UBC Wiki - User contributions [en]

Course:MATH110/Archive/2010-2011/003/Teams/Schaffhausen/Homework 13

2011-02-04T06:14:13Z

JoseTorresTorija:

=Homework 13 Group Work=

The Logarithmic Scale and How it is applied to the Richter Scale

The logarithmic scale is used generally when there is a very wide range of values. It is important to note that the change of a specific value on according to a logarithmic scale does not depend on "the size of the change is proportion to the value of it self"(http://mathforum.org/library/drmath/view/55574.html) This can be explained easier if one looks at the difference between a linear and logarithmic scale. A linear scale is used "if adding 1 to a value is just a big as a change whether the original value was 1 or 1000"(http://mathforum.org/library/drmath/view/55574.html). In other words a linear scale is used when one can see on a graph the difference of an increase in 1(or any reasonably small number)

GRAPH OF LINEAR EQUATION

http://upload.wikimedia.org/wikimedia/en-labs/0/0a/Y_equals_x_plus_2.PNG

A logarithmic scale is used when the, "doubling of a value is just as big as a change whether it is from 1 to 2, or 1000 to 2000(http://mathforum.org/library/drmath/view/55574.html). Therefore the logarithmic scale was created and along the Y-Axis the number increase exponentially by ten each increasing value.

GRAPH OF LOGARITHMIC SCALE

http://cryptodox.com/images/c/cd/600px-LogLogScale.png

In summation the logarithmic scale is used when there is a large range of values, and it does so by making each increasing value tenfold of the value preceding it. The logarithmic scale is used in a few instances including the Richter Scale.

===The Richter magnitude scale===

The Richter magnitude scale is used for the assigning of a numerical value to the seismic energy that is released by an earthquake. it is a base 10 logarithmic scale obtained by the calculation of the combined horizontal amplitude of the largest displacement from zero on a particular type of seismometer. (http://en.wikipedia.org/wiki/Richter_magnitude_scale)

The equation for the scale is
:<math>M_\mathrm{L} = \log_{10} A - \log_{10} A_\mathrm{0}(\delta)\ </math>

:<math> (A)\ </math> derived from the the maximum excursion of the Wood-Anderson seismograph. The seismograph is used to measure the frequency of shaking the earth is going through while an earthquake is in progress.

:<math> (\delta)\ </math> is the epicentra distance of the station which recoded the magnitude of the earthquake. The way that this distance is calculated is by measuring the time between P-waves and S-Waves at different stations that are located around the world. Once you have enough stations that have recorded these times one is able to triangulate the epicenter of the earthquake. (http://en.wikipedia.org/wiki/Epicentral_distance#Epicentral_distance)

Since this is a logarithmic function we know that an increase from 2 to 3 means that it is roughly 10 times stronger than its predecessor, in terms of earthquakes that means that the raw power that is released in that jump from 2 to 3 is about 31.6 times the energy released.

===Conclusion===

As stated above in the section regarding the Logarithmic Scale it does not depend on the absolute size of the change but according to the size of the change proportional to the initial value. So by applying this logic to that of earthquakes we have gathered information that shows how the power between a size 1 and a size 2 earthquake is about 31.6 times larger. This is jump in power can only be described in a logarithmic matter because as the size of the earthquake increases its power will always increase proportional to that of it.

Course:MATH110/Archive/2010-2011/003/Teams/Schaffhausen/Homework 13

2011-02-04T06:06:34Z

JoseTorresTorija:

Course:MATH110/Archive/2010-2011/003/Teams/Schaffhausen/Homework 13

2011-02-04T06:04:21Z

JoseTorresTorija:

=Homework 13 Group Work=

The Logarithmic Scale and How it is applied to the Richter Scale

The logarithmic scale is used generally when there is a very wide range of values. It is important to note that the change of a specific value on according to a logarithmic scale does not depend on "the size of the change is proportion to the value of it self"(http://mathforum.org/library/drmath/view/55574.html) This can be explained easier if one looks at the difference between a linear and logarithmic scale. A linear scale is used "if adding 1 to a value is just a big as a change whether the original value was 1 or 1000"(http://mathforum.org/library/drmath/view/55574.html). In other words a linear scale is used when one can see on a graph the difference of an increase in 1(or any reasonably small number)

GRAPH OF LINEAR EQUATION

http://upload.wikimedia.org/wikimedia/en-labs/0/0a/Y_equals_x_plus_2.PNG

A logarithmic scale is used when the, "doubling of a value is just as big as a change whether it is from 1 to 2, or 1000 to 2000(http://mathforum.org/library/drmath/view/55574.html). Therefore the logarithmic scale was created and along the Y-Axis the number increase exponentially by ten each increasing value.

GRAPH OF LOGARITHMIC SCALE

http://cryptodox.com/images/c/cd/600px-LogLogScale.png

In summation the logarithmic scale is used when there is a large range of values, and it does so by making each increasing value tenfold of the value preceding it. The logarithmic scale is used in a few instances including the Richter Scale.

===The Richter magnitude scale===

The Richter magnitude scale is used for the assigning of a numerical value to the seismic energy that is released by an earthquake. it is a base 10 logarithmic scale obtained by the calculation of the combined horizontal amplitude of the largest displacement from zero on a particular type of seismometer. (http://en.wikipedia.org/wiki/Richter_magnitude_scale)

The equation for the scale is
:<math>M_\mathrm{L} = \log_{10} A - \log_{10} A_\mathrm{0}(\delta)\ </math>

A derived from the the maximum excursion of the Wood-Anderson seismograph. The seismograph is used to measure the frequency of shaking the earth is going through while an earthquake is in progress.

:<math> (\delta)\ </math> is the epicentra distance of the station which recoded the magnitude of the earthquake. The way that this distance is calculated is by measuring the time between P-waves and S-Waves at different stations that are located around the world. Once you have enough stations that have recorded these times one is able to triangulate the epicenter of the earthquake. (http://en.wikipedia.org/wiki/Epicentral_distance#Epicentral_distance)

Since this is a logarithmic function we know that an increase from 2 to 3 means that it is roughly 10 times stronger than its predecessor, in terms of earthquakes that means that the raw power that is released in that jump from 2 to 3 is about 31.6 times the energy released.

Course:MATH110/Archive/2010-2011/003/Teams/Schaffhausen/Homework 13

2011-02-04T05:51:38Z

JoseTorresTorija:

Homework 13 Group Work

The Logarithmic Scale and How it is applied to the Richter Scale

The logarithmic scale is used generally when there is a very wide range of values. It is important to note that the change of a specific value on according to a logarithmic scale does not depend on "the size of the change is proportion to the value of it self"(http://mathforum.org/library/drmath/view/55574.html) This can be explained easier if one looks at the difference between a linear and logarithmic scale. A linear scale is used "if adding 1 to a value is just a big as a change whether the original value was 1 or 1000"(http://mathforum.org/library/drmath/view/55574.html). In other words a linear scale is used when one can see on a graph the difference of an increase in 1(or any reasonably small number)

GRAPH OF LINEAR EQUATION

http://upload.wikimedia.org/wikimedia/en-labs/0/0a/Y_equals_x_plus_2.PNG

A logarithmic scale is used when the, "doubling of a value is just as big as a change whether it is from 1 to 2, or 1000 to 2000(http://mathforum.org/library/drmath/view/55574.html). Therefore the logarithmic scale was created and along the Y-Axis the number increase exponentially by ten each increasing value.

GRAPH OF LOGARITHMIC SCALE

http://cryptodox.com/images/c/cd/600px-LogLogScale.png

In summation the logarithmic scale is used when there is a large range of values, and it does so by making each increasing value tenfold of the value preceding it. The logarithmic scale is used in a few instances including the Richter Scale.

User:JoseTorresTorija

2011-01-28T09:17:29Z

JoseTorresTorija:

=Pythagorean Theorem=

The Pythagorean Theorem is a formula which involves the solution to finding sides of a right triangle. This theorem is only applicable in situations where the triangle which is being measured is a right triangle which means that one side of the triangle is a 90◦ angle. If the triangle has a 90◦ and you have 2 of the sides of the triangle you can use the equation c^2=a^2+b^2. With this equation one is able to input the corresponding sides of the triangle and get a solution either equaling a missing side or the hypotenuse, the hypotenuse is always the longest side of the triangle is always going to be C in the equation. This equation can be used in multiple situations other than just finding the missing sides of a triangle, for instance when it can be used to find the length of a radius inside of a circle if you know the distance between points in the circle or used in real life scenarios when you need to know the length of a diagonal object.

=Application of Calculus in Psychology and Psychiatry=

The use of calculus in my psychology is actually something that is considerably very important in many sectors. The aspect of psychology is the bases of the study of people but to conclude things especially concerning a wide range of people you must conduct experiments which give you values and statistics. In these experiments people are either asked simple questions or are timed or just simply observed, but the results are generally compiled into numbers and sets. With these number sets a researcher is able to create a hypothesis concerning the topic that has been studied, these hypothesis are then tested again and again by comparing the different numerical values that are achieved every time the experiment is redone. These numerical experiments are usually based around standard deviation or average or specific formulas and trends which must be used so that the knowledge that is gathered can be applicable to all the people in a given geographic. The dependency on numerical values is important because with these numbers more authenticity is given to each experiment; it is like the saying says “the numbers don’t lie.” With authenticity given to experiments psychologists are able to start applying theorems and hypothesis on the general people. Also like psychology, psychiatry is very dependent on numbers. Since psychiatry is more medical based one must know proportions to ratios to be able to compute the use of medication depending on people’s bodies and different variables that must be taken into account. With the introduction of medications one must also know of outcomes and the chances of those outcomes affecting your patients, these outcomes could not be possible without numerical experiments that test the reaction in a group of people and compute the normal deviation of people in the group then apply it to the population of the world.

Course:MATH110/Archive/2010-2011/003/Teams/Schaffhausen/Homework 12

2011-01-28T07:47:10Z

JoseTorresTorija:

==Homework 12==

Starting with the function:
:<math>P(t) = \frac{1}{1+e^{-t}}</math>

-Change the height of the horizontal asymptote on the right and denote it by K.

-Change the y-intercept to any number between 0 and K. BONUS - Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical.

===='''Solution 1'''====
We saw that there are two possible ways to change the horizontal asymptote in this equation we can either replace the numerator by K which will then become the horizontal asymptote so for example like this.
:<math>P(t) = \frac{K}{1+e^{-t}}</math> So any what ever value K becomes we will gain a new horizontal asymptote.

We also saw that if we place a K value under the 1 in the denominator that also changed the value of the horizontal asymptote so it would look like this.
:<math>P(t) = \frac{1}{1/k+e^{-t}}</math>

As we moved around the K value we noticed that the y-intercept was constantly moving depending what we did with our K values, and we observed a pattern that when we used the first equation to change the horizontal asymptote that the y-intercept was usually about half of the value of K. So if we had imputed 6 into our K value we would thus get a y-intercept of 3. So if we wanted the y-intercept to be between 0 and K it we would just have to plug a value for K that is greater than 0.

'''Bonus'''

Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical. (If you graph it, it should be quite clear)

Well firstly we noticed that no matter how big we made our K value the equation would always reach the horizontal asymptote about the same t value which was around 12-13. Which means that as we increase the K value the slope normally also increases with it in order to keep up with a constant encroachment of the horizontal asymptote. Another way we noticed to change the slope of the curved part was to add an additional value before the T, by doing so the rate at which it increased was also increased so for example.

:<math>P(t) = \frac{1}{1+e^{-4t}}</math> by adding this 4 in front of the T, we increased the rate at which the slope reaches the horizontal asymptote, normally this slope would reach its asymptote of 1 at around T= 13 but when we add the 4 in font the value at which it reaches the asymptote is T= 4. Thus by inserting this new value we can increase the slope of the curved part.

Once you'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'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're taking your data from (real data or imagined data), what it is that you're modelling and how you are doing the math to answer a predictive question.

So an example of this would be that of exponential growth of a colony of bacteria that is growing on a rock. Lets say that initially we have 1,000 bacteria growing on this rock but their is only enough space and nutrients to support 5,000 bacteria on the rock. To calculate the time it would take for the bacteria to get to 95% would look like this.

5,000 x .95 = 4750

:<math>P(t) = \frac{5,000}{1+e^{-t}}</math>

:<math> 4750 = \frac{5,000}{1+e^{-t}}</math>

:<math> 5,000 = 4750 \left(1+e^-t)\right)</math>

:<math> 5,000 = \left(4750+4750e^-t)\right)</math>

:<math> 5,000 -4750 = \left(4750e^-t)\right)</math>

:<math> \frac{250}{4750}= \left(e^-t)\right)</math>

:<math> ln(.053) = \left(lne^-t)\right)</math>

:<math> (-2.94) = \left(-t)\right)</math>

:<math> 2.94 = \left(t)\right) </math>

So it would take about 2.94 days for the bacteria to grow to the population of 4750 which would be at 95% of its carrying capacity.

Course:MATH110/Archive/2010-2011/003/Teams/Schaffhausen

2011-01-19T10:26:22Z

JoseTorresTorija:

{{
Infobox MATH110 Teams
| team name = Schaffhausen
| member 1 = Jose Torres-Torija Cubillas
| member 2 = Kazi Ahmed
| member 3 = Matthew Robinson
| member 4 = Trevor Shumka
}}
In workshop J.

='''Homework 11'''=

==Model==
The information that we have been giving states that our current production rate is 20, our current total cost is $100 and our marginal cost is $7. With this information we can create an equation.

We use 20 which stands for units of production as an X value and cost which is $100 as a Y value.

F(x)=y, F(20)=$100 we know that function to be true now to create a formula we put that into a y=mx + b format plugging in 20 for x and 100 for y, this gives us 100=20m + b, our slope being our marginal cost which is the cost to produce 1 more unit, so our new equation looks like 100= 7(20) + b after we substitute 7 in. Now we solve for b which is,

100= 7(20) + b

100 = 140+b

100-140=b

-40 = b

So our final equation for this model is, y=7x - 40 if x≥20

====Now we must find the cost to produce 150 flags using this model====

we do this by plugging in 150 into x

y=7(150) -40

y=1050- 40

y= 1010, which is the cost to produce 150 flags using the model above.

====Average cost of this model at 150 units====

At the quantity of 150 flags we get the total cost to be $1010, to find the average cost of this we must divide $1010 by 150, when we do this we get that the average cost of each flag at this point in production is 6.74 which is higher than the average cost of the production of 20 flags which was an average cost of $5 per flag. So using this trend we can see that average cost increases as you produce more.

==Other Models==
-The average cost remains constant as production increases.

If we wanted to create a model which had a average cost remain the same them we would need a linear model with no B so when we try to get an average we always get the same outcome, so the model will look like y=mx, where m is any constant.

-For the average cost to remain constant to marginal cost must also remain constant a model that exemplifies this would be C(f)=MC where MC= Marginal

Cost. So, if MC=10 the formula will be C(f)=10(f). Now if flags increase by 10 products the formula will look like C(10)=10(10)=100, if it increases by C(20)=10(20)=200.

-The average cost diminishes as production increases.

We need to have a model which has a decreasing average cost as we increase the production so only way is to have a constant slope and no b value, we also need to make it so that as you increase the X value the Y value decreases. Thus this means that x and y are inversely proportional which is only attainable by having a situation which has y = m/x so as X increases and M stays the same we will get a smaller and smaller Y value.

-The average cost increases as production increases.

For this to happen we can not have a linear model, so it is not a linear model it must be a quadratic model which is can then be described as, y = x^a where a is any positive constant that remains the same for the specific model.

-You obtain an economy of scale. This means that starting at some specific production level, the marginal cost is always less than the average cost.

When one has an economy of scale that means that this company has an advantage at producing large number of quantities which thus lower its price at that range but when it comes down to small production its cost are quite larger. So for example we could have a company that for it to produce flags under a quantity of 600 flags its model would look like, y=7x-37 : x<600, but when it started producing at 600 or above it the cost of flags would be some thing fixed like, f(x) = 75 x≥ 600

-Any other interesting properties that you can think of and create a model for. Bonus points can be obtained for very interesting ideas.

Course:MATH110/Archive/2010-2011/003/Teams/Schaffhausen

2011-01-19T04:50:04Z

JoseTorresTorija:

Course:MATH110/Archive/2010-2011/003/Teams/Schaffhausen

2011-01-19T04:47:13Z

JoseTorresTorija:

Course:MATH110/Archive/2010-2011/003/Math Forum/Webwork Week 8

2010-10-29T02:41:05Z

JoseTorresTorija: /* Problem 8 */

==Problem 8==
I am having difficulty getting this one:

h(x) = <math> sqrt{(9x/x^2-361)} </math>

Given that -19 and 19 for x would cause the denominator to equal 0, I thought this was the answer:

(-infinity,-19)U(-19,19)U(19,infinity)

but that is incorrect. Maybe I am thinking about this in the wrong way. Anyways, I am stuck on this one and would appreciate a hint.
--[[User:SteffanyChwedoruk|SteffanyChwedoruk]] 20:05, 28 October 2010 (UTC)

I also have the same values as you, and what I have seen is that -19 and 19 don't work, but numbers between 1 through 18 don't work either because the bottom will be a negative number meaning the overall thing will be negative. So I have tried (-infinity, -19)U(-19,0]U(19,infinity) but that is also incorrect. I think it might be something about me excluding numbers between 1 through 18. I still haven't found the answer, but if that helps you find it.
--[[User:JoseTorresTorija|JoseTorresTorija]]

==Q6==
Was Question 6 removed from the Webwork?

I think so, I realized one of the questions was missing.
==Problem 4==
Can anyone help me out with question 4?

Problem 4 is tricky. Here is my hint for solving it. Think of each job as a different option. So you have two options which can be represented by two different formulas (I will let you work out the formulas). The key to answering this is the wording "at least as good." Another way of saying at least as good is saying "equal." Once you have your two formulas, think of what you do with the formulas to solve this...
--[[User:SteffanyChwedoruk|SteffanyChwedoruk]] 20:05, 28 October 2010 (UTC)

Course:MATH110/Archive/2010-2011/003/Groups/Group 07

2010-10-20T05:35:33Z

JoseTorresTorija: /* Homework 4 */

{{Infobox MATH110 Groups
| group number = 7
| member 1 = Byron Godoy Pinzon
| member 2 = Saloumeh Hassanzadeh
| member 3 = Michelle Little
| member 4 = Emily Oates
| member 5 = [[User:JoseTorresTorija|Jose Torres-Torija Cubillas]]
| member 6 =
}}

=Homework 4=

==Question #1==
Ok, so first we assigned the cat to Tosh, the frog to Bianca, the parrot to Jaela, and the snake to Jun. Then, we knew that the cat was named was named Jun and the frog was named Suzan. The owner of the turtle was the name of the owner of Tosh. So, we discarded Jun (because she was already named after the cat) and Suzan (because she was the name of the frog). That left us with Jaela. So, that means that Tosh is the parrot (because Jaela owns the parrot) and the turtle's name is Jaela. So that left us with Jun naming the snake after Bianca. And we know that Suzan's mother's name is Jun because the question states that her mother's pet's name is Bianca.

In conclusion:
Tosh owns a cat (named Jun)
Bianca owns a frog (named Suzan)
Jaela owns a parrot (named Tosh)
Suzan owns a turtle (named Jaela)
Jun owns a snake (named Bianca)

==Question #2==

Bohao, Stewart, Dylan, Tim and Chan are the five players of a basketball team. Two are left handed and three right handed, Two are over 2m tall and three are under 2m, Bohao and Dylan are of the same handedness, whereas Tim and Chan use different hands. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges. If you know that the one playing centre is over 2m tall and is left handed, can you guess his name?

Tim

So, this problem is solved by writing out the information:
2 people are left handed, 3 are right handed, 2 are over 2m and 3 are under 2m; Bohao and Dylan = same hand, Tim =/= Chan in hand; Stewart and Chan = same height, Dylan =/= Tim (height)
since Bohao and Dylan have the same hand and Tim and Chan don't, that means that Bohao and Dylan and ONE OF Tim or Chan are right handed.
since Stewart and Chan have the same height and Dylan and Tim don't, that means that Stewart and Chan and ONE OF Tim and Dylan are under 2m tall.

so when correlating this information with each player, you get:

Bahao: right handed, over 2 m tall
Stewart: left handed, under 2m tall
Dylan: right handed, either over / under 2m tall
Tim: either right / left handed, either over / under 2m tall
Chan: either right / left handed, under 2m tall

If the one playing centre is OVER 2m tall and is LEFT handed that takes out Bahao, Stewart, Dylan and Chan by eliminating the factors, so the centre must be Tim.

==Question #3==
Determine the positions of each player on the baseball team.
Outfielders:
center fielder -- Sung
left fielder -- Ed
right fielder -- Bobo
Infielders:
1st baseman -- Pascal
2nd baseman -- Charles
3rd baseman -- Adam
Shortstop -- Jason
Battery:
Pitcher -- Hassan
Catcher -- Mathieu

We determined this by process of elimination and by using the clues to determine what position the players for sure were NOT.

-Clue #1: Adam does not like the catcher --> hence, Adam is NOT the catcher

-Clue #2: Ed's sister is engaged to the second baseman --> the 2nd baseman is NOT single

-Clue #3: The centre fielder is taller than the right fielder --> upon figuring out with later clues that Bobo was the right fielder, it made sense that Sung was center
fielder

-Clue #4: Hassan and the third baseman live in the same building --> Hassan is NOT the third baseman

-Clue #5: Pascal and Charles each won $20 from the pitcher at a poker game --> Neither Pascal or Charles is the pitcher

-Clue #6: Ed and the outfielders play cards during their free time --> thus, Ed is NOT an outfielder

-Clue #7: The pitcher's wife is the third baseman's sister --> the pitcher is married

-Clue #8: All the battery and infield except Charles, Hassan and Adam are shorter than Sung --> Thus, Charles, Hassan, and Adam cannot be in the outfield

-Clue #9: Pascal, Adam and the shortstop lost $100 each at the race track -->Pascal and Adam cannot be the shortstop

-Clue #10: The second baseman beat Pascal, Hassan, Bobo and the catcher at billiards -->Pascal, Hassan nor Bobo can be the catcher OR the 2nd baseman

-Clue #11: Sung is in the process of getting a divorce --> Thus, Sung is SINGLE

-Clue #12: The catcher and the third baseman each have two legitimate children --> the catcher and the 3rd basemen are married

-Clue #13: Ed, Pascal Jason, the right fielder and the centre fielder are bachelors, the others are all married --> Ed, Pascal, Jason, the right fielder and the center
fielder must not be married

-Clue #14: The shortstop, the third baseman and Bobo all attended the fight --> Bobo cannot be the shortstop or the 3rd baseman

-Clue #15: Mathieu is the shortest player of the team --> he cannot be outfield

Through the process of elimination, we determined the results shown above.
We concluded the following facts about each player:
Adam: could NOT be the catcher, 2nd baseman, right, left, or center outfielder or the shortstop
Bobo: could NOT be the shortstop, 3rd baseman, the catcher, or 2nd baseman
Charles: could NOT be the pitcher, right, left, or center outfielder.
Ed: could NOT be right fielder, center fielder, catcher, 3rd baseman, pitcher or 2nd baseman
Hassan: could NOT be 2nd baseman, catcher, right, left or center outfielder, or 3rd baseman
Jason: could NOT be right fielder, center fielder, 3red baseman, catcher, or pitcher
Mathieu: could NOT be right, left or center outfielder
Pascal: could NOT be pitcher, 2nd baseman, catcher, right fielder, center fielder, 3rd baseman, or shortstop
Sung: single, and tall

From these conclusions about each person, we put each possible candidate next to each position, and by crossing off each name we had found the position for, we solved the puzzle:

Right fielder: Bobo
Left fielder: Ed, Jason, Pascal, Sung, Bobo
Center fielder: Bobo, Sung
1st baseman: Adam, Bobo, Charles, Hassan, Jason, Mathieu, Pascal, Sung
2nd baseman: Charles, Ed, Jason, Mathieu, Sung
3rd baseman: Adam, Charles, Mathieu, Sung
Shortstop: Charles, Hassan, Jason, Mathieu, Sung
Pitcher: Adam, Bobo, Hassan, Mathieu
Catcher: Charles, Mathieu, Sung

==Question #4==
Since you do not know the actual line up of the tournament you must create scenarios were every competitor must play a game versus each other only once. By doing this we can create random matches with players we know are not the “main” attraction and eventually find out who Fernanda will play on the fifth day.

-On the first day, we know that Carla played Petra then that means Fernanda could have played against Sandra, Janet, or Li. So we can set up matches as is Carla vs. Petra, Fernanda vs. Li, Janet vs. Sandra
C.vs.P F.vs.L J.vs.S

-On the second day, we know that Carla played Janet then that means Fernanda could have played Petra, Sandra, or Li. Since on the first day Fernanda and Li already played each other they cannot play each other again so we must put her with Sandra. So the matches are Carla vs. Janet, Fernanda vs. Sandra, and Petra vs. Li
C.vs.J F.vs.S P.vs.L

-On the third day, If Janet played Li then that means Fernanda could have played Petra, Sandra or Carla. Since we know that Fernanda has already played against Sandra we cannot match her up with her so she must play against Petra. So matches are Janet vs. Li, Fernanda vs. Petra, Carla vs. Sandra
J.vs.L F.vs.P C.vs.S

-On the fourth day, If Petra played Sandra then that means Fernanda could have played Carla, Janet or Li. But sense Fernanda already played against Petra and Li she must play Janet because otherwise Carla would play against the same person. Petra vs. Sandra, Fernanda vs. Janet, Carla vs. Li
P.vs.S F.vs.J C.vs.L

-On the firth day since each player only plays each of the others once, the only matches left of are those that involve people not playing each other before hand so the matches are Petra vs. Janet, Li vs. Sandra and Fernanda vs. Carla.
P.vs.J F.vs.C L.vs.S

So on the 5th day Fernanda must play against Sandra

==Question #5==
Homer finally had a week off from his job at the nuclear power plant and intended to spend all nine days of his vacation (Saturday through the following Sunday) sleeping late. But his plans were foiled by some of the people who work in his neighbourhood.
On Saturday, his first morning off, Homer was wakened by the doorbell; it was a salesman of magazine subscriptions.
On Sunday, the barking of the neighbour's dog abruptly ended Homer's sleep.
On Monday, he was again wakened by the persistent salesman but was able to fall asleep again, only to be disturbed by the construction workers next door.
In fact, the salesman, the neighbour's dog and the construction workers combined to wake Homer at least once each day of his vacation, with only one exception.
The salesman woke him again on Wednesday; the construction workers on the second Saturday; the dog on Wednesday and on the final Sunday.
No one of the three noisemakers was quiet for three consecutive days; but yet, no pair of them made noise on more than one day during Homer's vacation. On which day of his holiday was Homer actually able to sleep late?

Salesman = S
Dog = D
Construction Worker = CW
Homer had a vacation from Saturday through to the next Sunday

Saturday -> S

Sunday -> D

Monday -> S+CW

Tuesday ->

Wednesday -> S+D

Thursday -> CW+D

Friday ->

Saturday -> CW+S

Sunday -> D

Homer would get sleep on Tuesday and Friday during his vacation. This will work if every noise maker made noise on the 3rd day of the non-consecutive day period. The Salesmen would have to show up the first Saturday, Monday, Wednesday and then the second Saturday. The dog then makes noise on Sunday, Wednesday, Thursday and Sunday the dog makes noise on 2 consecutive days so that he can pair up with the construction worker. So then the construction worker makes noise on Monday, Thursday and Saturday.

Course:MATH110/Archive/2010-2011/003/Groups/Group 07

2010-10-20T05:27:12Z

JoseTorresTorija:

{{Infobox MATH110 Groups
| group number = 7
| member 1 = Byron Godoy Pinzon
| member 2 = Saloumeh Hassanzadeh
| member 3 = Michelle Little
| member 4 = Emily Oates
| member 5 = [[User:JoseTorresTorija|Jose Torres-Torija Cubillas]]
| member 6 =
}}

=Homework 4=

==Question #2==

Bohao, Stewart, Dylan, Tim and Chan are the five players of a basketball team. Two are left handed and three right handed, Two are over 2m tall and three are under 2m, Bohao and Dylan are of the same handedness, whereas Tim and Chan use different hands. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges. If you know that the one playing centre is over 2m tall and is left handed, can you guess his name?

Tim

So, this problem is solved by writing out the information:
2 people are left handed, 3 are right handed, 2 are over 2m and 3 are under 2m; Bohao and Dylan = same hand, Tim =/= Chan in hand; Stewart and Chan = same height, Dylan =/= Tim (height)
since Bohao and Dylan have the same hand and Tim and Chan don't, that means that Bohao and Dylan and ONE OF Tim or Chan are right handed.
since Stewart and Chan have the same height and Dylan and Tim don't, that means that Stewart and Chan and ONE OF Tim and Dylan are under 2m tall.

so when correlating this information with each player, you get:

Bahao: right handed, over 2 m tall
Stewart: left handed, under 2m tall
Dylan: right handed, either over / under 2m tall
Tim: either right / left handed, either over / under 2m tall
Chan: either right / left handed, under 2m tall

If the one playing centre is OVER 2m tall and is LEFT handed that takes out Bahao, Stewart, Dylan and Chan by eliminating the factors, so the centre must be Tim.

==Question #3==
Determine the positions of each player on the baseball team.
Outfielders:
center fielder -- Sung
left fielder -- Ed
right fielder -- Bobo
Infielders:
1st baseman -- Pascal
2nd baseman -- Charles
3rd baseman -- Adam
Shortstop -- Jason
Battery:
Pitcher -- Hassan
Catcher -- Mathieu

We determined this by process of elimination and by using the clues to determine what position the players for sure were NOT.

-Clue #1: Adam does not like the catcher --> hence, Adam is NOT the catcher

-Clue #2: Ed's sister is engaged to the second baseman --> the 2nd baseman is NOT single

-Clue #3: The centre fielder is taller than the right fielder --> upon figuring out with later clues that Bobo was the right fielder, it made sense that Sung was center
fielder

-Clue #4: Hassan and the third baseman live in the same building --> Hassan is NOT the third baseman

-Clue #5: Pascal and Charles each won $20 from the pitcher at a poker game --> Neither Pascal or Charles is the pitcher

-Clue #6: Ed and the outfielders play cards during their free time --> thus, Ed is NOT an outfielder

-Clue #7: The pitcher's wife is the third baseman's sister --> the pitcher is married

-Clue #8: All the battery and infield except Charles, Hassan and Adam are shorter than Sung --> Thus, Charles, Hassan, and Adam cannot be in the outfield

-Clue #9: Pascal, Adam and the shortstop lost $100 each at the race track -->Pascal and Adam cannot be the shortstop

-Clue #10: The second baseman beat Pascal, Hassan, Bobo and the catcher at billiards -->Pascal, Hassan nor Bobo can be the catcher OR the 2nd baseman

-Clue #11: Sung is in the process of getting a divorce --> Thus, Sung is SINGLE

-Clue #12: The catcher and the third baseman each have two legitimate children --> the catcher and the 3rd basemen are married

-Clue #13: Ed, Pascal Jason, the right fielder and the centre fielder are bachelors, the others are all married --> Ed, Pascal, Jason, the right fielder and the center
fielder must not be married

-Clue #14: The shortstop, the third baseman and Bobo all attended the fight --> Bobo cannot be the shortstop or the 3rd baseman

-Clue #15: Mathieu is the shortest player of the team --> he cannot be outfield

Through the process of elimination, we determined the results shown above.
We concluded the following facts about each player:
Adam: could NOT be the catcher, 2nd baseman, right, left, or center outfielder or the shortstop
Bobo: could NOT be the shortstop, 3rd baseman, the catcher, or 2nd baseman
Charles: could NOT be the pitcher, right, left, or center outfielder.
Ed: could NOT be right fielder, center fielder, catcher, 3rd baseman, pitcher or 2nd baseman
Hassan: could NOT be 2nd baseman, catcher, right, left or center outfielder, or 3rd baseman
Jason: could NOT be right fielder, center fielder, 3red baseman, catcher, or pitcher
Mathieu: could NOT be right, left or center outfielder
Pascal: could NOT be pitcher, 2nd baseman, catcher, right fielder, center fielder, 3rd baseman, or shortstop
Sung: single, and tall

From these conclusions about each person, we put each possible candidate next to each position, and by crossing off each name we had found the position for, we solved the puzzle:

Right fielder: Bobo
Left fielder: Ed, Jason, Pascal, Sung, Bobo
Center fielder: Bobo, Sung
1st baseman: Adam, Bobo, Charles, Hassan, Jason, Mathieu, Pascal, Sung
2nd baseman: Charles, Ed, Jason, Mathieu, Sung
3rd baseman: Adam, Charles, Mathieu, Sung
Shortstop: Charles, Hassan, Jason, Mathieu, Sung
Pitcher: Adam, Bobo, Hassan, Mathieu
Catcher: Charles, Mathieu, Sung

==Question #4==
Since you do not know the actual line up of the tournament you must create scenarios were every competitor must play a game versus each other only once. By doing this we can create random matches with players we know are not the “main” attraction and eventually find out who Fernanda will play on the fifth day.

-On the first day, we know that Carla played Petra then that means Fernanda could have played against Sandra, Janet, or Li. So we can set up matches as is Carla vs. Petra, Fernanda vs. Li, Janet vs. Sandra
C.vs.P F.vs.L J.vs.S

-On the second day, we know that Carla played Janet then that means Fernanda could have played Petra, Sandra, or Li. Since on the first day Fernanda and Li already played each other they cannot play each other again so we must put her with Sandra. So the matches are Carla vs. Janet, Fernanda vs. Sandra, and Petra vs. Li
C.vs.J F.vs.S P.vs.L

-On the third day, If Janet played Li then that means Fernanda could have played Petra, Sandra or Carla. Since we know that Fernanda has already played against Sandra we cannot match her up with her so she must play against Petra. So matches are Janet vs. Li, Fernanda vs. Petra, Carla vs. Sandra
J.vs.L F.vs.P C.vs.S

-On the fourth day, If Petra played Sandra then that means Fernanda could have played Carla, Janet or Li. But sense Fernanda already played against Petra and Li she must play Janet because otherwise Carla would play against the same person. Petra vs. Sandra, Fernanda vs. Janet, Carla vs. Li
P.vs.S F.vs.J C.vs.L

-On the firth day since each player only plays each of the others once, the only matches left of are those that involve people not playing each other before hand so the matches are Petra vs. Janet, Li vs. Sandra and Fernanda vs. Carla.
P.vs.J F.vs.C L.vs.S

So on the 5th day Fernanda must play against Sandra

==Question #5==
Homer finally had a week off from his job at the nuclear power plant and intended to spend all nine days of his vacation (Saturday through the following Sunday) sleeping late. But his plans were foiled by some of the people who work in his neighbourhood.
On Saturday, his first morning off, Homer was wakened by the doorbell; it was a salesman of magazine subscriptions.
On Sunday, the barking of the neighbour's dog abruptly ended Homer's sleep.
On Monday, he was again wakened by the persistent salesman but was able to fall asleep again, only to be disturbed by the construction workers next door.
In fact, the salesman, the neighbour's dog and the construction workers combined to wake Homer at least once each day of his vacation, with only one exception.
The salesman woke him again on Wednesday; the construction workers on the second Saturday; the dog on Wednesday and on the final Sunday.
No one of the three noisemakers was quiet for three consecutive days; but yet, no pair of them made noise on more than one day during Homer's vacation. On which day of his holiday was Homer actually able to sleep late?

Salesman = S
Dog = D
Construction Worker = CW
Homer had a vacation from Saturday through to the next Sunday

Saturday -> S

Sunday -> D

Monday -> S+CW

Tuesday ->

Wednesday -> S+D

Thursday -> CW+D

Friday ->

Saturday -> CW+S

Sunday -> D

Homer would get sleep on Tuesday and Friday during his vacation. This will work if every noise maker made noise on the 3rd day of the non-consecutive day period. The Salesmen would have to show up the first Saturday, Monday, Wednesday and then the second Saturday. The dog then makes noise on Sunday, Wednesday, Thursday and Sunday the dog makes noise on 2 consecutive days so that he can pair up with the construction worker. So then the construction worker makes noise on Monday, Thursday and Saturday.

User:JoseTorresTorija

2010-09-20T06:18:41Z

JoseTorresTorija: Created page with 'The Pythagorean Theorem is a formula which involves the solution to finding sides of a right triangle. This theorem is only applicable in situations where the triangle which is b…'

The Pythagorean Theorem is a formula which involves the solution to finding sides of a right triangle. This theorem is only applicable in situations where the triangle which is being measured is a right triangle which means that one side of the triangle is a 90◦ angle. If the triangle has a 90◦ and you have 2 of the sides of the triangle you can use the equation c^2=a^2+b^2. With this equation one is able to input the corresponding sides of the triangle and get a solution either equaling a missing side or the hypotenuse, the hypotenuse is always the longest side of the triangle is always going to be C in the equation. This equation can be used in multiple situations other than just finding the missing sides of a triangle, for instance when it can be used to find the length of a radius inside of a circle if you know the distance between points in the circle or used in real life scenarios when you need to know the length of a diagonal object.