&gt;
&lt;br /&gt;
x=9, y=3.9991&lt;br /&gt;
&lt;br /&gt;
x=10, y=3.9997&lt;br /&gt;
&lt;br /&gt;
x=11, y=3.9999&lt;br /&gt;
&lt;br /&gt;
x=12, y=4&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
* Here we can see that by placing &#039;b&#039; into the equation we can alter the y-intercept but still maintain the same horizontal asymptote for both funtions!&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
* In order to change the slope of the curved part so that the slope can go from very close to zero to almost vertical we must add yet another variable which we can call &#039;a&#039; to the function:&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
a(1/(1+e^-)), where &#039;a&#039; cannot be equal to zero.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
* If we make &#039;a&#039; a very large number such as 1,000,000 the slope becomes almost vertical. If we make &#039;a&#039; a very small number such as 1/10000000, the slope gets close to zero. This is an example of a vertical strech of the function by a dilation factor of &#039;a&#039;.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Once you&#039;ve played with the function enough, try to find an application of the graph to model something. It can be anything which starts at a value and then goes to another one (think for a population, it goes from 0 to it&#039;s carrying capacity). Explain what you are modelling and how you decide to attribute a numerical value to each of the 2 or 3 parameters that you researched just above. Then use the model to make a prediction. For example, if your model is suppose to describe a population for which you have its initial population and carrying capacity (potentially its rate of increase if you solved the bonus part), then use that data to make a prediction for the population in 20 years, or use the model to predict when will the population reach 95% of its carrying capacity). When doing this last part, explain well where you&#039;re taking your data from (real data or imagined data), what it is that you&#039;re modelling and how you are doing the math to answer a predictive question.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
CAN SOMEONE IN THE GROUP DO THIS PART!!!!!!!!!!!!! &lt;br /&gt;
&lt;br /&gt;
If we use the example of having two fish initially and if they have babies, the graph shows the increase over time of fish in the aquarium. If we only have enough food to feed four fish then, we can see the four fish only live and the amount of fish in the aquarium plateaus.&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_12&amp;diff=72712</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework 12</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_12&amp;diff=72712"/>
		<updated>2011-01-26T07:13:52Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;===Homework 12===&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
We are given the function:    &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
1/(1+e^-t)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
* In order to change the height of the horizontal asymptote on the right, we must add a variable &#039;c&#039; to the equation:&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
(1/(1+e^-t))+ c    &lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
* Here &#039;c&#039; is the factor which vertically translates the graph and this therefore changes the horizontal asymptote from its original position by a value of &#039;c&#039;. The horizontal asymptote of the original function is a y=1. Therefore we can say that K = c + 1.    &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
* If we substitute in a value of 3 for &#039;c&#039; the right asymptote changes to y=4, which can also be found by using the equation:&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
K = c + 1&lt;br /&gt;
 &lt;br /&gt;
K= 3+1 =4&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
* We can see the following pattern occur by changing the value that that function is vertically translated by the value of &#039;c&#039; using this formula.    &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
* To change the y-intercept to any number between 0 and K (which we chose to be 4 in this case) we can do this by adding another variable &#039;b&#039; into the equation:     &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
(1/(1+(&#039;&#039;&#039;b&#039;&#039;&#039;)e^-t))+3 , where &#039;b&#039; cannot be equal to zero   &lt;br /&gt;
&lt;br /&gt;
* Here &#039;b&#039; provides us with a horizontal stretch with a dilation factor of &#039;b&#039;.    &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
The values of the original function are plotted as follows:&lt;br /&gt;
&lt;br /&gt;
x=-1, y=3.26&lt;br /&gt;
&lt;br /&gt;
x=0, y=3.5&lt;br /&gt;
&lt;br /&gt;
x=1, y=3.73&lt;br /&gt;
&lt;br /&gt;
x=2, y=3.88&lt;br /&gt;
&lt;br /&gt;
x=3, y=3.95&lt;br /&gt;
&lt;br /&gt;
x=4, y=3.982&lt;br /&gt;
&lt;br /&gt;
x=5, y=3.9933&lt;br /&gt;
&lt;br /&gt;
x=6, y=3.9975&lt;br /&gt;
&lt;br /&gt;
x=7, y=3.9991&lt;br /&gt;
&lt;br /&gt;
x=8, y=3.9997&lt;br /&gt;
&lt;br /&gt;
x=9, y=3.9999&lt;br /&gt;
&lt;br /&gt;
x=10, y=4&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we let b=7, then (1/(1+(7)e^-t))+3, then the values are shown as:&lt;br /&gt;
&lt;br /&gt;
x=-1, y=3.04&lt;br /&gt;
&lt;br /&gt;
x=0, y=3.125&lt;br /&gt;
&lt;br /&gt;
x=1, y=3.2797&lt;br /&gt;
&lt;br /&gt;
x=2, y=3.51&lt;br /&gt;
&lt;br /&gt;
x=3, y=3.74&lt;br /&gt;
&lt;br /&gt;
x=4, y=3.8864&lt;br /&gt;
&lt;br /&gt;
x=5, y=3.955&lt;br /&gt;
&lt;br /&gt;
x=6, y=3.9829&lt;br /&gt;
&lt;br /&gt;
x=7, y=3.9937&lt;br /&gt;
&lt;br /&gt;
x=8, y=3.9977&lt;br /&gt;
&lt;br /&gt;
x=9, y=3.9991&lt;br /&gt;
&lt;br /&gt;
x=10, y=3.9997&lt;br /&gt;
&lt;br /&gt;
x=11, y=3.9999&lt;br /&gt;
&lt;br /&gt;
x=12, y=4&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
* Here we can see that by placing &#039;b&#039; into the equation we can alter the y-intercept but still maintain the same horizontal asymptote for both funtions!&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
* In order to change the slope of the curved part so that the slope can go from very close to zero to almost vertical we must add yet another variable which we can call &#039;a&#039; to the function:&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
a(1/(1+e^-)), where &#039;a&#039; cannot be equal to zero.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
* If we make &#039;a&#039; a very large number such as 1,000,000 the slope becomes almost vertical. If we make &#039;a&#039; a very small number such as 1/10000000, the slope gets close to zero. This is an example of a vertical strech of the function by a dilation factor of &#039;a&#039;.  &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Once you&#039;ve played with the function enough, try to find an application of the graph to model something. It can be anything which starts at a value and then goes to another one (think for a population, it goes from 0 to it&#039;s carrying capacity). Explain what you are modelling and how you decide to attribute a numerical value to each of the 2 or 3 parameters that you researched just above. Then use the model to make a prediction. For example, if your model is suppose to describe a population for which you have its initial population and carrying capacity (potentially its rate of increase if you solved the bonus part), then use that data to make a prediction for the population in 20 years, or use the model to predict when will the population reach 95% of its carrying capacity). When doing this last part, explain well where you&#039;re taking your data from (real data or imagined data), what it is that you&#039;re modelling and how you are doing the math to answer a predictive question.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
CAN SOMEONE IN THE GROUP DO THIS PART!!!!!!!!!!!!! &lt;br /&gt;
&lt;br /&gt;
If we use the example of having two fish initially and if they have babies, the graph shows the increase over time of fish in the aquarium. If we only have enough food to feed four fish then, we can see the four fish only live and the amount of fish in the aquarium plateaus.&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_12&amp;diff=72711</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework 12</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_12&amp;diff=72711"/>
		<updated>2011-01-26T07:06:30Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;===Homework 12===&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
We are given the function:  &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
1/(1+e^-t)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
* In order to change the height of the horizontal asymptote on the right, we must add a variable &#039;c&#039; to the equation:&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
(1/(1+e^-t))+ c  &lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
* Here &#039;c&#039; is the factor which vertically translates the graph and this therefore changes the horizontal asymptote from its original position by a value of &#039;c&#039;. The horizontal asymptote of the original function is a y=1. Therefore we can say that K = c + 1.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
* If we substitute in a value of 3 for &#039;c&#039; the right asymptote changes to y=4, which can also be found by using the equation:&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
K = c + 1&lt;br /&gt;
 &lt;br /&gt;
K= 3+1 =4&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
* We can see the following pattern occur by changing the value that that function is vertically translated by the value of &#039;c&#039; using this formula.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
* To change  the y-intercept to any number between 0 and K (which we chose to be 4 in this case) we can do this by adding another variable &#039;b&#039; into the equation: &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
(1/(1+(&#039;&#039;&#039;b&#039;&#039;&#039;)e^-t))+3  &lt;br /&gt;
&lt;br /&gt;
* Here &#039;b&#039; provides us with a horizontal stretch with a dilation factor of &#039;b&#039;.  &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
The values of the original function are plotted as follows:&lt;br /&gt;
&lt;br /&gt;
x=-1, y=3.26&lt;br /&gt;
&lt;br /&gt;
x=0, y=3.5&lt;br /&gt;
&lt;br /&gt;
x=1, y=3.73&lt;br /&gt;
&lt;br /&gt;
x=2, y=3.88&lt;br /&gt;
&lt;br /&gt;
x=3, y=3.95&lt;br /&gt;
&lt;br /&gt;
x=4, y=3.982&lt;br /&gt;
&lt;br /&gt;
x=5, y=3.9933&lt;br /&gt;
&lt;br /&gt;
x=6, y=3.9975&lt;br /&gt;
&lt;br /&gt;
x=7, y=3.9991&lt;br /&gt;
&lt;br /&gt;
x=8, y=3.9997&lt;br /&gt;
&lt;br /&gt;
x=9, y=3.9999&lt;br /&gt;
&lt;br /&gt;
x=10, y=4&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we let b=7, then (1/(1+(7)e^-t))+3, then the values are shown as:&lt;br /&gt;
&lt;br /&gt;
x=-1, y=3.04&lt;br /&gt;
&lt;br /&gt;
x=0, y=3.125&lt;br /&gt;
&lt;br /&gt;
x=1, y=3.2797&lt;br /&gt;
&lt;br /&gt;
x=2, y=3.51&lt;br /&gt;
&lt;br /&gt;
x=3, y=3.74&lt;br /&gt;
&lt;br /&gt;
x=4, y=3.8864&lt;br /&gt;
&lt;br /&gt;
x=5, y=3.955&lt;br /&gt;
&lt;br /&gt;
x=6, y=3.9829&lt;br /&gt;
&lt;br /&gt;
x=7, y=3.9937&lt;br /&gt;
&lt;br /&gt;
x=8, y=3.9977&lt;br /&gt;
&lt;br /&gt;
x=9, y=3.9991&lt;br /&gt;
&lt;br /&gt;
x=10, y=3.9997&lt;br /&gt;
&lt;br /&gt;
x=11, y=3.9999&lt;br /&gt;
&lt;br /&gt;
x=12, y=4&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
* Here we can see that by placing &#039;b&#039; into the equation we can alter the y-intercept but still maintain the same horizontal asymptote for both funtions!&lt;br /&gt;
&lt;br /&gt;
If we use the example of having two fish initially and if they have babies, the graph shows the increase over time of fish in the aquarium. If we only have enough food to feed four fish then, we can see the four fish only live and the amount of fish in the aquarium plateaus.&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_12&amp;diff=72706</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework 12</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_12&amp;diff=72706"/>
		<updated>2011-01-26T06:55:56Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;===Homework 12===&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
We are given the function:  &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
1/(1+e^-t)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
* In order to change the height of the horizontal asymptote on the right, we must add a variable &#039;c&#039; to the equation:&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
(1/(1+e^-t))+ c  &lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
* Here &#039;c&#039; is the factor which vertically translates the graph and this therefore changes the horizontal asymptote from its original position by a value of &#039;c&#039;. The horizontal asymptote of the original function is a y=1. Therefore we can say that K = c + 1.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
* If we substitute in a value of 3 for &#039;c&#039; the right asymptote changes to y=4, which can also be found by using the equation:&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
K = c + 1&lt;br /&gt;
 &lt;br /&gt;
K= 3+1 =4&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
* We can see the following pattern occur by changing the value that that function is vertically translated by the value of &#039;c&#039; using this formula.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
* To change  the y-intercept to any number between 0 and K (which we chose to be 4 in this case) we can do this by adding another variable &#039;b&#039; into the equation: &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
(1/(1+(&#039;&#039;&#039;b&#039;&#039;&#039;)e^-t))+3  &lt;br /&gt;
&lt;br /&gt;
* Here &#039;b&#039; provides us with a horizontal stretch with a dilation factor of &#039;b&#039;.  &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
The values of the original function are plotted as follows:&lt;br /&gt;
&lt;br /&gt;
x=-1, y=3.26&lt;br /&gt;
&lt;br /&gt;
x=0, y=3.5&lt;br /&gt;
&lt;br /&gt;
x=1, y=3.73&lt;br /&gt;
&lt;br /&gt;
x=2, y=3.88&lt;br /&gt;
&lt;br /&gt;
x=3, y=3.95&lt;br /&gt;
&lt;br /&gt;
x=4, y=3.982&lt;br /&gt;
&lt;br /&gt;
x=5, y=3.9933&lt;br /&gt;
&lt;br /&gt;
x=6, y=3.9975&lt;br /&gt;
&lt;br /&gt;
x=7, y=3.9991&lt;br /&gt;
&lt;br /&gt;
x=8, y=3.9997&lt;br /&gt;
&lt;br /&gt;
x=9, y=3.9999&lt;br /&gt;
&lt;br /&gt;
x=10, y=4&lt;br /&gt;
&lt;br /&gt;
If we let b=7, then (1/(1+(7)e^-t))+3, then the values are shown as:&lt;br /&gt;
&lt;br /&gt;
x=-1, y=3.04&lt;br /&gt;
&lt;br /&gt;
x=0, y=3.125&lt;br /&gt;
&lt;br /&gt;
x=1, y=3.2797&lt;br /&gt;
&lt;br /&gt;
x=2, y=3.51&lt;br /&gt;
&lt;br /&gt;
x=3, y=3.74&lt;br /&gt;
&lt;br /&gt;
x=4, y=3.8864&lt;br /&gt;
&lt;br /&gt;
x=5, y=3.955&lt;br /&gt;
&lt;br /&gt;
x=6, y=3.9829&lt;br /&gt;
&lt;br /&gt;
x=7, y=3.9937&lt;br /&gt;
&lt;br /&gt;
x=8, y=3.9977&lt;br /&gt;
&lt;br /&gt;
x=9, y=3.9991&lt;br /&gt;
&lt;br /&gt;
x=10, y=3.9997&lt;br /&gt;
&lt;br /&gt;
x=11, y=3.9999&lt;br /&gt;
&lt;br /&gt;
x=12, y=4&lt;br /&gt;
&lt;br /&gt;
If we use the example of having two fish initially and if they have babies, the graph shows the increase over time of fish in the aquarium. If we only have enough food to feed four fish then, we can see the four fish only live and the amount of fish in the aquarium plateaus.&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_12&amp;diff=72705</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework 12</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_12&amp;diff=72705"/>
		<updated>2011-01-26T06:47:18Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;===Homework 12===&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
We are given the function:  &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
1/(1+e^-t)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
* In order to change the height of the horizontal asymptote on the right, we must add a variable &#039;c&#039; to the equation:&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
(1/(1+e^-t))+ c  &lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
* Here &#039;c&#039; is the factor which vertically translates the graph and this therefore changes the horizontal asymptote from its original position by a value of &#039;c&#039;. The horizontal asymptote of the original function is a y=1. Therefore we can say that K = c + 1.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
* If we substitute in a value of 3 for &#039;c&#039; the right asymptote changes to y=4, which can also be found by using the equation:&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
K = c + 1&lt;br /&gt;
 &lt;br /&gt;
K= 3+1 =4&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
* We can see the following pattern occur by changing the value that that function is vertically translated by the value of &#039;c&#039; using this formula.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
* To change  the y-intercept to any number between 0 and K (which we chose to be 4 in this case) we can do this by adding another variable &#039;b&#039; into the equation: &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
(1/(1+(b)e^-t))+3  , here &#039;b&#039; provides us with a horizontal stretch with a dilation factor of &#039;b&#039;.  &lt;br /&gt;
&lt;br /&gt;
Originally then the values are:&lt;br /&gt;
x=-1, y=3.26&lt;br /&gt;
&lt;br /&gt;
x=0, y=3.5&lt;br /&gt;
&lt;br /&gt;
x=1, y=3.73&lt;br /&gt;
&lt;br /&gt;
x=2, y=3.88&lt;br /&gt;
&lt;br /&gt;
x=3, y=3.95&lt;br /&gt;
&lt;br /&gt;
x=4, y=3.982&lt;br /&gt;
&lt;br /&gt;
x=5, y=3.9933&lt;br /&gt;
&lt;br /&gt;
x=6, y=3.9975&lt;br /&gt;
&lt;br /&gt;
x=7, y=3.9991&lt;br /&gt;
&lt;br /&gt;
x=8, y=3.9997&lt;br /&gt;
&lt;br /&gt;
x=9, y=3.9999&lt;br /&gt;
&lt;br /&gt;
x=10, y=4&lt;br /&gt;
&lt;br /&gt;
If we let b=7, then (1)/(((1+(7)e^-t))+3), then the values are shown as:&lt;br /&gt;
&lt;br /&gt;
x=-1, y=3.04&lt;br /&gt;
x=0, y=3.125&lt;br /&gt;
x=1, y=3.2797&lt;br /&gt;
x=2, y=3.51&lt;br /&gt;
x=3, y=3.74&lt;br /&gt;
x=4, y=3.8864&lt;br /&gt;
x=5, y=3.955&lt;br /&gt;
x=6, y=3.9829&lt;br /&gt;
x=7, y=3.9937&lt;br /&gt;
x=8, y=3.9977&lt;br /&gt;
x=9, y=3.9991&lt;br /&gt;
x=10, y=3.9997&lt;br /&gt;
x=11, y=3.9999&lt;br /&gt;
x=12, y=4&lt;br /&gt;
&lt;br /&gt;
If we use the example of having two fish initially and if they have babies, the graph shows the increase over time of fish in the aquarium. If we only have enough food to feed four fish then, we can see the four fish only live and the amount of fish in the aquarium plateaus.&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_12&amp;diff=72704</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework 12</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_12&amp;diff=72704"/>
		<updated>2011-01-26T06:20:36Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Homework 12&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
We are given the function: &lt;br /&gt;
&lt;br /&gt;
1/(1+e^-t)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In order to change the height of the horizontal asymptote on the right, we add c to the denominator giving the equation: (1)/(1+e^-t)+c, here c acts as the factor that gives the right asymptote denoted K. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Originally the right asymptote of the function was at y=1. However, if we try our given equation for instance when c=3, the right asymptote is around K=4. We can see the following pattern occur by changing the values that it moves up the value of c, K=c+1. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Given the value we found for K as 4, we can change the y-intercept to any number between 0 and 4. This can be done by using the value b, which will show the equation as: (1)/(((1+(b)e^-t))+3). &lt;br /&gt;
&lt;br /&gt;
Originally then the values are:&lt;br /&gt;
x=-1, y=3.26&lt;br /&gt;
&lt;br /&gt;
x=0, y=3.5&lt;br /&gt;
&lt;br /&gt;
x=1, y=3.73&lt;br /&gt;
&lt;br /&gt;
x=2, y=3.88&lt;br /&gt;
&lt;br /&gt;
x=3, y=3.95&lt;br /&gt;
&lt;br /&gt;
x=4, y=3.982&lt;br /&gt;
&lt;br /&gt;
x=5, y=3.9933&lt;br /&gt;
&lt;br /&gt;
x=6, y=3.9975&lt;br /&gt;
&lt;br /&gt;
x=7, y=3.9991&lt;br /&gt;
&lt;br /&gt;
x=8, y=3.9997&lt;br /&gt;
&lt;br /&gt;
x=9, y=3.9999&lt;br /&gt;
&lt;br /&gt;
x=10, y=4&lt;br /&gt;
&lt;br /&gt;
If we let b=7, then (1)/(((1+(7)e^-t))+3), then the values are shown as:&lt;br /&gt;
&lt;br /&gt;
x=-1, y=3.04&lt;br /&gt;
x=0, y=3.125&lt;br /&gt;
x=1, y=3.2797&lt;br /&gt;
x=2, y=3.51&lt;br /&gt;
x=3, y=3.74&lt;br /&gt;
x=4, y=3.8864&lt;br /&gt;
x=5, y=3.955&lt;br /&gt;
x=6, y=3.9829&lt;br /&gt;
x=7, y=3.9937&lt;br /&gt;
x=8, y=3.9977&lt;br /&gt;
x=9, y=3.9991&lt;br /&gt;
x=10, y=3.9997&lt;br /&gt;
x=11, y=3.9999&lt;br /&gt;
x=12, y=4&lt;br /&gt;
&lt;br /&gt;
If we use the example of having two fish initially and if they have babies, the graph shows the increase over time of fish in the aquarium. If we only have enough food to feed four fish then, we can see the four fish only live and the amount of fish in the aquarium plateaus.&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_12&amp;diff=72703</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework 12</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz/Homework_12&amp;diff=72703"/>
		<updated>2011-01-26T06:18:45Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Homework 12&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Given the function: (1)/(1+e^-t)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In order to change the height of the horizontal asymptote on the right, we add c to the denominator giving the equation: (1)/(1+e^-t)+c, here c acts as the factor that gives the right asymptote denoted K. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Originally the right asymptote of the function was at y=1. However, if we try our given equation for instance when c=3, the right asymptote is around K=4. We can see the following pattern occur by changing the values that it moves up the value of c, K=c+1. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Given the value we found for K as 4, we can change the y-intercept to any number between 0 and 4. This can be done by using the value b, which will show the equation as: (1)/(((1+(b)e^-t))+3). &lt;br /&gt;
&lt;br /&gt;
Originally then the values are:&lt;br /&gt;
x=-1, y=3.26&lt;br /&gt;
&lt;br /&gt;
x=0, y=3.5&lt;br /&gt;
&lt;br /&gt;
x=1, y=3.73&lt;br /&gt;
&lt;br /&gt;
x=2, y=3.88&lt;br /&gt;
&lt;br /&gt;
x=3, y=3.95&lt;br /&gt;
&lt;br /&gt;
x=4, y=3.982&lt;br /&gt;
&lt;br /&gt;
x=5, y=3.9933&lt;br /&gt;
&lt;br /&gt;
x=6, y=3.9975&lt;br /&gt;
&lt;br /&gt;
x=7, y=3.9991&lt;br /&gt;
&lt;br /&gt;
x=8, y=3.9997&lt;br /&gt;
&lt;br /&gt;
x=9, y=3.9999&lt;br /&gt;
&lt;br /&gt;
x=10, y=4&lt;br /&gt;
&lt;br /&gt;
If we let b=7, then (1)/(((1+(7)e^-t))+3), then the values are shown as:&lt;br /&gt;
&lt;br /&gt;
x=-1, y=3.04&lt;br /&gt;
x=0, y=3.125&lt;br /&gt;
x=1, y=3.2797&lt;br /&gt;
x=2, y=3.51&lt;br /&gt;
x=3, y=3.74&lt;br /&gt;
x=4, y=3.8864&lt;br /&gt;
x=5, y=3.955&lt;br /&gt;
x=6, y=3.9829&lt;br /&gt;
x=7, y=3.9937&lt;br /&gt;
x=8, y=3.9977&lt;br /&gt;
x=9, y=3.9991&lt;br /&gt;
x=10, y=3.9997&lt;br /&gt;
x=11, y=3.9999&lt;br /&gt;
x=12, y=4&lt;br /&gt;
&lt;br /&gt;
If we use the example of having two fish initially and if they have babies, the graph shows the increase over time of fish in the aquarium. If we only have enough food to feed four fish then, we can see the four fish only live and the amount of fish in the aquarium plateaus.&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:CassandraTravlos&amp;diff=72701</id>
		<title>User:CassandraTravlos</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:CassandraTravlos&amp;diff=72701"/>
		<updated>2011-01-26T06:17:15Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==About Me...==&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
http://wiki.ubc.ca/User:CassandraTravlos/About_me_...     &lt;br /&gt;
 &lt;br /&gt;
----&lt;br /&gt;
==Essay: Calculus and the Faculty of Human Kinetics at UBC==&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
 &lt;br /&gt;
http://wiki.ubc.ca/User:CassandraTravlos/Essay: Calculus and the Faculty of Human Kinetics at UBC&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:CassandraTravlos&amp;diff=72699</id>
		<title>User:CassandraTravlos</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:CassandraTravlos&amp;diff=72699"/>
		<updated>2011-01-26T06:16:47Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==About Me...==&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
http://wiki.ubc.ca/User:CassandraTravlos/About_me_...     &lt;br /&gt;
 &lt;br /&gt;
----&lt;br /&gt;
==Essay: Calculus and the Faculty of Human Kinetics at UBC==&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
 &lt;br /&gt;
http://wiki.ubc.ca/User:CassandraTravlos/Essay:Calculus and the Faculty of Human Kinetics at UBC&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:CassandraTravlos/Essay&amp;diff=72697</id>
		<title>User:CassandraTravlos/Essay</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:CassandraTravlos/Essay&amp;diff=72697"/>
		<updated>2011-01-26T06:15:44Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: Created page with &amp;quot; ==Calculus in The Faculty of Human Kinetics==  The program that I am taking at the University of British Columbia is the Faculty of Human Kinetics. Calculus applies quite direct...&amp;quot;&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
==Calculus in The Faculty of Human Kinetics==&lt;br /&gt;
&lt;br /&gt;
The program that I am taking at the University of British Columbia is the Faculty of Human Kinetics. Calculus applies quite directly to this field of study, particularly in the courses of biomechanics, physiology, chemistry and biology. These are required courses in the Human Kinetics program. Calculus is a branch of mathematics and is closely correlated to the sciences.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In biomechanics, most of the concepts and calculations involve the use of calculus. Aspects of this course require you to find the velocities of projectiles as well as their accelerations. Velocity is found by finding the time derivative of displacement and acceleration is found by calculating the time derivative of velocity, which is also known as the second derivative. Calculus has life applications in this field through such aspects as calculating the acceleration of a baseball being thrown by a pitcher and then calculating the velocity of the ball. Calculus can also be used to find the moment of inertia of your arm in elbow flexion during a curl up in the field of biomechanics. Here you can see that calculus is closely related to sports. Being able to do such calculations facilitates coaches and trainers to improve and assess athletic ability and technique in various sports, in a safe manner. The desired result of this would be to reach the optimum levels of athletic ability for each sport. For example in shot put the optimal result would be to thrown the shot put as far as you could without causing injury.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In biomechanics we also use Newton’s Laws. These laws were created through the application of calculus. Newton’s second law applies the concept of calculus termed “the rate of change” which is the derivative. This is expressed as a formula, force= mass x acceleration. It involves calculus because acceleration is the time derivative of velocity. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Calculus is also used in physiology. Here we can use calculus to determine flow rates of pumping fluid out of the heart as well as looking as the effect of various exercises on cardiac flow. We can also measure to see if the valve flow in the heart is appropriate, we can predict the onset of epileptic seizure by examining the electrical activity of the brain or predict the onset of cardiac arrest by examining the electrical activity of the heart. This is a very important application of calculus as this allows us to assess human health. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In chemistry calculus is mainly used to compute reaction rates, which uses the derivative. We can assess the rate of reaction that acetic acid has with sodium hydroxide. In biology we use it to make models. We can model population growth, half-lives of medications, and the carrying capacities of deer populations in B.C. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
As you can see, calculus is very significant and important to the field of Human Kinetics. Without it we would not have the ability to assess physical human health and understand why different techniques in sports are more effective.&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:CassandraTravlos&amp;diff=72696</id>
		<title>User:CassandraTravlos</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:CassandraTravlos&amp;diff=72696"/>
		<updated>2011-01-26T06:15:07Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;[[About me..]]&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
http://wiki.ubc.ca/User:CassandraTravlos/About_me_...     &lt;br /&gt;
 &lt;br /&gt;
[[Essay: Calculus and the Faculty of Human Kinetics at UBC]]&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
http://wiki.ubc.ca/User:CassandraTravlos/Essay: Calculus and the Faculty of Human Kinetics at UBC&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:CassandraTravlos/About_me&amp;diff=72695</id>
		<title>User:CassandraTravlos/About me</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:CassandraTravlos/About_me&amp;diff=72695"/>
		<updated>2011-01-26T06:14:05Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: Created page with &amp;quot; Hello, my name is Cassandra Travlos. I am a first year student at UBC taking the Human Kinetic undergraduate program. I live in West Vancouver and I have my own horse which I sh...&amp;quot;&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
Hello, my name is Cassandra Travlos. I am a first year student at UBC taking the Human Kinetic undergraduate program. I live in West Vancouver and I have my own horse which I show competively in Canada.&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:CassandraTravlos&amp;diff=72694</id>
		<title>User:CassandraTravlos</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:CassandraTravlos&amp;diff=72694"/>
		<updated>2011-01-26T06:12:19Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;[[About me..]]&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
http://wiki.ubc.ca/User:CassandraTravlos/About_me_...  &lt;br /&gt;
 &lt;br /&gt;
[[Essay: Calculus and the Faculty of Human Kinetics at UBC]]&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:CassandraTravlos&amp;diff=72693</id>
		<title>User:CassandraTravlos</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:CassandraTravlos&amp;diff=72693"/>
		<updated>2011-01-26T06:05:58Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;[[About me..]]&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
[[Essay: Calculus and the Faculty of Human Kinetics at UBC]]&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:CassandraTravlos&amp;diff=72692</id>
		<title>User:CassandraTravlos</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:CassandraTravlos&amp;diff=72692"/>
		<updated>2011-01-26T06:05:34Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;[[About me..]]&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
wiki.ubc.ca/User:CassandraTravlos/About_me_...&lt;br /&gt;
 &lt;br /&gt;
[[Essay: Calculus and the Faculty of Human Kinetics at UBC]]&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Thread:User_talk:CassandraTravlos/Your_new_pages/reply&amp;diff=72691</id>
		<title>Thread:User talk:CassandraTravlos/Your new pages/reply</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Thread:User_talk:CassandraTravlos/Your_new_pages/reply&amp;diff=72691"/>
		<updated>2011-01-26T06:04:47Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: Reply to Your new pages&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;I have no clue how to do that..&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz&amp;diff=72690</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Schwytz</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz&amp;diff=72690"/>
		<updated>2011-01-26T06:04:17Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{&lt;br /&gt;
Infobox MATH110 Teams&lt;br /&gt;
| team name = Schwytz&lt;br /&gt;
| member 1 = Angus McWhirter&lt;br /&gt;
| member 2 = Anna Koniuhova&lt;br /&gt;
| member 3 = Cassandra Travlos&lt;br /&gt;
| member 4 = Gracie Mann&lt;br /&gt;
}}&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
==Key words that describe us==&lt;br /&gt;
Gracie- perseverance      &lt;br /&gt;
Cassandra- indecisive      &lt;br /&gt;
Anna- fun&lt;br /&gt;
Angus- mia&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
== Team Schwytz Subpages ==&lt;br /&gt;
&lt;br /&gt;
==Homework #11- Team Schwytz==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Write a linear model to predict the cost of producing flags of your teams Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items the cost is $100. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Our model is described by a linear function.&lt;br /&gt;
* We know that our model has to pass through the point (20,100) because the question tells us that the current production level of 20 items costs $100 (where the cost in dollar is the y-axis and the number of items is the x-axis) therefore we can use this point to formulate a function using the standard equation formula (y=mx +b)&lt;br /&gt;
* We can use 20 items for ‘x’ and 100 dollars for ‘y’&lt;br /&gt;
* Now we must find ‘m’&lt;br /&gt;
* To do this we need to use the formula: y2-y1/ x2-x1&lt;br /&gt;
To find a point on the graph:&lt;br /&gt;
&lt;br /&gt;
* We know that 20 items cost 100 dollars and the marginal cost is $7 per unit&lt;br /&gt;
* Therefore we know that 21 items costs $107 and then 22 items costs $114&lt;br /&gt;
* Now we can set this up in the above equation (where the cost is ‘y’ and the items is ‘x’:&lt;br /&gt;
114-107/ 22-21 = 7&lt;br /&gt;
&lt;br /&gt;
* Therefore:&lt;br /&gt;
&lt;br /&gt;
m=7&lt;br /&gt;
&lt;br /&gt;
* Now we need to find &#039;b&#039;:&lt;br /&gt;
&lt;br /&gt;
y=mx + b&lt;br /&gt;
&lt;br /&gt;
107 = 7 (21) +b&lt;br /&gt;
&lt;br /&gt;
b = -40&lt;br /&gt;
&lt;br /&gt;
* Therefore the equation of the line is:&lt;br /&gt;
&lt;br /&gt;
y= 7x - 40 , which is a linear model       &lt;br /&gt;
&lt;br /&gt;
* If we were to graph this model the function would cross the y-axis at -40.&lt;br /&gt;
&lt;br /&gt;
* However we can also determine the value of ‘b’ by understanding that it costs $100 to produce 20 flags which averages to $5 per flag. If the flags were $7 each it would have cost $140. The difference between these two costs is $40. Therefore to take into account the saving of $40 we make ‘b’ -40 in the formula y=mx+b&lt;br /&gt;
&lt;br /&gt;
* To relate this formula to the specific terminology of the question we will replace &#039;x&#039; with &#039;f&#039; and let &#039;f&#039; represent the number of flags produced (where &#039;f&#039; must be greater than 20). We will replace &#039;y&#039; with C(f) and let C(f) represent the cost to produce the flags:&lt;br /&gt;
C(f) = 7f - 40 , where f ≥ 20         &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;What does your model predict for production of a 150 items?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* In order to predict the total cost of the production of 150 items we can substitute into our model 150 for &#039;f&#039; . Which gives us the equation:         &lt;br /&gt;
&#039;&#039;C(f)=7(150)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=1010&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;According to your model, what happens when the average cost per item, as production levels increase?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
*In order to determine the average cost per item as production increases we must first find the total cost of three consecutive units (e.g the total cost of producing 23, 24, and 25 flags) and divide each total you get but the number of flags produced as shown below:          &lt;br /&gt;
The average cost equation: total cost/number of flags&lt;br /&gt;
&lt;br /&gt;
1. Let f=23&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(23)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=121&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 23 units is : 121/23=5.26&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Let f=24&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(24)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=128&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 24 units is : 128/24 =5.33&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Let f=25&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(25)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=135&#039;&#039;         &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 25 units is: 135/25=5.40&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* Based on the information above we can see that the averge cost per item increases as the production levels increase.       &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Finally, find some other models ( not necessarily linear) for which you get other behaviours such as&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
1. The average cost remains constant as production increases:         &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y=3x&#039;&#039;&#039;      &lt;br /&gt;
&lt;br /&gt;
* This model &#039;&#039;&#039;increases&#039;&#039;&#039; at a &#039;&#039;&#039;constant rate&#039;&#039;&#039;, because the slope of the function is the same everywhere. This model shows that the average cost will therefore remain constant as the production increases.      &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2. The average cost diminishes as the production increases:         &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= x^1/2&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* For this model when you plug in any postive numbers in for &#039;x&#039;, you will see that as &#039;x&#039; or the number of items produced increases, the average cost decreases:      &lt;br /&gt;
&lt;br /&gt;
ex.      &lt;br /&gt;
&lt;br /&gt;
y= x^1/2 , let x = 100 and let y represent the total cost of x&lt;br /&gt;
&lt;br /&gt;
y= (100^1/2)/ 100&lt;br /&gt;
&lt;br /&gt;
y= 0.1&lt;br /&gt;
&lt;br /&gt;
Now let x= 1000000&lt;br /&gt;
&lt;br /&gt;
y= (1000000^1/2)/1000000&lt;br /&gt;
&lt;br /&gt;
y= 0.001      &lt;br /&gt;
&lt;br /&gt;
* Here you can see that the average cost is diminishing as the production (&#039;x&#039;) increases.         &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
3. The average cost increases as production increases:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= e^x&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* In this model as the production increases so does the slope of the model which causes the average cost to also therefore increase.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
4. You obtain an economy of scale. This means that staring at some specific production level, the marginal cost is always less than the average cost:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= x^1/2&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* For this model, we can see that the marginal cost is less than the average cost&lt;br /&gt;
&lt;br /&gt;
ex.&lt;br /&gt;
&lt;br /&gt;
let x=3&lt;br /&gt;
&lt;br /&gt;
y=3^1/2&lt;br /&gt;
&lt;br /&gt;
y= 1.73&lt;br /&gt;
&lt;br /&gt;
* now we must divide by 3 to get the average      &lt;br /&gt;
&lt;br /&gt;
y= 1.73/3&lt;br /&gt;
&lt;br /&gt;
y= 0.577 , this means that it costs an average of $0.58 to produce 3 items         &lt;br /&gt;
&lt;br /&gt;
* And when x=2, y= 1.41&lt;br /&gt;
&lt;br /&gt;
* Therefore if we calculate the difference ($1.73-$1.41) we get 0.318 or $0.32. This means that it only cost $0.32 to produce the third item but it cost an average of $0.58 per item (of the 3 total items). This model therefore obtains an economy of scale.&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz&amp;diff=72689</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Schwytz</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz&amp;diff=72689"/>
		<updated>2011-01-26T05:59:52Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{&lt;br /&gt;
Infobox MATH110 Teams&lt;br /&gt;
| team name = Schwytz&lt;br /&gt;
| member 1 = Angus McWhirter&lt;br /&gt;
| member 2 = Anna Koniuhova&lt;br /&gt;
| member 3 = Cassandra Travlos&lt;br /&gt;
| member 4 = Gracie Mann&lt;br /&gt;
}}&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;dpl&amp;gt;&lt;br /&gt;
titlematch={{PAGENAME}}/%&lt;br /&gt;
namespace={{NAMESPACE}}&lt;br /&gt;
shownamespace=false&lt;br /&gt;
&amp;lt;/dpl&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Key words that describe us==&lt;br /&gt;
Gracie- perseverance     &lt;br /&gt;
Cassandra- indecisive     &lt;br /&gt;
Anna- fun&lt;br /&gt;
Angus- mia&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
== Team Schwytz Subpages ==&lt;br /&gt;
&lt;br /&gt;
==Homework #11- Team Schwytz==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Write a linear model to predict the cost of producing flags of your teams Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items the cost is $100. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Our model is described by a linear function.&lt;br /&gt;
* We know that our model has to pass through the point (20,100) because the question tells us that the current production level of 20 items costs $100 (where the cost in dollar is the y-axis and the number of items is the x-axis) therefore we can use this point to formulate a function using the standard equation formula (y=mx +b)&lt;br /&gt;
* We can use 20 items for ‘x’ and 100 dollars for ‘y’&lt;br /&gt;
* Now we must find ‘m’&lt;br /&gt;
* To do this we need to use the formula: y2-y1/ x2-x1&lt;br /&gt;
To find a point on the graph:&lt;br /&gt;
&lt;br /&gt;
* We know that 20 items cost 100 dollars and the marginal cost is $7 per unit&lt;br /&gt;
* Therefore we know that 21 items costs $107 and then 22 items costs $114&lt;br /&gt;
* Now we can set this up in the above equation (where the cost is ‘y’ and the items is ‘x’:&lt;br /&gt;
114-107/ 22-21 = 7&lt;br /&gt;
&lt;br /&gt;
* Therefore:&lt;br /&gt;
&lt;br /&gt;
m=7&lt;br /&gt;
&lt;br /&gt;
* Now we need to find &#039;b&#039;:&lt;br /&gt;
&lt;br /&gt;
y=mx + b&lt;br /&gt;
&lt;br /&gt;
107 = 7 (21) +b&lt;br /&gt;
&lt;br /&gt;
b = -40&lt;br /&gt;
&lt;br /&gt;
* Therefore the equation of the line is:&lt;br /&gt;
&lt;br /&gt;
y= 7x - 40 , which is a linear model      &lt;br /&gt;
&lt;br /&gt;
* If we were to graph this model the function would cross the y-axis at -40.&lt;br /&gt;
&lt;br /&gt;
* However we can also determine the value of ‘b’ by understanding that it costs $100 to produce 20 flags which averages to $5 per flag. If the flags were $7 each it would have cost $140. The difference between these two costs is $40. Therefore to take into account the saving of $40 we make ‘b’ -40 in the formula y=mx+b&lt;br /&gt;
&lt;br /&gt;
* To relate this formula to the specific terminology of the question we will replace &#039;x&#039; with &#039;f&#039; and let &#039;f&#039; represent the number of flags produced (where &#039;f&#039; must be greater than 20). We will replace &#039;y&#039; with C(f) and let C(f) represent the cost to produce the flags:&lt;br /&gt;
C(f) = 7f - 40 , where f ≥ 20        &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;What does your model predict for production of a 150 items?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* In order to predict the total cost of the production of 150 items we can substitute into our model 150 for &#039;f&#039; . Which gives us the equation:        &lt;br /&gt;
&#039;&#039;C(f)=7(150)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=1010&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;According to your model, what happens when the average cost per item, as production levels increase?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
*In order to determine the average cost per item as production increases we must first find the total cost of three consecutive units (e.g the total cost of producing 23, 24, and 25 flags) and divide each total you get but the number of flags produced as shown below:         &lt;br /&gt;
The average cost equation: total cost/number of flags&lt;br /&gt;
&lt;br /&gt;
1. Let f=23&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(23)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=121&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 23 units is : 121/23=5.26&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Let f=24&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(24)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=128&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 24 units is : 128/24 =5.33&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Let f=25&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(25)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=135&#039;&#039;        &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 25 units is: 135/25=5.40&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* Based on the information above we can see that the averge cost per item increases as the production levels increase.      &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Finally, find some other models ( not necessarily linear) for which you get other behaviours such as&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
1. The average cost remains constant as production increases:        &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y=3x&#039;&#039;&#039;     &lt;br /&gt;
&lt;br /&gt;
* This model &#039;&#039;&#039;increases&#039;&#039;&#039; at a &#039;&#039;&#039;constant rate&#039;&#039;&#039;, because the slope of the function is the same everywhere. This model shows that the average cost will therefore remain constant as the production increases.     &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2. The average cost diminishes as the production increases:        &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= x^1/2&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* For this model when you plug in any postive numbers in for &#039;x&#039;, you will see that as &#039;x&#039; or the number of items produced increases, the average cost decreases:     &lt;br /&gt;
&lt;br /&gt;
ex.     &lt;br /&gt;
&lt;br /&gt;
y= x^1/2 , let x = 100 and let y represent the total cost of x&lt;br /&gt;
&lt;br /&gt;
y= (100^1/2)/ 100&lt;br /&gt;
&lt;br /&gt;
y= 0.1&lt;br /&gt;
&lt;br /&gt;
Now let x= 1000000&lt;br /&gt;
&lt;br /&gt;
y= (1000000^1/2)/1000000&lt;br /&gt;
&lt;br /&gt;
y= 0.001     &lt;br /&gt;
&lt;br /&gt;
* Here you can see that the average cost is diminishing as the production (&#039;x&#039;) increases.        &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
3. The average cost increases as production increases:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= e^x&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* In this model as the production increases so does the slope of the model which causes the average cost to also therefore increase.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
4. You obtain an economy of scale. This means that staring at some specific production level, the marginal cost is always less than the average cost:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= x^1/2&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* For this model, we can see that the marginal cost is less than the average cost&lt;br /&gt;
&lt;br /&gt;
ex.&lt;br /&gt;
&lt;br /&gt;
let x=3&lt;br /&gt;
&lt;br /&gt;
y=3^1/2&lt;br /&gt;
&lt;br /&gt;
y= 1.73&lt;br /&gt;
&lt;br /&gt;
* now we must divide by 3 to get the average     &lt;br /&gt;
&lt;br /&gt;
y= 1.73/3&lt;br /&gt;
&lt;br /&gt;
y= 0.577 , this means that it costs an average of $0.58 to produce 3 items        &lt;br /&gt;
&lt;br /&gt;
* And when x=2, y= 1.41&lt;br /&gt;
&lt;br /&gt;
* Therefore if we calculate the difference ($1.73-$1.41) we get 0.318 or $0.32. This means that it only cost $0.32 to produce the third item but it cost an average of $0.58 per item (of the 3 total items). This model therefore obtains an economy of scale.&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz&amp;diff=72688</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Schwytz</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz&amp;diff=72688"/>
		<updated>2011-01-26T05:59:32Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: /* Team Schwytz Subpages */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{&lt;br /&gt;
Infobox MATH110 Teams&lt;br /&gt;
| team name = Schwytz&lt;br /&gt;
| member 1 = Angus McWhirter&lt;br /&gt;
| member 2 = Anna Koniuhova&lt;br /&gt;
| member 3 = Cassandra Travlos&lt;br /&gt;
| member 4 = Gracie Mann&lt;br /&gt;
}}&lt;br /&gt;
&lt;br /&gt;
&amp;lt;dpl&amp;gt; titlematch={{Course:MATH110/003/Teams/Schwytz}}/% namespace={{NAMESPACE}} shownamespace=false &amp;lt;/dpl&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Key words that describe us==&lt;br /&gt;
Gracie- perseverance    &lt;br /&gt;
Cassandra- indecisive    &lt;br /&gt;
Anna- fun&lt;br /&gt;
Angus- mia&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
== Team Schwytz Subpages ==&lt;br /&gt;
&lt;br /&gt;
==Homework #11- Team Schwytz==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Write a linear model to predict the cost of producing flags of your teams Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items the cost is $100. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Our model is described by a linear function.&lt;br /&gt;
* We know that our model has to pass through the point (20,100) because the question tells us that the current production level of 20 items costs $100 (where the cost in dollar is the y-axis and the number of items is the x-axis) therefore we can use this point to formulate a function using the standard equation formula (y=mx +b)&lt;br /&gt;
* We can use 20 items for ‘x’ and 100 dollars for ‘y’&lt;br /&gt;
* Now we must find ‘m’&lt;br /&gt;
* To do this we need to use the formula: y2-y1/ x2-x1&lt;br /&gt;
To find a point on the graph:&lt;br /&gt;
&lt;br /&gt;
* We know that 20 items cost 100 dollars and the marginal cost is $7 per unit&lt;br /&gt;
* Therefore we know that 21 items costs $107 and then 22 items costs $114&lt;br /&gt;
* Now we can set this up in the above equation (where the cost is ‘y’ and the items is ‘x’:&lt;br /&gt;
114-107/ 22-21 = 7&lt;br /&gt;
&lt;br /&gt;
* Therefore:&lt;br /&gt;
&lt;br /&gt;
m=7&lt;br /&gt;
&lt;br /&gt;
* Now we need to find &#039;b&#039;:&lt;br /&gt;
&lt;br /&gt;
y=mx + b&lt;br /&gt;
&lt;br /&gt;
107 = 7 (21) +b&lt;br /&gt;
&lt;br /&gt;
b = -40&lt;br /&gt;
&lt;br /&gt;
* Therefore the equation of the line is:&lt;br /&gt;
&lt;br /&gt;
y= 7x - 40 , which is a linear model     &lt;br /&gt;
&lt;br /&gt;
* If we were to graph this model the function would cross the y-axis at -40.&lt;br /&gt;
&lt;br /&gt;
* However we can also determine the value of ‘b’ by understanding that it costs $100 to produce 20 flags which averages to $5 per flag. If the flags were $7 each it would have cost $140. The difference between these two costs is $40. Therefore to take into account the saving of $40 we make ‘b’ -40 in the formula y=mx+b&lt;br /&gt;
&lt;br /&gt;
* To relate this formula to the specific terminology of the question we will replace &#039;x&#039; with &#039;f&#039; and let &#039;f&#039; represent the number of flags produced (where &#039;f&#039; must be greater than 20). We will replace &#039;y&#039; with C(f) and let C(f) represent the cost to produce the flags:&lt;br /&gt;
C(f) = 7f - 40 , where f ≥ 20       &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;What does your model predict for production of a 150 items?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* In order to predict the total cost of the production of 150 items we can substitute into our model 150 for &#039;f&#039; . Which gives us the equation:       &lt;br /&gt;
&#039;&#039;C(f)=7(150)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=1010&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;According to your model, what happens when the average cost per item, as production levels increase?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
*In order to determine the average cost per item as production increases we must first find the total cost of three consecutive units (e.g the total cost of producing 23, 24, and 25 flags) and divide each total you get but the number of flags produced as shown below:        &lt;br /&gt;
The average cost equation: total cost/number of flags&lt;br /&gt;
&lt;br /&gt;
1. Let f=23&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(23)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=121&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 23 units is : 121/23=5.26&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Let f=24&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(24)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=128&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 24 units is : 128/24 =5.33&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Let f=25&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(25)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=135&#039;&#039;       &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 25 units is: 135/25=5.40&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* Based on the information above we can see that the averge cost per item increases as the production levels increase.     &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Finally, find some other models ( not necessarily linear) for which you get other behaviours such as&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
1. The average cost remains constant as production increases:       &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y=3x&#039;&#039;&#039;    &lt;br /&gt;
&lt;br /&gt;
* This model &#039;&#039;&#039;increases&#039;&#039;&#039; at a &#039;&#039;&#039;constant rate&#039;&#039;&#039;, because the slope of the function is the same everywhere. This model shows that the average cost will therefore remain constant as the production increases.    &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2. The average cost diminishes as the production increases:       &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= x^1/2&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* For this model when you plug in any postive numbers in for &#039;x&#039;, you will see that as &#039;x&#039; or the number of items produced increases, the average cost decreases:    &lt;br /&gt;
&lt;br /&gt;
ex.    &lt;br /&gt;
&lt;br /&gt;
y= x^1/2 , let x = 100 and let y represent the total cost of x&lt;br /&gt;
&lt;br /&gt;
y= (100^1/2)/ 100&lt;br /&gt;
&lt;br /&gt;
y= 0.1&lt;br /&gt;
&lt;br /&gt;
Now let x= 1000000&lt;br /&gt;
&lt;br /&gt;
y= (1000000^1/2)/1000000&lt;br /&gt;
&lt;br /&gt;
y= 0.001    &lt;br /&gt;
&lt;br /&gt;
* Here you can see that the average cost is diminishing as the production (&#039;x&#039;) increases.       &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
3. The average cost increases as production increases:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= e^x&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* In this model as the production increases so does the slope of the model which causes the average cost to also therefore increase.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
4. You obtain an economy of scale. This means that staring at some specific production level, the marginal cost is always less than the average cost:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= x^1/2&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* For this model, we can see that the marginal cost is less than the average cost&lt;br /&gt;
&lt;br /&gt;
ex.&lt;br /&gt;
&lt;br /&gt;
let x=3&lt;br /&gt;
&lt;br /&gt;
y=3^1/2&lt;br /&gt;
&lt;br /&gt;
y= 1.73&lt;br /&gt;
&lt;br /&gt;
* now we must divide by 3 to get the average    &lt;br /&gt;
&lt;br /&gt;
y= 1.73/3&lt;br /&gt;
&lt;br /&gt;
y= 0.577 , this means that it costs an average of $0.58 to produce 3 items       &lt;br /&gt;
&lt;br /&gt;
* And when x=2, y= 1.41&lt;br /&gt;
&lt;br /&gt;
* Therefore if we calculate the difference ($1.73-$1.41) we get 0.318 or $0.32. This means that it only cost $0.32 to produce the third item but it cost an average of $0.58 per item (of the 3 total items). This model therefore obtains an economy of scale.&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz&amp;diff=72687</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Schwytz</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz&amp;diff=72687"/>
		<updated>2011-01-26T05:59:13Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{&lt;br /&gt;
Infobox MATH110 Teams&lt;br /&gt;
| team name = Schwytz&lt;br /&gt;
| member 1 = Angus McWhirter&lt;br /&gt;
| member 2 = Anna Koniuhova&lt;br /&gt;
| member 3 = Cassandra Travlos&lt;br /&gt;
| member 4 = Gracie Mann&lt;br /&gt;
}}&lt;br /&gt;
&lt;br /&gt;
&amp;lt;dpl&amp;gt; titlematch={{Course:MATH110/003/Teams/Schwytz}}/% namespace={{NAMESPACE}} shownamespace=false &amp;lt;/dpl&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Key words that describe us==&lt;br /&gt;
Gracie- perseverance    &lt;br /&gt;
Cassandra- indecisive    &lt;br /&gt;
Anna- fun&lt;br /&gt;
Angus- mia&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
== Team Schwytz Subpages ==&lt;br /&gt;
&lt;br /&gt;
&amp;lt;dpl&amp;gt;&lt;br /&gt;
titlematch={{PAGENAME}}/%&lt;br /&gt;
namespace={{NAMESPACE}}&lt;br /&gt;
shownamespace=false&lt;br /&gt;
&amp;lt;/dpl&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Homework #11- Team Schwytz==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Write a linear model to predict the cost of producing flags of your teams Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items the cost is $100. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Our model is described by a linear function.&lt;br /&gt;
* We know that our model has to pass through the point (20,100) because the question tells us that the current production level of 20 items costs $100 (where the cost in dollar is the y-axis and the number of items is the x-axis) therefore we can use this point to formulate a function using the standard equation formula (y=mx +b)&lt;br /&gt;
* We can use 20 items for ‘x’ and 100 dollars for ‘y’&lt;br /&gt;
* Now we must find ‘m’&lt;br /&gt;
* To do this we need to use the formula: y2-y1/ x2-x1&lt;br /&gt;
To find a point on the graph:&lt;br /&gt;
&lt;br /&gt;
* We know that 20 items cost 100 dollars and the marginal cost is $7 per unit&lt;br /&gt;
* Therefore we know that 21 items costs $107 and then 22 items costs $114&lt;br /&gt;
* Now we can set this up in the above equation (where the cost is ‘y’ and the items is ‘x’:&lt;br /&gt;
114-107/ 22-21 = 7&lt;br /&gt;
&lt;br /&gt;
* Therefore:&lt;br /&gt;
&lt;br /&gt;
m=7&lt;br /&gt;
&lt;br /&gt;
* Now we need to find &#039;b&#039;:&lt;br /&gt;
&lt;br /&gt;
y=mx + b&lt;br /&gt;
&lt;br /&gt;
107 = 7 (21) +b&lt;br /&gt;
&lt;br /&gt;
b = -40&lt;br /&gt;
&lt;br /&gt;
* Therefore the equation of the line is:&lt;br /&gt;
&lt;br /&gt;
y= 7x - 40 , which is a linear model     &lt;br /&gt;
&lt;br /&gt;
* If we were to graph this model the function would cross the y-axis at -40.&lt;br /&gt;
&lt;br /&gt;
* However we can also determine the value of ‘b’ by understanding that it costs $100 to produce 20 flags which averages to $5 per flag. If the flags were $7 each it would have cost $140. The difference between these two costs is $40. Therefore to take into account the saving of $40 we make ‘b’ -40 in the formula y=mx+b&lt;br /&gt;
&lt;br /&gt;
* To relate this formula to the specific terminology of the question we will replace &#039;x&#039; with &#039;f&#039; and let &#039;f&#039; represent the number of flags produced (where &#039;f&#039; must be greater than 20). We will replace &#039;y&#039; with C(f) and let C(f) represent the cost to produce the flags:&lt;br /&gt;
C(f) = 7f - 40 , where f ≥ 20       &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;What does your model predict for production of a 150 items?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* In order to predict the total cost of the production of 150 items we can substitute into our model 150 for &#039;f&#039; . Which gives us the equation:       &lt;br /&gt;
&#039;&#039;C(f)=7(150)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=1010&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;According to your model, what happens when the average cost per item, as production levels increase?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
*In order to determine the average cost per item as production increases we must first find the total cost of three consecutive units (e.g the total cost of producing 23, 24, and 25 flags) and divide each total you get but the number of flags produced as shown below:        &lt;br /&gt;
The average cost equation: total cost/number of flags&lt;br /&gt;
&lt;br /&gt;
1. Let f=23&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(23)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=121&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 23 units is : 121/23=5.26&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Let f=24&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(24)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=128&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 24 units is : 128/24 =5.33&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Let f=25&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(25)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=135&#039;&#039;       &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 25 units is: 135/25=5.40&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* Based on the information above we can see that the averge cost per item increases as the production levels increase.     &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Finally, find some other models ( not necessarily linear) for which you get other behaviours such as&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
1. The average cost remains constant as production increases:       &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y=3x&#039;&#039;&#039;    &lt;br /&gt;
&lt;br /&gt;
* This model &#039;&#039;&#039;increases&#039;&#039;&#039; at a &#039;&#039;&#039;constant rate&#039;&#039;&#039;, because the slope of the function is the same everywhere. This model shows that the average cost will therefore remain constant as the production increases.    &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2. The average cost diminishes as the production increases:       &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= x^1/2&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* For this model when you plug in any postive numbers in for &#039;x&#039;, you will see that as &#039;x&#039; or the number of items produced increases, the average cost decreases:    &lt;br /&gt;
&lt;br /&gt;
ex.    &lt;br /&gt;
&lt;br /&gt;
y= x^1/2 , let x = 100 and let y represent the total cost of x&lt;br /&gt;
&lt;br /&gt;
y= (100^1/2)/ 100&lt;br /&gt;
&lt;br /&gt;
y= 0.1&lt;br /&gt;
&lt;br /&gt;
Now let x= 1000000&lt;br /&gt;
&lt;br /&gt;
y= (1000000^1/2)/1000000&lt;br /&gt;
&lt;br /&gt;
y= 0.001    &lt;br /&gt;
&lt;br /&gt;
* Here you can see that the average cost is diminishing as the production (&#039;x&#039;) increases.       &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
3. The average cost increases as production increases:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= e^x&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* In this model as the production increases so does the slope of the model which causes the average cost to also therefore increase.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
4. You obtain an economy of scale. This means that staring at some specific production level, the marginal cost is always less than the average cost:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= x^1/2&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* For this model, we can see that the marginal cost is less than the average cost&lt;br /&gt;
&lt;br /&gt;
ex.&lt;br /&gt;
&lt;br /&gt;
let x=3&lt;br /&gt;
&lt;br /&gt;
y=3^1/2&lt;br /&gt;
&lt;br /&gt;
y= 1.73&lt;br /&gt;
&lt;br /&gt;
* now we must divide by 3 to get the average    &lt;br /&gt;
&lt;br /&gt;
y= 1.73/3&lt;br /&gt;
&lt;br /&gt;
y= 0.577 , this means that it costs an average of $0.58 to produce 3 items       &lt;br /&gt;
&lt;br /&gt;
* And when x=2, y= 1.41&lt;br /&gt;
&lt;br /&gt;
* Therefore if we calculate the difference ($1.73-$1.41) we get 0.318 or $0.32. This means that it only cost $0.32 to produce the third item but it cost an average of $0.58 per item (of the 3 total items). This model therefore obtains an economy of scale.&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz&amp;diff=72686</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Schwytz</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz&amp;diff=72686"/>
		<updated>2011-01-26T05:58:27Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{&lt;br /&gt;
Infobox MATH110 Teams&lt;br /&gt;
| team name = Schwytz&lt;br /&gt;
| member 1 = Angus McWhirter&lt;br /&gt;
| member 2 = Anna Koniuhova&lt;br /&gt;
| member 3 = Cassandra Travlos&lt;br /&gt;
| member 4 = Gracie Mann&lt;br /&gt;
}}&lt;br /&gt;
&lt;br /&gt;
&amp;lt;dpl&amp;gt; titlematch={{PAGENAME}}/% namespace={{NAMESPACE}} shownamespace=false &amp;lt;/dpl&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Key words that describe us==&lt;br /&gt;
Gracie- perseverance   &lt;br /&gt;
Cassandra- indecisive   &lt;br /&gt;
Anna- fun&lt;br /&gt;
Angus- mia&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
== Team Schwytz Subpages ==&lt;br /&gt;
&lt;br /&gt;
&amp;lt;dpl&amp;gt;&lt;br /&gt;
titlematch={{PAGENAME}}/%&lt;br /&gt;
namespace={{NAMESPACE}}&lt;br /&gt;
shownamespace=false&lt;br /&gt;
&amp;lt;/dpl&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Homework #11- Team Schwytz==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Write a linear model to predict the cost of producing flags of your teams Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items the cost is $100. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Our model is described by a linear function.&lt;br /&gt;
* We know that our model has to pass through the point (20,100) because the question tells us that the current production level of 20 items costs $100 (where the cost in dollar is the y-axis and the number of items is the x-axis) therefore we can use this point to formulate a function using the standard equation formula (y=mx +b)&lt;br /&gt;
* We can use 20 items for ‘x’ and 100 dollars for ‘y’&lt;br /&gt;
* Now we must find ‘m’&lt;br /&gt;
* To do this we need to use the formula: y2-y1/ x2-x1&lt;br /&gt;
To find a point on the graph:&lt;br /&gt;
&lt;br /&gt;
* We know that 20 items cost 100 dollars and the marginal cost is $7 per unit&lt;br /&gt;
* Therefore we know that 21 items costs $107 and then 22 items costs $114&lt;br /&gt;
* Now we can set this up in the above equation (where the cost is ‘y’ and the items is ‘x’:&lt;br /&gt;
114-107/ 22-21 = 7&lt;br /&gt;
&lt;br /&gt;
* Therefore:&lt;br /&gt;
&lt;br /&gt;
m=7&lt;br /&gt;
&lt;br /&gt;
* Now we need to find &#039;b&#039;:&lt;br /&gt;
&lt;br /&gt;
y=mx + b&lt;br /&gt;
&lt;br /&gt;
107 = 7 (21) +b&lt;br /&gt;
&lt;br /&gt;
b = -40&lt;br /&gt;
&lt;br /&gt;
* Therefore the equation of the line is:&lt;br /&gt;
&lt;br /&gt;
y= 7x - 40 , which is a linear model    &lt;br /&gt;
&lt;br /&gt;
* If we were to graph this model the function would cross the y-axis at -40.&lt;br /&gt;
&lt;br /&gt;
* However we can also determine the value of ‘b’ by understanding that it costs $100 to produce 20 flags which averages to $5 per flag. If the flags were $7 each it would have cost $140. The difference between these two costs is $40. Therefore to take into account the saving of $40 we make ‘b’ -40 in the formula y=mx+b&lt;br /&gt;
&lt;br /&gt;
* To relate this formula to the specific terminology of the question we will replace &#039;x&#039; with &#039;f&#039; and let &#039;f&#039; represent the number of flags produced (where &#039;f&#039; must be greater than 20). We will replace &#039;y&#039; with C(f) and let C(f) represent the cost to produce the flags:&lt;br /&gt;
C(f) = 7f - 40 , where f ≥ 20      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;What does your model predict for production of a 150 items?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* In order to predict the total cost of the production of 150 items we can substitute into our model 150 for &#039;f&#039; . Which gives us the equation:      &lt;br /&gt;
&#039;&#039;C(f)=7(150)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=1010&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;According to your model, what happens when the average cost per item, as production levels increase?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
*In order to determine the average cost per item as production increases we must first find the total cost of three consecutive units (e.g the total cost of producing 23, 24, and 25 flags) and divide each total you get but the number of flags produced as shown below:       &lt;br /&gt;
The average cost equation: total cost/number of flags&lt;br /&gt;
&lt;br /&gt;
1. Let f=23&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(23)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=121&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 23 units is : 121/23=5.26&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Let f=24&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(24)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=128&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 24 units is : 128/24 =5.33&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Let f=25&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(25)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=135&#039;&#039;      &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 25 units is: 135/25=5.40&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* Based on the information above we can see that the averge cost per item increases as the production levels increase.    &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Finally, find some other models ( not necessarily linear) for which you get other behaviours such as&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
1. The average cost remains constant as production increases:      &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y=3x&#039;&#039;&#039;   &lt;br /&gt;
&lt;br /&gt;
* This model &#039;&#039;&#039;increases&#039;&#039;&#039; at a &#039;&#039;&#039;constant rate&#039;&#039;&#039;, because the slope of the function is the same everywhere. This model shows that the average cost will therefore remain constant as the production increases.   &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2. The average cost diminishes as the production increases:      &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= x^1/2&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* For this model when you plug in any postive numbers in for &#039;x&#039;, you will see that as &#039;x&#039; or the number of items produced increases, the average cost decreases:   &lt;br /&gt;
&lt;br /&gt;
ex.   &lt;br /&gt;
&lt;br /&gt;
y= x^1/2 , let x = 100 and let y represent the total cost of x&lt;br /&gt;
&lt;br /&gt;
y= (100^1/2)/ 100&lt;br /&gt;
&lt;br /&gt;
y= 0.1&lt;br /&gt;
&lt;br /&gt;
Now let x= 1000000&lt;br /&gt;
&lt;br /&gt;
y= (1000000^1/2)/1000000&lt;br /&gt;
&lt;br /&gt;
y= 0.001   &lt;br /&gt;
&lt;br /&gt;
* Here you can see that the average cost is diminishing as the production (&#039;x&#039;) increases.      &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
3. The average cost increases as production increases:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= e^x&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* In this model as the production increases so does the slope of the model which causes the average cost to also therefore increase.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
4. You obtain an economy of scale. This means that staring at some specific production level, the marginal cost is always less than the average cost:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= x^1/2&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* For this model, we can see that the marginal cost is less than the average cost&lt;br /&gt;
&lt;br /&gt;
ex.&lt;br /&gt;
&lt;br /&gt;
let x=3&lt;br /&gt;
&lt;br /&gt;
y=3^1/2&lt;br /&gt;
&lt;br /&gt;
y= 1.73&lt;br /&gt;
&lt;br /&gt;
* now we must divide by 3 to get the average   &lt;br /&gt;
&lt;br /&gt;
y= 1.73/3&lt;br /&gt;
&lt;br /&gt;
y= 0.577 , this means that it costs an average of $0.58 to produce 3 items      &lt;br /&gt;
&lt;br /&gt;
* And when x=2, y= 1.41&lt;br /&gt;
&lt;br /&gt;
* Therefore if we calculate the difference ($1.73-$1.41) we get 0.318 or $0.32. This means that it only cost $0.32 to produce the third item but it cost an average of $0.58 per item (of the 3 total items). This model therefore obtains an economy of scale.&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz&amp;diff=72685</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Schwytz</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz&amp;diff=72685"/>
		<updated>2011-01-26T05:56:24Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: /* Key words that describe us */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{&lt;br /&gt;
Infobox MATH110 Teams&lt;br /&gt;
| team name = Schwytz&lt;br /&gt;
| member 1 = Angus McWhirter&lt;br /&gt;
| member 2 = Anna Koniuhova&lt;br /&gt;
| member 3 = Cassandra Travlos&lt;br /&gt;
| member 4 = Gracie Mann&lt;br /&gt;
}}&lt;br /&gt;
In workshop J.&lt;br /&gt;
==Key words that describe us==&lt;br /&gt;
Gracie- perseverance  &lt;br /&gt;
Cassandra- indecisive  &lt;br /&gt;
Anna- fun&lt;br /&gt;
Angus- mia&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
== Team Schwytz Subpages ==&lt;br /&gt;
&lt;br /&gt;
&amp;lt;dpl&amp;gt;&lt;br /&gt;
titlematch={{PAGENAME}}/%&lt;br /&gt;
namespace={{NAMESPACE}}&lt;br /&gt;
shownamespace=false&lt;br /&gt;
&amp;lt;/dpl&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Homework #11- Team Schwytz==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Write a linear model to predict the cost of producing flags of your teams Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items the cost is $100. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Our model is described by a linear function.&lt;br /&gt;
* We know that our model has to pass through the point (20,100) because the question tells us that the current production level of 20 items costs $100 (where the cost in dollar is the y-axis and the number of items is the x-axis) therefore we can use this point to formulate a function using the standard equation formula (y=mx +b)&lt;br /&gt;
* We can use 20 items for ‘x’ and 100 dollars for ‘y’&lt;br /&gt;
* Now we must find ‘m’&lt;br /&gt;
* To do this we need to use the formula: y2-y1/ x2-x1&lt;br /&gt;
To find a point on the graph:&lt;br /&gt;
&lt;br /&gt;
* We know that 20 items cost 100 dollars and the marginal cost is $7 per unit&lt;br /&gt;
* Therefore we know that 21 items costs $107 and then 22 items costs $114&lt;br /&gt;
* Now we can set this up in the above equation (where the cost is ‘y’ and the items is ‘x’:&lt;br /&gt;
114-107/ 22-21 = 7&lt;br /&gt;
&lt;br /&gt;
* Therefore:&lt;br /&gt;
&lt;br /&gt;
m=7&lt;br /&gt;
&lt;br /&gt;
* Now we need to find &#039;b&#039;:&lt;br /&gt;
&lt;br /&gt;
y=mx + b&lt;br /&gt;
&lt;br /&gt;
107 = 7 (21) +b&lt;br /&gt;
&lt;br /&gt;
b = -40&lt;br /&gt;
&lt;br /&gt;
* Therefore the equation of the line is:&lt;br /&gt;
&lt;br /&gt;
y= 7x - 40 , which is a linear model   &lt;br /&gt;
&lt;br /&gt;
* If we were to graph this model the function would cross the y-axis at -40.&lt;br /&gt;
&lt;br /&gt;
* However we can also determine the value of ‘b’ by understanding that it costs $100 to produce 20 flags which averages to $5 per flag. If the flags were $7 each it would have cost $140. The difference between these two costs is $40. Therefore to take into account the saving of $40 we make ‘b’ -40 in the formula y=mx+b&lt;br /&gt;
&lt;br /&gt;
* To relate this formula to the specific terminology of the question we will replace &#039;x&#039; with &#039;f&#039; and let &#039;f&#039; represent the number of flags produced (where &#039;f&#039; must be greater than 20). We will replace &#039;y&#039; with C(f) and let C(f) represent the cost to produce the flags:&lt;br /&gt;
C(f) = 7f - 40 , where f ≥ 20     &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;What does your model predict for production of a 150 items?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* In order to predict the total cost of the production of 150 items we can substitute into our model 150 for &#039;f&#039; . Which gives us the equation:     &lt;br /&gt;
&#039;&#039;C(f)=7(150)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=1010&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;According to your model, what happens when the average cost per item, as production levels increase?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
*In order to determine the average cost per item as production increases we must first find the total cost of three consecutive units (e.g the total cost of producing 23, 24, and 25 flags) and divide each total you get but the number of flags produced as shown below:      &lt;br /&gt;
The average cost equation: total cost/number of flags&lt;br /&gt;
&lt;br /&gt;
1. Let f=23&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(23)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=121&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 23 units is : 121/23=5.26&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Let f=24&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(24)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=128&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 24 units is : 128/24 =5.33&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Let f=25&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(25)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=135&#039;&#039;     &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 25 units is: 135/25=5.40&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* Based on the information above we can see that the averge cost per item increases as the production levels increase.   &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Finally, find some other models ( not necessarily linear) for which you get other behaviours such as&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
1. The average cost remains constant as production increases:     &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y=3x&#039;&#039;&#039;  &lt;br /&gt;
&lt;br /&gt;
* This model &#039;&#039;&#039;increases&#039;&#039;&#039; at a &#039;&#039;&#039;constant rate&#039;&#039;&#039;, because the slope of the function is the same everywhere. This model shows that the average cost will therefore remain constant as the production increases.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2. The average cost diminishes as the production increases:     &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= x^1/2&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* For this model when you plug in any postive numbers in for &#039;x&#039;, you will see that as &#039;x&#039; or the number of items produced increases, the average cost decreases:  &lt;br /&gt;
&lt;br /&gt;
ex.  &lt;br /&gt;
&lt;br /&gt;
y= x^1/2 , let x = 100 and let y represent the total cost of x&lt;br /&gt;
&lt;br /&gt;
y= (100^1/2)/ 100&lt;br /&gt;
&lt;br /&gt;
y= 0.1&lt;br /&gt;
&lt;br /&gt;
Now let x= 1000000&lt;br /&gt;
&lt;br /&gt;
y= (1000000^1/2)/1000000&lt;br /&gt;
&lt;br /&gt;
y= 0.001  &lt;br /&gt;
&lt;br /&gt;
* Here you can see that the average cost is diminishing as the production (&#039;x&#039;) increases.     &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
3. The average cost increases as production increases:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= e^x&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* In this model as the production increases so does the slope of the model which causes the average cost to also therefore increase.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
4. You obtain an economy of scale. This means that staring at some specific production level, the marginal cost is always less than the average cost:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= x^1/2&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* For this model, we can see that the marginal cost is less than the average cost&lt;br /&gt;
&lt;br /&gt;
ex.&lt;br /&gt;
&lt;br /&gt;
let x=3&lt;br /&gt;
&lt;br /&gt;
y=3^1/2&lt;br /&gt;
&lt;br /&gt;
y= 1.73&lt;br /&gt;
&lt;br /&gt;
* now we must divide by 3 to get the average  &lt;br /&gt;
&lt;br /&gt;
y= 1.73/3&lt;br /&gt;
&lt;br /&gt;
y= 0.577 , this means that it costs an average of $0.58 to produce 3 items     &lt;br /&gt;
&lt;br /&gt;
* And when x=2, y= 1.41&lt;br /&gt;
&lt;br /&gt;
* Therefore if we calculate the difference ($1.73-$1.41) we get 0.318 or $0.32. This means that it only cost $0.32 to produce the third item but it cost an average of $0.58 per item (of the 3 total items). This model therefore obtains an economy of scale.&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz&amp;diff=72684</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Schwytz</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz&amp;diff=72684"/>
		<updated>2011-01-26T05:54:28Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: /* Key words that describe us */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{&lt;br /&gt;
Infobox MATH110 Teams&lt;br /&gt;
| team name = Schwytz&lt;br /&gt;
| member 1 = Angus McWhirter&lt;br /&gt;
| member 2 = Anna Koniuhova&lt;br /&gt;
| member 3 = Cassandra Travlos&lt;br /&gt;
| member 4 = Gracie Mann&lt;br /&gt;
}}&lt;br /&gt;
In workshop J.&lt;br /&gt;
==Key words that describe us==&lt;br /&gt;
Gracie- perseverance &lt;br /&gt;
Cassandra- indecisive &lt;br /&gt;
Anna- fun&lt;br /&gt;
Angus- mia&lt;br /&gt;
&lt;br /&gt;
==Homework #11- Team Schwytz==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Write a linear model to predict the cost of producing flags of your teams Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items the cost is $100. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Our model is described by a linear function.&lt;br /&gt;
* We know that our model has to pass through the point (20,100) because the question tells us that the current production level of 20 items costs $100 (where the cost in dollar is the y-axis and the number of items is the x-axis) therefore we can use this point to formulate a function using the standard equation formula (y=mx +b)&lt;br /&gt;
* We can use 20 items for ‘x’ and 100 dollars for ‘y’&lt;br /&gt;
* Now we must find ‘m’&lt;br /&gt;
* To do this we need to use the formula: y2-y1/ x2-x1&lt;br /&gt;
To find a point on the graph:&lt;br /&gt;
&lt;br /&gt;
* We know that 20 items cost 100 dollars and the marginal cost is $7 per unit&lt;br /&gt;
* Therefore we know that 21 items costs $107 and then 22 items costs $114&lt;br /&gt;
* Now we can set this up in the above equation (where the cost is ‘y’ and the items is ‘x’:&lt;br /&gt;
114-107/ 22-21 = 7&lt;br /&gt;
&lt;br /&gt;
* Therefore:&lt;br /&gt;
&lt;br /&gt;
m=7&lt;br /&gt;
&lt;br /&gt;
* Now we need to find &#039;b&#039;:&lt;br /&gt;
&lt;br /&gt;
y=mx + b&lt;br /&gt;
&lt;br /&gt;
107 = 7 (21) +b&lt;br /&gt;
&lt;br /&gt;
b = -40&lt;br /&gt;
&lt;br /&gt;
* Therefore the equation of the line is:&lt;br /&gt;
&lt;br /&gt;
y= 7x - 40 , which is a linear model   &lt;br /&gt;
&lt;br /&gt;
* If we were to graph this model the function would cross the y-axis at -40.&lt;br /&gt;
&lt;br /&gt;
* However we can also determine the value of ‘b’ by understanding that it costs $100 to produce 20 flags which averages to $5 per flag. If the flags were $7 each it would have cost $140. The difference between these two costs is $40. Therefore to take into account the saving of $40 we make ‘b’ -40 in the formula y=mx+b&lt;br /&gt;
&lt;br /&gt;
* To relate this formula to the specific terminology of the question we will replace &#039;x&#039; with &#039;f&#039; and let &#039;f&#039; represent the number of flags produced (where &#039;f&#039; must be greater than 20). We will replace &#039;y&#039; with C(f) and let C(f) represent the cost to produce the flags:&lt;br /&gt;
C(f) = 7f - 40 , where f ≥ 20     &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;What does your model predict for production of a 150 items?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* In order to predict the total cost of the production of 150 items we can substitute into our model 150 for &#039;f&#039; . Which gives us the equation:     &lt;br /&gt;
&#039;&#039;C(f)=7(150)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=1010&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;According to your model, what happens when the average cost per item, as production levels increase?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
*In order to determine the average cost per item as production increases we must first find the total cost of three consecutive units (e.g the total cost of producing 23, 24, and 25 flags) and divide each total you get but the number of flags produced as shown below:      &lt;br /&gt;
The average cost equation: total cost/number of flags&lt;br /&gt;
&lt;br /&gt;
1. Let f=23&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(23)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=121&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 23 units is : 121/23=5.26&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Let f=24&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(24)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=128&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 24 units is : 128/24 =5.33&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Let f=25&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(25)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=135&#039;&#039;     &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 25 units is: 135/25=5.40&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* Based on the information above we can see that the averge cost per item increases as the production levels increase.   &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Finally, find some other models ( not necessarily linear) for which you get other behaviours such as&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
1. The average cost remains constant as production increases:     &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y=3x&#039;&#039;&#039;  &lt;br /&gt;
&lt;br /&gt;
* This model &#039;&#039;&#039;increases&#039;&#039;&#039; at a &#039;&#039;&#039;constant rate&#039;&#039;&#039;, because the slope of the function is the same everywhere. This model shows that the average cost will therefore remain constant as the production increases.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2. The average cost diminishes as the production increases:     &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= x^1/2&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* For this model when you plug in any postive numbers in for &#039;x&#039;, you will see that as &#039;x&#039; or the number of items produced increases, the average cost decreases:  &lt;br /&gt;
&lt;br /&gt;
ex.  &lt;br /&gt;
&lt;br /&gt;
y= x^1/2 , let x = 100 and let y represent the total cost of x&lt;br /&gt;
&lt;br /&gt;
y= (100^1/2)/ 100&lt;br /&gt;
&lt;br /&gt;
y= 0.1&lt;br /&gt;
&lt;br /&gt;
Now let x= 1000000&lt;br /&gt;
&lt;br /&gt;
y= (1000000^1/2)/1000000&lt;br /&gt;
&lt;br /&gt;
y= 0.001  &lt;br /&gt;
&lt;br /&gt;
* Here you can see that the average cost is diminishing as the production (&#039;x&#039;) increases.     &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
3. The average cost increases as production increases:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= e^x&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* In this model as the production increases so does the slope of the model which causes the average cost to also therefore increase.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
4. You obtain an economy of scale. This means that staring at some specific production level, the marginal cost is always less than the average cost:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= x^1/2&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* For this model, we can see that the marginal cost is less than the average cost&lt;br /&gt;
&lt;br /&gt;
ex.&lt;br /&gt;
&lt;br /&gt;
let x=3&lt;br /&gt;
&lt;br /&gt;
y=3^1/2&lt;br /&gt;
&lt;br /&gt;
y= 1.73&lt;br /&gt;
&lt;br /&gt;
* now we must divide by 3 to get the average  &lt;br /&gt;
&lt;br /&gt;
y= 1.73/3&lt;br /&gt;
&lt;br /&gt;
y= 0.577 , this means that it costs an average of $0.58 to produce 3 items     &lt;br /&gt;
&lt;br /&gt;
* And when x=2, y= 1.41&lt;br /&gt;
&lt;br /&gt;
* Therefore if we calculate the difference ($1.73-$1.41) we get 0.318 or $0.32. This means that it only cost $0.32 to produce the third item but it cost an average of $0.58 per item (of the 3 total items). This model therefore obtains an economy of scale.&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz&amp;diff=72683</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Schwytz</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz&amp;diff=72683"/>
		<updated>2011-01-26T05:53:48Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: /* Key words that describe us */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{&lt;br /&gt;
Infobox MATH110 Teams&lt;br /&gt;
| team name = Schwytz&lt;br /&gt;
| member 1 = Angus McWhirter&lt;br /&gt;
| member 2 = Anna Koniuhova&lt;br /&gt;
| member 3 = Cassandra Travlos&lt;br /&gt;
| member 4 = Gracie Mann&lt;br /&gt;
}}&lt;br /&gt;
In workshop J.&lt;br /&gt;
==Key words that describe us==&lt;br /&gt;
Gracie- Perseverance &lt;br /&gt;
Cassandra- indecisive &lt;br /&gt;
Anna- fun&lt;br /&gt;
Angus- mia&lt;br /&gt;
&lt;br /&gt;
&amp;lt;dpl&amp;gt;&lt;br /&gt;
titlematch={{Course: Mathh110/003/Teams/Schwytz}}/%&lt;br /&gt;
namespace={{NAMESPACE}}&lt;br /&gt;
shownamespace=false&lt;br /&gt;
&amp;lt;/dpl&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Homework #11- Team Schwytz==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Write a linear model to predict the cost of producing flags of your teams Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items the cost is $100. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Our model is described by a linear function.&lt;br /&gt;
* We know that our model has to pass through the point (20,100) because the question tells us that the current production level of 20 items costs $100 (where the cost in dollar is the y-axis and the number of items is the x-axis) therefore we can use this point to formulate a function using the standard equation formula (y=mx +b)&lt;br /&gt;
* We can use 20 items for ‘x’ and 100 dollars for ‘y’&lt;br /&gt;
* Now we must find ‘m’&lt;br /&gt;
* To do this we need to use the formula: y2-y1/ x2-x1&lt;br /&gt;
To find a point on the graph:&lt;br /&gt;
&lt;br /&gt;
* We know that 20 items cost 100 dollars and the marginal cost is $7 per unit&lt;br /&gt;
* Therefore we know that 21 items costs $107 and then 22 items costs $114&lt;br /&gt;
* Now we can set this up in the above equation (where the cost is ‘y’ and the items is ‘x’:&lt;br /&gt;
114-107/ 22-21 = 7&lt;br /&gt;
&lt;br /&gt;
* Therefore:&lt;br /&gt;
&lt;br /&gt;
m=7&lt;br /&gt;
&lt;br /&gt;
* Now we need to find &#039;b&#039;:&lt;br /&gt;
&lt;br /&gt;
y=mx + b&lt;br /&gt;
&lt;br /&gt;
107 = 7 (21) +b&lt;br /&gt;
&lt;br /&gt;
b = -40&lt;br /&gt;
&lt;br /&gt;
* Therefore the equation of the line is:&lt;br /&gt;
&lt;br /&gt;
y= 7x - 40 , which is a linear model   &lt;br /&gt;
&lt;br /&gt;
* If we were to graph this model the function would cross the y-axis at -40.&lt;br /&gt;
&lt;br /&gt;
* However we can also determine the value of ‘b’ by understanding that it costs $100 to produce 20 flags which averages to $5 per flag. If the flags were $7 each it would have cost $140. The difference between these two costs is $40. Therefore to take into account the saving of $40 we make ‘b’ -40 in the formula y=mx+b&lt;br /&gt;
&lt;br /&gt;
* To relate this formula to the specific terminology of the question we will replace &#039;x&#039; with &#039;f&#039; and let &#039;f&#039; represent the number of flags produced (where &#039;f&#039; must be greater than 20). We will replace &#039;y&#039; with C(f) and let C(f) represent the cost to produce the flags:&lt;br /&gt;
C(f) = 7f - 40 , where f ≥ 20     &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;What does your model predict for production of a 150 items?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* In order to predict the total cost of the production of 150 items we can substitute into our model 150 for &#039;f&#039; . Which gives us the equation:     &lt;br /&gt;
&#039;&#039;C(f)=7(150)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=1010&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;According to your model, what happens when the average cost per item, as production levels increase?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
*In order to determine the average cost per item as production increases we must first find the total cost of three consecutive units (e.g the total cost of producing 23, 24, and 25 flags) and divide each total you get but the number of flags produced as shown below:      &lt;br /&gt;
The average cost equation: total cost/number of flags&lt;br /&gt;
&lt;br /&gt;
1. Let f=23&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(23)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=121&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 23 units is : 121/23=5.26&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Let f=24&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(24)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=128&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 24 units is : 128/24 =5.33&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Let f=25&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(25)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=135&#039;&#039;     &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 25 units is: 135/25=5.40&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* Based on the information above we can see that the averge cost per item increases as the production levels increase.   &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Finally, find some other models ( not necessarily linear) for which you get other behaviours such as&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
1. The average cost remains constant as production increases:     &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y=3x&#039;&#039;&#039;  &lt;br /&gt;
&lt;br /&gt;
* This model &#039;&#039;&#039;increases&#039;&#039;&#039; at a &#039;&#039;&#039;constant rate&#039;&#039;&#039;, because the slope of the function is the same everywhere. This model shows that the average cost will therefore remain constant as the production increases.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2. The average cost diminishes as the production increases:     &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= x^1/2&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* For this model when you plug in any postive numbers in for &#039;x&#039;, you will see that as &#039;x&#039; or the number of items produced increases, the average cost decreases:  &lt;br /&gt;
&lt;br /&gt;
ex.  &lt;br /&gt;
&lt;br /&gt;
y= x^1/2 , let x = 100 and let y represent the total cost of x&lt;br /&gt;
&lt;br /&gt;
y= (100^1/2)/ 100&lt;br /&gt;
&lt;br /&gt;
y= 0.1&lt;br /&gt;
&lt;br /&gt;
Now let x= 1000000&lt;br /&gt;
&lt;br /&gt;
y= (1000000^1/2)/1000000&lt;br /&gt;
&lt;br /&gt;
y= 0.001  &lt;br /&gt;
&lt;br /&gt;
* Here you can see that the average cost is diminishing as the production (&#039;x&#039;) increases.     &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
3. The average cost increases as production increases:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= e^x&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* In this model as the production increases so does the slope of the model which causes the average cost to also therefore increase.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
4. You obtain an economy of scale. This means that staring at some specific production level, the marginal cost is always less than the average cost:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= x^1/2&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* For this model, we can see that the marginal cost is less than the average cost&lt;br /&gt;
&lt;br /&gt;
ex.&lt;br /&gt;
&lt;br /&gt;
let x=3&lt;br /&gt;
&lt;br /&gt;
y=3^1/2&lt;br /&gt;
&lt;br /&gt;
y= 1.73&lt;br /&gt;
&lt;br /&gt;
* now we must divide by 3 to get the average  &lt;br /&gt;
&lt;br /&gt;
y= 1.73/3&lt;br /&gt;
&lt;br /&gt;
y= 0.577 , this means that it costs an average of $0.58 to produce 3 items     &lt;br /&gt;
&lt;br /&gt;
* And when x=2, y= 1.41&lt;br /&gt;
&lt;br /&gt;
* Therefore if we calculate the difference ($1.73-$1.41) we get 0.318 or $0.32. This means that it only cost $0.32 to produce the third item but it cost an average of $0.58 per item (of the 3 total items). This model therefore obtains an economy of scale.&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz&amp;diff=72681</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Schwytz</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz&amp;diff=72681"/>
		<updated>2011-01-26T05:52:50Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: /* Homework #11- Team Schwytz */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{&lt;br /&gt;
Infobox MATH110 Teams&lt;br /&gt;
| team name = Schwytz&lt;br /&gt;
| member 1 = Angus McWhirter&lt;br /&gt;
| member 2 = Anna Koniuhova&lt;br /&gt;
| member 3 = Cassandra Travlos&lt;br /&gt;
| member 4 = Gracie Mann&lt;br /&gt;
}}&lt;br /&gt;
In workshop J.&lt;br /&gt;
==Key words that describe us==&lt;br /&gt;
Gracie- Perseverance &lt;br /&gt;
Cassandra- indecisive &lt;br /&gt;
Anna- fun&lt;br /&gt;
Angus- mia&lt;br /&gt;
&lt;br /&gt;
&amp;lt;dpl&amp;gt;&lt;br /&gt;
titlematch={{PAGENAME}}/%&lt;br /&gt;
namespace={{NAMESPACE}}&lt;br /&gt;
shownamespace=false&lt;br /&gt;
&amp;lt;/dpl&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Homework #11- Team Schwytz==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Write a linear model to predict the cost of producing flags of your teams Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items the cost is $100. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Our model is described by a linear function.&lt;br /&gt;
* We know that our model has to pass through the point (20,100) because the question tells us that the current production level of 20 items costs $100 (where the cost in dollar is the y-axis and the number of items is the x-axis) therefore we can use this point to formulate a function using the standard equation formula (y=mx +b)&lt;br /&gt;
* We can use 20 items for ‘x’ and 100 dollars for ‘y’&lt;br /&gt;
* Now we must find ‘m’&lt;br /&gt;
* To do this we need to use the formula: y2-y1/ x2-x1&lt;br /&gt;
To find a point on the graph:&lt;br /&gt;
&lt;br /&gt;
* We know that 20 items cost 100 dollars and the marginal cost is $7 per unit&lt;br /&gt;
* Therefore we know that 21 items costs $107 and then 22 items costs $114&lt;br /&gt;
* Now we can set this up in the above equation (where the cost is ‘y’ and the items is ‘x’:&lt;br /&gt;
114-107/ 22-21 = 7&lt;br /&gt;
&lt;br /&gt;
* Therefore:&lt;br /&gt;
&lt;br /&gt;
m=7&lt;br /&gt;
&lt;br /&gt;
* Now we need to find &#039;b&#039;:&lt;br /&gt;
&lt;br /&gt;
y=mx + b&lt;br /&gt;
&lt;br /&gt;
107 = 7 (21) +b&lt;br /&gt;
&lt;br /&gt;
b = -40&lt;br /&gt;
&lt;br /&gt;
* Therefore the equation of the line is:&lt;br /&gt;
&lt;br /&gt;
y= 7x - 40 , which is a linear model   &lt;br /&gt;
&lt;br /&gt;
* If we were to graph this model the function would cross the y-axis at -40.&lt;br /&gt;
&lt;br /&gt;
* However we can also determine the value of ‘b’ by understanding that it costs $100 to produce 20 flags which averages to $5 per flag. If the flags were $7 each it would have cost $140. The difference between these two costs is $40. Therefore to take into account the saving of $40 we make ‘b’ -40 in the formula y=mx+b&lt;br /&gt;
&lt;br /&gt;
* To relate this formula to the specific terminology of the question we will replace &#039;x&#039; with &#039;f&#039; and let &#039;f&#039; represent the number of flags produced (where &#039;f&#039; must be greater than 20). We will replace &#039;y&#039; with C(f) and let C(f) represent the cost to produce the flags:&lt;br /&gt;
C(f) = 7f - 40 , where f ≥ 20     &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;What does your model predict for production of a 150 items?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* In order to predict the total cost of the production of 150 items we can substitute into our model 150 for &#039;f&#039; . Which gives us the equation:     &lt;br /&gt;
&#039;&#039;C(f)=7(150)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=1010&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;According to your model, what happens when the average cost per item, as production levels increase?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
*In order to determine the average cost per item as production increases we must first find the total cost of three consecutive units (e.g the total cost of producing 23, 24, and 25 flags) and divide each total you get but the number of flags produced as shown below:      &lt;br /&gt;
The average cost equation: total cost/number of flags&lt;br /&gt;
&lt;br /&gt;
1. Let f=23&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(23)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=121&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 23 units is : 121/23=5.26&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Let f=24&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(24)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=128&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 24 units is : 128/24 =5.33&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Let f=25&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(25)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=135&#039;&#039;     &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 25 units is: 135/25=5.40&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* Based on the information above we can see that the averge cost per item increases as the production levels increase.   &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Finally, find some other models ( not necessarily linear) for which you get other behaviours such as&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
1. The average cost remains constant as production increases:     &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y=3x&#039;&#039;&#039;  &lt;br /&gt;
&lt;br /&gt;
* This model &#039;&#039;&#039;increases&#039;&#039;&#039; at a &#039;&#039;&#039;constant rate&#039;&#039;&#039;, because the slope of the function is the same everywhere. This model shows that the average cost will therefore remain constant as the production increases.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2. The average cost diminishes as the production increases:     &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= x^1/2&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* For this model when you plug in any postive numbers in for &#039;x&#039;, you will see that as &#039;x&#039; or the number of items produced increases, the average cost decreases:  &lt;br /&gt;
&lt;br /&gt;
ex.  &lt;br /&gt;
&lt;br /&gt;
y= x^1/2 , let x = 100 and let y represent the total cost of x&lt;br /&gt;
&lt;br /&gt;
y= (100^1/2)/ 100&lt;br /&gt;
&lt;br /&gt;
y= 0.1&lt;br /&gt;
&lt;br /&gt;
Now let x= 1000000&lt;br /&gt;
&lt;br /&gt;
y= (1000000^1/2)/1000000&lt;br /&gt;
&lt;br /&gt;
y= 0.001  &lt;br /&gt;
&lt;br /&gt;
* Here you can see that the average cost is diminishing as the production (&#039;x&#039;) increases.     &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
3. The average cost increases as production increases:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= e^x&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* In this model as the production increases so does the slope of the model which causes the average cost to also therefore increase.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
4. You obtain an economy of scale. This means that staring at some specific production level, the marginal cost is always less than the average cost:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= x^1/2&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* For this model, we can see that the marginal cost is less than the average cost&lt;br /&gt;
&lt;br /&gt;
ex.&lt;br /&gt;
&lt;br /&gt;
let x=3&lt;br /&gt;
&lt;br /&gt;
y=3^1/2&lt;br /&gt;
&lt;br /&gt;
y= 1.73&lt;br /&gt;
&lt;br /&gt;
* now we must divide by 3 to get the average  &lt;br /&gt;
&lt;br /&gt;
y= 1.73/3&lt;br /&gt;
&lt;br /&gt;
y= 0.577 , this means that it costs an average of $0.58 to produce 3 items     &lt;br /&gt;
&lt;br /&gt;
* And when x=2, y= 1.41&lt;br /&gt;
&lt;br /&gt;
* Therefore if we calculate the difference ($1.73-$1.41) we get 0.318 or $0.32. This means that it only cost $0.32 to produce the third item but it cost an average of $0.58 per item (of the 3 total items). This model therefore obtains an economy of scale.&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:CassandraTravlos&amp;diff=72626</id>
		<title>User:CassandraTravlos</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:CassandraTravlos&amp;diff=72626"/>
		<updated>2011-01-26T03:34:55Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;[[About me..]]&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
[[Essay: Calculus and the Faculty of Human Kinetics at UBC]]&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:CassandraTravlos&amp;diff=72624</id>
		<title>User:CassandraTravlos</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:CassandraTravlos&amp;diff=72624"/>
		<updated>2011-01-26T03:31:37Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;[[wiki.ubc.ca/User:CassandraTravlos/About me..]]&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
==Homework #12==&lt;br /&gt;
 &lt;br /&gt;
[[wiki.ubc.ca/User:CassandraTravlos/Essay: Calculus and the Faculty of Human Kinetics at UBC]]&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:CassandraTravlos&amp;diff=72623</id>
		<title>User:CassandraTravlos</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:CassandraTravlos&amp;diff=72623"/>
		<updated>2011-01-26T03:30:57Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;wiki.ubc.ca/User:CassandraTravlos/About me..&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
==Homework #12==&lt;br /&gt;
 &lt;br /&gt;
wiki.ubc.ca/User:CassandraTravlos/Essay: Calculus and the Faculty of Human Kinetics at UBC]]&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:CassandraTravlos&amp;diff=72108</id>
		<title>User:CassandraTravlos</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:CassandraTravlos&amp;diff=72108"/>
		<updated>2011-01-25T07:20:23Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;[[About me..]]&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
==Homework #12==&lt;br /&gt;
 &lt;br /&gt;
[[Essay: Calculus and the Faculty of Human Kinetics at UBC]]&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:CassandraTravlos&amp;diff=72106</id>
		<title>User:CassandraTravlos</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:CassandraTravlos&amp;diff=72106"/>
		<updated>2011-01-25T07:19:22Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;[[About me..]]&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
[[Essay: Calculus and the Faculty of Human Kinetics at UBC]]&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:CassandraTravlos&amp;diff=72098</id>
		<title>User:CassandraTravlos</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:CassandraTravlos&amp;diff=72098"/>
		<updated>2011-01-25T07:13:49Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;About me..&lt;br /&gt;
&lt;br /&gt;
Hello, my name is Cassandra Travlos. I am a first year student at UBC taking the Human Kinetic undergraduate program. I live in West Vancouver and I have my own horse which I show competively in B.C.&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
[[Essay: Calculus and the Faculty of Human Kinetics at UBC]]&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz&amp;diff=70814</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Schwytz</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz&amp;diff=70814"/>
		<updated>2011-01-19T06:01:19Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: /* Homework #11- Team Schwytz */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{&lt;br /&gt;
Infobox MATH110 Teams&lt;br /&gt;
| team name = Schwytz&lt;br /&gt;
| member 1 = Angus McWhirter&lt;br /&gt;
| member 2 = Anna Koniuhova&lt;br /&gt;
| member 3 = Cassandra Travlos&lt;br /&gt;
| member 4 = Gracie Mann&lt;br /&gt;
}}&lt;br /&gt;
In workshop J.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Homework #11- Team Schwytz==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Write a linear model to predict the cost of producing flags of your teams Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items the cost is $100. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Our model is described by a linear function.&lt;br /&gt;
* We know that our model has to pass through the point (20,100) because the question tells us that the current production level of 20 items costs $100 (where the cost in dollar is the y-axis and the number of items is the x-axis) therefore we can use this point to formulate a function using the standard equation formula (y=mx +b)&lt;br /&gt;
* We can use 20 items for ‘x’ and 100 dollars for ‘y’&lt;br /&gt;
* Now we must find ‘m’&lt;br /&gt;
* To do this we need to use the formula: y2-y1/ x2-x1&lt;br /&gt;
To find a point on the graph:&lt;br /&gt;
&lt;br /&gt;
* We know that 20 items cost 100 dollars and the marginal cost is $7 per unit&lt;br /&gt;
* Therefore we know that 21 items costs $107 and then 22 items costs $114&lt;br /&gt;
* Now we can set this up in the above equation (where the cost is ‘y’ and the items is ‘x’:&lt;br /&gt;
114-107/ 22-21 = 7&lt;br /&gt;
&lt;br /&gt;
* Therefore:&lt;br /&gt;
&lt;br /&gt;
m=7&lt;br /&gt;
&lt;br /&gt;
* Now we need to find &#039;b&#039;:&lt;br /&gt;
&lt;br /&gt;
y=mx + b&lt;br /&gt;
&lt;br /&gt;
107 = 7 (21) +b&lt;br /&gt;
&lt;br /&gt;
b = -40&lt;br /&gt;
&lt;br /&gt;
* Therefore the equation of the line is:&lt;br /&gt;
&lt;br /&gt;
y= 7x - 40 , which is a linear model  &lt;br /&gt;
&lt;br /&gt;
* If we were to graph this model the function would cross the y-axis at -40.&lt;br /&gt;
&lt;br /&gt;
* However we can also determine the value of ‘b’ by understanding that it costs $100 to produce 20 flags which averages to $5 per flag. If the flags were $7 each it would have cost $140. The difference between these two costs is $40. Therefore to take into account the saving of $40 we make ‘b’ -40 in the formula y=mx+b&lt;br /&gt;
&lt;br /&gt;
* To relate this formula to the specific terminology of the question we will replace &#039;x&#039; with &#039;f&#039; and let &#039;f&#039; represent the number of flags produced (where &#039;f&#039; must be greater than 20). We will replace &#039;y&#039; with C(f) and let C(f) represent the cost to produce the flags:&lt;br /&gt;
C(f) = 7f - 40 , where f ≥ 20    &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;What does your model predict for production of a 150 items?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* In order to predict the total cost of the production of 150 items we can substitute into our model 150 for &#039;f&#039; . Which gives us the equation:    &lt;br /&gt;
&#039;&#039;C(f)=7(150)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=1010&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;According to your model, what happens when the average cost per item, as production levels increase?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
*In order to determine the average cost per item as production increases we must first find the total cost of three consecutive units (e.g the total cost of producing 23, 24, and 25 flags) and divide each total you get but the number of flags produced as shown below:     &lt;br /&gt;
The average cost equation: total cost/number of flags&lt;br /&gt;
&lt;br /&gt;
1. Let f=23&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(23)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=121&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 23 units is : 121/23=5.26&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Let f=24&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(24)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=128&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 24 units is : 128/24 =5.33&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Let f=25&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(25)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=135&#039;&#039;    &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 25 units is: 135/25=5.40&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* Based on the information above we can see that the averge cost per item increases as the production levels increase.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Finally, find some other models ( not necessarily linear) for which you get other behaviours such as&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
1. The average cost remains constant as production increases:    &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y=3x&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
* This model &#039;&#039;&#039;increases&#039;&#039;&#039; at a &#039;&#039;&#039;constant rate&#039;&#039;&#039;, because the slope of the function is the same everywhere. This model shows that the average cost will therefore remain constant as the production increases. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2. The average cost diminishes as the production increases:    &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= x^1/2&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* For this model when you plug in any postive numbers in for &#039;x&#039;, you will see that as &#039;x&#039; or the number of items produced increases, the average cost decreases: &lt;br /&gt;
&lt;br /&gt;
ex. &lt;br /&gt;
&lt;br /&gt;
y= x^1/2 , let x = 100 and let y represent the total cost of x&lt;br /&gt;
&lt;br /&gt;
y= (100^1/2)/ 100&lt;br /&gt;
&lt;br /&gt;
y= 0.1&lt;br /&gt;
&lt;br /&gt;
Now let x= 1000000&lt;br /&gt;
&lt;br /&gt;
y= (1000000^1/2)/1000000&lt;br /&gt;
&lt;br /&gt;
y= 0.001 &lt;br /&gt;
&lt;br /&gt;
* Here you can see that the average cost is diminishing as the production (&#039;x&#039;) increases.    &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
3. The average cost increases as production increases:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= e^x&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* In this model as the production increases so does the slope of the model which causes the average cost to also therefore increase.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
4. You obtain an economy of scale. This means that staring at some specific production level, the marginal cost is always less than the average cost:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= x^1/2&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* For this model, we can see that the marginal cost is less than the average cost&lt;br /&gt;
&lt;br /&gt;
ex.&lt;br /&gt;
&lt;br /&gt;
let x=3&lt;br /&gt;
&lt;br /&gt;
y=3^1/2&lt;br /&gt;
&lt;br /&gt;
y= 1.73&lt;br /&gt;
&lt;br /&gt;
* now we must divide by 3 to get the average &lt;br /&gt;
&lt;br /&gt;
y= 1.73/3&lt;br /&gt;
&lt;br /&gt;
y= 0.577   , this means that it costs an average of $0.58 to produce 3 items &lt;br /&gt;
&lt;br /&gt;
* And when x=2, y= 1.41&lt;br /&gt;
&lt;br /&gt;
* Therefore if we calculate the difference ($1.73-$1.41) we get 0.318 or $0.32. This means that it only cost $0.32 to produce the third item but it cost an average of $0.58 per item (of the 3 total items). This model therefore obtains an economy of scale.&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz&amp;diff=70813</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Schwytz</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz&amp;diff=70813"/>
		<updated>2011-01-19T05:58:22Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: /* Homework #11- Team Schwytz */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{&lt;br /&gt;
Infobox MATH110 Teams&lt;br /&gt;
| team name = Schwytz&lt;br /&gt;
| member 1 = Angus McWhirter&lt;br /&gt;
| member 2 = Anna Koniuhova&lt;br /&gt;
| member 3 = Cassandra Travlos&lt;br /&gt;
| member 4 = Gracie Mann&lt;br /&gt;
}}&lt;br /&gt;
In workshop J.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Homework #11- Team Schwytz==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Write a linear model to predict the cost of producing flags of your teams Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items the cost is $100. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Our model is described by a linear function.&lt;br /&gt;
* We know that our model has to pass through the point (20,100) because the question tells us that the current production level of 20 items costs $100 (where the cost in dollar is the y-axis and the number of items is the x-axis) therefore we can use this point to formulate a function using the standard equation formula (y=mx +b)&lt;br /&gt;
* We can use 20 items for ‘x’ and 100 dollars for ‘y’&lt;br /&gt;
* Now we must find ‘m’&lt;br /&gt;
* To do this we need to use the formula: y2-y1/ x2-x1&lt;br /&gt;
To find a point on the graph:&lt;br /&gt;
&lt;br /&gt;
* We know that 20 items cost 100 dollars and the marginal cost is $7 per unit&lt;br /&gt;
* Therefore we know that 21 items costs $107 and then 22 items costs $114&lt;br /&gt;
* Now we can set this up in the above equation (where the cost is ‘y’ and the items is ‘x’:&lt;br /&gt;
114-107/ 22-21 = 7&lt;br /&gt;
&lt;br /&gt;
* Therefore:&lt;br /&gt;
&lt;br /&gt;
m=7&lt;br /&gt;
&lt;br /&gt;
* Now we need to find &#039;b&#039;:&lt;br /&gt;
&lt;br /&gt;
y=mx + b&lt;br /&gt;
&lt;br /&gt;
107 = 7 (21) +b&lt;br /&gt;
&lt;br /&gt;
b = -40&lt;br /&gt;
&lt;br /&gt;
* Therefore the equation of the line is:&lt;br /&gt;
&lt;br /&gt;
y= 7x - 40 , which is a linear model  &lt;br /&gt;
&lt;br /&gt;
* If we were to graph this model the function would cross the y-axis at -40.&lt;br /&gt;
&lt;br /&gt;
* However we can also determine the value of ‘b’ by understanding that it costs $100 to produce 20 flags which averages to $5 per flag. If the flags were $7 each it would have cost $140. The difference between these two costs is $40. Therefore to take into account the saving of $40 we make ‘b’ -40 in the formula y=mx+b&lt;br /&gt;
&lt;br /&gt;
* To relate this formula to the specific terminology of the question we will replace &#039;x&#039; with &#039;f&#039; and let &#039;f&#039; represent the number of flags produced (where &#039;f&#039; must be greater than 20). We will replace &#039;y&#039; with C(f) and let C(f) represent the cost to produce the flags:&lt;br /&gt;
C(f) = 7f - 40 , where f ≥ 20    &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;What does your model predict for production of a 150 items?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* In order to predict the total cost of the production of 150 items we can substitute into our model 150 for &#039;f&#039; . Which gives us the equation:    &lt;br /&gt;
&#039;&#039;C(f)=7(150)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=1010&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;According to your model, what happens when the average cost per item, as production levels increase?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
*In order to determine the average cost per item as production increases we must first find the total cost of three consecutive units (e.g the total cost of producing 23, 24, and 25 flags) and divide each total you get but the number of flags produced as shown below:     &lt;br /&gt;
The average cost equation: total cost/number of flags&lt;br /&gt;
&lt;br /&gt;
1. Let f=23&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(23)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=121&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 23 units is : 121/23=5.26&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Let f=24&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(24)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=128&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 24 units is : 128/24 =5.33&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Let f=25&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(25)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=135&#039;&#039;    &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 25 units is: 135/25=5.40&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* Based on the information above we can see that the averge cost per item increases as the production levels increase.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Finally, find some other models ( not necessarily linear) for which you get other behaviours such as&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
1. The average cost remains constant as production increases:    &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y=3x&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
* This model &#039;&#039;&#039;increases&#039;&#039;&#039; at a &#039;&#039;&#039;constant rate&#039;&#039;&#039;, because the slope of the function is the same everywhere. This model shows that the average cost will therefore remain constant as the production increases. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2. The average cost diminishes as the production increases:    &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= x^1/2&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* For this model when you plug in any postive numbers in for &#039;x&#039;, you will see that as &#039;x&#039; or the number of items produced increases, the average cost decreases: &lt;br /&gt;
&lt;br /&gt;
ex. &lt;br /&gt;
&lt;br /&gt;
y= x^1/2 , let x = 100 and let y represent the total cost of x&lt;br /&gt;
&lt;br /&gt;
y= (100^1/2)/ 100&lt;br /&gt;
&lt;br /&gt;
y= 0.1&lt;br /&gt;
&lt;br /&gt;
Now let x= 1000000&lt;br /&gt;
&lt;br /&gt;
y= (1000000^1/2)/1000000&lt;br /&gt;
&lt;br /&gt;
y= 0.001 &lt;br /&gt;
&lt;br /&gt;
* Here you can see that the average cost is diminishing as the production (&#039;x&#039;) increases.    &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
3. The average cost increases as production increases:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= e^x&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* In this model as the production increases so does the slope of the model which causes the average cost to also therefore increase.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
4. You obtain an economy of scale. This means that staring at some specific production level, the marginal cost is always less than the average cost:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= x^1/2&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* For this model, we can see that the marginal cost is less than the average cost&lt;br /&gt;
&lt;br /&gt;
ex.&lt;br /&gt;
&lt;br /&gt;
let x=3&lt;br /&gt;
&lt;br /&gt;
y=3^1/2&lt;br /&gt;
&lt;br /&gt;
y= 1.73&lt;br /&gt;
&lt;br /&gt;
* now we must divide by 3 to get the average &lt;br /&gt;
&lt;br /&gt;
y= 1.73/3&lt;br /&gt;
&lt;br /&gt;
y= 0.577   , this means that it costs an average of $0.58 to produce 3 items &lt;br /&gt;
&lt;br /&gt;
* And when x=2, y= 1.41&lt;br /&gt;
&lt;br /&gt;
* Therefore if we calculate the difference (1.73-1.41) we get 0.318 or $0.32. This means that it only cost $0.32 to produce the third item but it cost an average of $0.58 per item (of the 3 total items). This model therefore obtains an economy of scale.&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz&amp;diff=70812</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Schwytz</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz&amp;diff=70812"/>
		<updated>2011-01-19T05:57:14Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: /* Homework #11- Team Schwytz */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{&lt;br /&gt;
Infobox MATH110 Teams&lt;br /&gt;
| team name = Schwytz&lt;br /&gt;
| member 1 = Angus McWhirter&lt;br /&gt;
| member 2 = Anna Koniuhova&lt;br /&gt;
| member 3 = Cassandra Travlos&lt;br /&gt;
| member 4 = Gracie Mann&lt;br /&gt;
}}&lt;br /&gt;
In workshop J.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Homework #11- Team Schwytz==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Write a linear model to predict the cost of producing flags of your teams Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items the cost is $100. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Our model is described by a linear function.&lt;br /&gt;
* We know that our model has to pass through the point (20,100) because the question tells us that the current production level of 20 items costs $100 (where the cost in dollar is the y-axis and the number of items is the x-axis) therefore we can use this point to formulate a function using the standard equation formula (y=mx +b)&lt;br /&gt;
* We can use 20 items for ‘x’ and 100 dollars for ‘y’&lt;br /&gt;
* Now we must find ‘m’&lt;br /&gt;
* To do this we need to use the formula: y2-y1/ x2-x1&lt;br /&gt;
To find a point on the graph:&lt;br /&gt;
&lt;br /&gt;
* We know that 20 items cost 100 dollars and the marginal cost is $7 per unit&lt;br /&gt;
* Therefore we know that 21 items costs $107 and then 22 items costs $114&lt;br /&gt;
* Now we can set this up in the above equation (where the cost is ‘y’ and the items is ‘x’:&lt;br /&gt;
114-107/ 22-21 = 7&lt;br /&gt;
&lt;br /&gt;
* Therefore:&lt;br /&gt;
&lt;br /&gt;
m=7&lt;br /&gt;
&lt;br /&gt;
* Now we need to find &#039;b&#039;:&lt;br /&gt;
&lt;br /&gt;
y=mx + b&lt;br /&gt;
&lt;br /&gt;
107 = 7 (21) +b&lt;br /&gt;
&lt;br /&gt;
b = -40&lt;br /&gt;
&lt;br /&gt;
* Therefore the equation of the line is:&lt;br /&gt;
&lt;br /&gt;
y= 7x - 40 , which is a linear model  &lt;br /&gt;
&lt;br /&gt;
* If we were to graph this model the function would cross the y-axis at -40.&lt;br /&gt;
&lt;br /&gt;
* However we can also determine the value of ‘b’ by understanding that it costs $100 to produce 20 flags which averages to $5 per flag. If the flags were $7 each it would have cost $140. The difference between these two costs is $40. Therefore to take into account the saving of $40 we make ‘b’ -40 in the formula y=mx+b&lt;br /&gt;
&lt;br /&gt;
* To relate this formula to the specific terminology of the question we will replace &#039;x&#039; with &#039;f&#039; and let &#039;f&#039; represent the number of flags produced (where &#039;f&#039; must be greater than 20). We will replace &#039;y&#039; with C(f) and let C(f) represent the cost to produce the flags:&lt;br /&gt;
C(f) = 7f - 40 , where f ≥ 20    &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;What does your model predict for production of a 150 items?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* In order to predict the total cost of the production of 150 items we can substitute into our model 150 for &#039;f&#039; . Which gives us the equation:    &lt;br /&gt;
&#039;&#039;C(f)=7(150)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=1010&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;According to your model, what happens when the average cost per item, as production levels increase?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
*In order to determine the average cost per item as production increases we must first find the total cost of three consecutive units (e.g the total cost of producing 23, 24, and 25 flags) and divide each total you get but the number of flags produced as shown below:     &lt;br /&gt;
The average cost equation: total cost/number of flags&lt;br /&gt;
&lt;br /&gt;
1. Let f=23&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(23)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=121&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 23 units is : 121/23=5.26&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Let f=24&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(24)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=128&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 24 units is : 128/24 =5.33&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Let f=25&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(25)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=135&#039;&#039;    &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 25 units is: 135/25=5.40&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* Based on the information above we can see that the averge cost per item increases as the production levels increase.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Finally, find some other models ( not necessarily linear) for which you get other behaviours such as&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
1. The average cost remains constant as production increases:    &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y=3x&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
* This model &#039;&#039;&#039;increases&#039;&#039;&#039; at a &#039;&#039;&#039;constant rate&#039;&#039;&#039;, because the slope of the function is the same everywhere. This model shows that the average cost will therefore remain constant as the production increases. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2. The average cost diminishes as the production increases:    &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= x^1/2&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* For this model when you plug in any postive numbers in for &#039;x&#039;, you will see that as &#039;x&#039; or the number of items produced increases, the average cost decreases: &lt;br /&gt;
&lt;br /&gt;
ex. &lt;br /&gt;
&lt;br /&gt;
y= x^1/2 , let x = 100 and let y represent the total cost of x&lt;br /&gt;
&lt;br /&gt;
y= (100^1/2)/ 100&lt;br /&gt;
&lt;br /&gt;
y= 0.1&lt;br /&gt;
&lt;br /&gt;
Now let x= 1000000&lt;br /&gt;
&lt;br /&gt;
y= (1000000^1/2)/1000000&lt;br /&gt;
&lt;br /&gt;
y= 0.001 &lt;br /&gt;
&lt;br /&gt;
* Here you can see that the average cost is diminishing as the production (&#039;x&#039;) increases.    &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
3. The average cost increases as production increases:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= e^x&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* In this model as the production increases so does the slope of the model which causes the average cost to also therefore increase.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
4. You obtain an economy of scale. This means that staring at some specific production level, the marginal cost is always less than the average cost:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= x^1/2&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* For this model, we can see that the marginal cost is less than the average cost&lt;br /&gt;
&lt;br /&gt;
ex.&lt;br /&gt;
&lt;br /&gt;
let x=3&lt;br /&gt;
&lt;br /&gt;
y=3^1/2&lt;br /&gt;
&lt;br /&gt;
y= 1.73&lt;br /&gt;
&lt;br /&gt;
now we must divide by 3 to get the average &lt;br /&gt;
&lt;br /&gt;
y= 1.73/3&lt;br /&gt;
&lt;br /&gt;
y= 0.577   , this means that it costs an average of $0.58 to produce 3 items &lt;br /&gt;
&lt;br /&gt;
And when x=2, y= 1.41&lt;br /&gt;
&lt;br /&gt;
Therefore if we calculate the difference (1.73-1.41) we get 0.318 or $0.32. This means that it only cost $0.32 to produce the third item but it cost an average of $0.58 per item (of the 3 total items). This model therefore obtains an economy of scale.&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz&amp;diff=70809</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Schwytz</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz&amp;diff=70809"/>
		<updated>2011-01-19T05:31:07Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: /* Homework #11- Team Schwytz */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{&lt;br /&gt;
Infobox MATH110 Teams&lt;br /&gt;
| team name = Schwytz&lt;br /&gt;
| member 1 = Angus McWhirter&lt;br /&gt;
| member 2 = Anna Koniuhova&lt;br /&gt;
| member 3 = Cassandra Travlos&lt;br /&gt;
| member 4 = Gracie Mann&lt;br /&gt;
}}&lt;br /&gt;
In workshop J.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Homework #11- Team Schwytz==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Write a linear model to predict the cost of producing flags of your teams Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items the cost is $100. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Our model is described by a linear function.&lt;br /&gt;
* We know that our model has to pass through the point (20,100) because the question tells us that the current production level of 20 items costs $100 (where the cost in dollar is the y-axis and the number of items is the x-axis) therefore we can use this point to formulate a function using the standard equation formula (y=mx +b)&lt;br /&gt;
* We can use 20 items for ‘x’ and 100 dollars for ‘y’&lt;br /&gt;
* Now we must find ‘m’&lt;br /&gt;
* To do this we need to use the formula: y2-y1/ x2-x1&lt;br /&gt;
To find a point on the graph:&lt;br /&gt;
&lt;br /&gt;
* We know that 20 items cost 100 dollars and the marginal cost is $7 per unit&lt;br /&gt;
* Therefore we know that 21 items costs $107 and then 22 items costs $114&lt;br /&gt;
* Now we can set this up in the above equation (where the cost is ‘y’ and the items is ‘x’:&lt;br /&gt;
114-107/ 22-21 = 7&lt;br /&gt;
&lt;br /&gt;
* Therefore:&lt;br /&gt;
&lt;br /&gt;
m=7&lt;br /&gt;
&lt;br /&gt;
* Now we need to find &#039;b&#039;:&lt;br /&gt;
&lt;br /&gt;
y=mx + b&lt;br /&gt;
&lt;br /&gt;
107 = 7 (21) +b&lt;br /&gt;
&lt;br /&gt;
b = -40&lt;br /&gt;
&lt;br /&gt;
* Therefore the equation of the line is:&lt;br /&gt;
&lt;br /&gt;
y= 7x - 40 , which is a linear model  &lt;br /&gt;
&lt;br /&gt;
* If we were to graph this model the function would cross the y-axis at -40.&lt;br /&gt;
&lt;br /&gt;
* However we can also determine the value of ‘b’ by understanding that it costs $100 to produce 20 flags which averages to $5 per flag. If the flags were $7 each it would have cost $140. The difference between these two costs is $40. Therefore to take into account the saving of $40 we make ‘b’ -40 in the formula y=mx+b&lt;br /&gt;
&lt;br /&gt;
* To relate this formula to the specific terminology of the question we will replace &#039;x&#039; with &#039;f&#039; and let &#039;f&#039; represent the number of flags produced (where &#039;f&#039; must be greater than 20). We will replace &#039;y&#039; with C(f) and let C(f) represent the cost to produce the flags:&lt;br /&gt;
C(f) = 7f - 40 , where f ≥ 20    &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;What does your model predict for production of a 150 items?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* In order to predict the total cost of the production of 150 items we can substitute into our model 150 for &#039;f&#039; . Which gives us the equation:    &lt;br /&gt;
&#039;&#039;C(f)=7(150)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=1010&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;According to your model, what happens when the average cost per item, as production levels increase?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
*In order to determine the average cost per item as production increases we must first find the total cost of three consecutive units (e.g the total cost of producing 23, 24, and 25 flags) and divide each total you get but the number of flags produced as shown below:     &lt;br /&gt;
The average cost equation: total cost/number of flags&lt;br /&gt;
&lt;br /&gt;
1. Let f=23&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(23)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=121&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 23 units is : 121/23=5.26&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Let f=24&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(24)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=128&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 24 units is : 128/24 =5.33&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Let f=25&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(25)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=135&#039;&#039;    &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 25 units is: 135/25=5.4&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* Based on the information above we can see that the averge cost per item increases as the production levels increase.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Finally, find some other models ( not necessarily linear) for which you get other behaviours such as&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
1. The average cost remains constant as production increases:    &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y=3x&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
* This model &#039;&#039;&#039;increases&#039;&#039;&#039; at a &#039;&#039;&#039;constant rate&#039;&#039;&#039;, because the slope of the function is the same everywhere. This model shows that the average cost will therefore remain constant as the production increases. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2. The average cost diminishes as the production increases:    &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= x^1/2&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* For this model when you plug in any postive numbers in for &#039;x&#039;, you will see that as &#039;x&#039; or the number of items produced increases, the average cost decreases: &lt;br /&gt;
&lt;br /&gt;
ex. &lt;br /&gt;
&lt;br /&gt;
y= x^1/2 , let x = 100 and let y represent the total cost of x&lt;br /&gt;
&lt;br /&gt;
y= (100^1/2)/ 100&lt;br /&gt;
&lt;br /&gt;
y= 0.1&lt;br /&gt;
&lt;br /&gt;
Now let x= 1000000&lt;br /&gt;
&lt;br /&gt;
y= (1000000^1/2)/1000000&lt;br /&gt;
&lt;br /&gt;
y= 0.001 &lt;br /&gt;
&lt;br /&gt;
* Here you can see that the average cost is diminishing as the production (&#039;x&#039;) increases.    &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
3. The average cost increases as production increases:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= e^x&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* In this model as the production increases so does the slope of the model which causes the average cost to also therefore increase.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
4. You obtain an economy of scale. This means that staring at some specific production level, the marginal cost is always less than the average cost:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y= x^1/2&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
*&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz&amp;diff=70808</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Schwytz</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Schwytz&amp;diff=70808"/>
		<updated>2011-01-19T05:30:13Z</updated>

		<summary type="html">&lt;p&gt;CassandraTravlos: /* Homework #11- Team Schwytz */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{&lt;br /&gt;
Infobox MATH110 Teams&lt;br /&gt;
| team name = Schwytz&lt;br /&gt;
| member 1 = Angus McWhirter&lt;br /&gt;
| member 2 = Anna Koniuhova&lt;br /&gt;
| member 3 = Cassandra Travlos&lt;br /&gt;
| member 4 = Gracie Mann&lt;br /&gt;
}}&lt;br /&gt;
In workshop J.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Homework #11- Team Schwytz==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Write a linear model to predict the cost of producing flags of your teams Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items the cost is $100. &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Our model is described by a linear function.&lt;br /&gt;
* We know that our model has to pass through the point (20,100) because the question tells us that the current production level of 20 items costs $100 (where the cost in dollar is the y-axis and the number of items is the x-axis) therefore we can use this point to formulate a function using the standard equation formula (y=mx +b)&lt;br /&gt;
* We can use 20 items for ‘x’ and 100 dollars for ‘y’&lt;br /&gt;
* Now we must find ‘m’&lt;br /&gt;
* To do this we need to use the formula: y2-y1/ x2-x1&lt;br /&gt;
To find a point on the graph:&lt;br /&gt;
&lt;br /&gt;
* We know that 20 items cost 100 dollars and the marginal cost is $7 per unit&lt;br /&gt;
* Therefore we know that 21 items costs $107 and then 22 items costs $114&lt;br /&gt;
* Now we can set this up in the above equation (where the cost is ‘y’ and the items is ‘x’:&lt;br /&gt;
114-107/ 22-21 = 7&lt;br /&gt;
&lt;br /&gt;
* Therefore:&lt;br /&gt;
&lt;br /&gt;
m=7&lt;br /&gt;
&lt;br /&gt;
* Now we need to find &#039;b&#039;:&lt;br /&gt;
&lt;br /&gt;
y=mx + b&lt;br /&gt;
&lt;br /&gt;
107 = 7 (21) +b&lt;br /&gt;
&lt;br /&gt;
b = -40&lt;br /&gt;
&lt;br /&gt;
* Therefore the equation of the line is:&lt;br /&gt;
&lt;br /&gt;
y= 7x - 40 , which is a linear model  &lt;br /&gt;
&lt;br /&gt;
* If we were to graph this model the function would cross the y-axis at -40.&lt;br /&gt;
&lt;br /&gt;
* However we can also determine the value of ‘b’ by understanding that it costs $100 to produce 20 flags which averages to $5 per flag. If the flags were $7 each it would have cost $140. The difference between these two costs is $40. Therefore to take into account the saving of $40 we make ‘b’ -40 in the formula y=mx+b&lt;br /&gt;
&lt;br /&gt;
* To relate this formula to the specific terminology of the question we will replace &#039;x&#039; with &#039;f&#039; and let &#039;f&#039; represent the number of flags produced (where &#039;f&#039; must be greater than 20). We will replace &#039;y&#039; with C(f) and let C(f) represent the cost to produce the flags:&lt;br /&gt;
C(f) = 7f - 40 , where f ≥ 20    &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;What does your model predict for production of a 150 items?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* In order to predict the total cost of the production of 150 items we can substitute into our model 150 for &#039;f&#039; . Which gives us the equation:    &lt;br /&gt;
&#039;&#039;C(f)=7(150)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=1010&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;According to your model, what happens when the average cost per item, as production levels increase?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
*In order to determine the average cost per item as production increases we must first find the total cost of three consecutive units (e.g the total cost of producing 23, 24, and 25 flags) and divide each total you get but the number of flags produced as shown below:     &lt;br /&gt;
The average cost equation: total cost/number of flags&lt;br /&gt;
&lt;br /&gt;
1. Let f=23&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(23)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=121&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 23 units is : 121/23=5.26&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
2. Let f=24&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(24)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=128&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 24 units is : 128/24 =5.33&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3. Let f=25&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=7(25)-40&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C(f)=135&#039;&#039;    &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Average cost of 25 units is: 135/25=5.4&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
* Based on the information above we can see that the averge cost per item increases as the production levels increase.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Finally, find some other models ( not necessarily linear) for which you get other behaviours such as&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
1. The average cost remains constant as production increases:    &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Model: y=3x&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
* This model &#039;&#039;&#039;increases&#039;&#039;&#039; at a &#039;&#039;&#039;constant rate&#039;&#039;&#039;, because the slope of the function is the same everywhere. This model shows that the average cost will therefore remain constant as the production increases. &lt;br /&gt;
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2. The average cost diminishes as the production increases:    &lt;br /&gt;
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&#039;&#039;&#039;Model: y= x^1/2&#039;&#039;&#039;&lt;br /&gt;
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* For this model when you plug in any postive numbers in for &#039;x&#039;, you will see that as &#039;x&#039; or the number of items produced increases, the average cost decreases: &lt;br /&gt;
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ex. &lt;br /&gt;
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y= x^1/2 , let x = 100 and let y represent the total cost of x&lt;br /&gt;
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y= (100^1/2)/ 100&lt;br /&gt;
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y= 0.1&lt;br /&gt;
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Now let x= 1000000&lt;br /&gt;
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y= (1000000^1/2)/1000000&lt;br /&gt;
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y= 0.001 &lt;br /&gt;
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* Here you can see that the average cost is diminishing as the production (&#039;x&#039;) increases.    &lt;br /&gt;
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3. The average cost increases as production increases:&lt;br /&gt;
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&#039;&#039;&#039;Model: y= e^x&#039;&#039;&#039;&lt;br /&gt;
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* In this model as the production increases so does the slope of the model which causes the average cost to also therefore increase.&lt;br /&gt;
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4. You obtain an economy of scale. This means that staring at some specific production level, the marginal cost is always less than the average cost:&lt;br /&gt;
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&#039;&#039;&#039;Model: y= x^1/2&#039;&#039;&#039;&lt;br /&gt;
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*&lt;/div&gt;</summary>
		<author><name>CassandraTravlos</name></author>
	</entry>
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