UBC Wiki - User contributions [en]

Thread:Course talk:POLI380JAN2011Owen/Survey/Environment--Citizens' sustainability choices/Final Questions (draft 1)/reply (19)

2011-02-08T18:50:49Z

ShannonLee: Reply to Final Questions (draft 1) & Final Draft

I completely understand where you are coming from with your critique of "during, after... my life time," my thought process, however, was that this measure would allow us to assess if the individual believed climate change was an issue that would impact their lives, or if they believed in it but still viewed it as a distant issue not relevant to their lives. I just think the year scale in question 4 is a little arbitrary and the results might now be that relevant. I think the age of the individuals is irrelevant as we are not measuring a time frame as much as we would be measuring direct impact on their lives.

Thread:Course talk:POLI380JAN2011Owen/Survey/Environment--Citizens' sustainability choices/Final Questions (draft 1)/reply (17)

2011-02-08T07:20:11Z

ShannonLee: Reply to Final Questions (draft 1)

Final draft of question- version 2

So this thread is getting increasingly difficult to navigate, I'll try to include as many as the revisions as possible. I am going to change Q5 to an answer scale in ranges rather than having 7 options.

Q2 '''OPINION ABOUT IMPACT OF INDIVIDUAL EFFORTS''' Individual efforts, including recycling, composting, biking, and using public transit, make substantial differences in environmental issues such as climate change? (1)strongly agree (2)somewhat agree (3)neutral (4)somewhat disagree (5)strongly disagree

Q5 '''SUSTAINABLE PRACTICES''' Which of the following sustainable practices do you do on a weekly basis: -recycle -use reusable food/beverage containers -use reusable bags -compost -take public or alternative modes of transportation -buy local products -use energy-efficient lighting and/or appliances (1)1-2 of the above (2)3-4 of the above (3) 5 or more of the above (4) not applicable .

Q1 '''COST''' "How important is cost in making choices that reduce your impact on the environment?" (1) Very Important (2) Important (3) Neutral (4) Not important (5) Not at all important (6) Not Applicable

Q3 '''SUGGESTED QUESTION FOR NUMBER 3'''"How much time do you spend learning about environmental issues per month: 1)10+ hours, 2)7-10 hours, 3) 4-7 hours, 4)1-3 hours, 5) less than one hour, 6) none.

Q4 '''BELIEF ABOUT THE EFFECT OF CLIMATE CHANGE ON CANADA''' Climate change will have a noticeable effect on Canada's environment in the next ___ a)25 years b)50 years c)100 years d)0 years - Climate change HAS ALREADY noticeably affected Canada's environment e)Not Applicable - Climate change will NOT have a lasting effect on Canada's environment.

I am also going to suggest that for Q4 we change the question to:

Climate change will have a noticeable effect on Canada's environment _____

(1)during my lifetime (2)after my lifetime (3)not applicable - climate change is HAS ALREADY noticeably affected Canada's environment (4)not applicable - climate change will not have a noticeable affect on Canada's environment

or something along these lines, just so we don't have such arbitrary time frames.

Thread:Course talk:POLI380JAN2011Owen/Survey/Environment--Citizens' sustainability choices/Q4. Do you think climate change will have an affect on Canada/Vancouver in.../reply (11)

2011-02-06T20:53:38Z

ShannonLee: Reply to [[Thread:Course talk:POLI380JAN2011Owen/Survey/Environment--Citizens' sustainability choices/Q4. Do you think climate change will have an affect on Canada/Vancouver in...|Q4. Do you think climate change will have an affect on Canada/Vancouve

I also support option B. Option A would be hard to interpret the results, and option C leaves too much open to interpretation. What some may describe as some effect others may define as neutral or strong.

Thread:Course talk:POLI380JAN2011Owen/Survey/Environment--Citizens' sustainability choices/Q5. Which of the following sustainable practices do you do on a regular basis:/reply

2011-02-06T00:45:38Z

ShannonLee: Reply to [[Thread:Course talk:POLI380JAN2011Owen/Survey/Environment--Citizens' sustainability choices/Q5. Which of the following sustainable practices do you do on a regular basis:|Q5. Which of the following sustainable practices do you do on a regul

if we were to use a question such as this maybe the options should be put into ranges,
1)none of the above 2)1-3 3)4-6 4)all of the above

Perhaps the question could be phrased:

How many of the following sustainable practices do you do on a regular basis?
(list the practices)

One of my concerns with this question is that people's differing opinions of "a regular basis." Another issue is the list of practices we choose to include, we may not include several key sustainable practices that people practice. It really depends on what we are trying to measure with this question, is it simply do people recycle? do people buy local? or is it how sustainable are people's behaviors? etc.

Thread:Course talk:POLI380JAN2011Owen/Survey/Environment--Citizens' sustainability choices/Q1. "Is cost a factor in your decision to make sustainable choices?" (1)Yes (2)No/reply (11)

2011-02-06T00:36:03Z

ShannonLee: Reply to [[Thread:Course talk:POLI380JAN2011Owen/Survey/Environment--Citizens' sustainability choices/Q1. "Is cost a factor in your decision to make sustainable choices?" (1)Yes (2)No|Q1. "Is cost a factor in your decision to make sustainable choices

Niel Chah, the last question your suggested is very well phrased, i was going to mention that we should include a not applicable option, so I am glad that you have done so. This could definitely be a final draft of this question.

Thread:Course talk:POLI380JAN2011Owen/Survey/Environment--Citizens' sustainability choices/Q4. Do you think climate change will have an affect on Canada/Vancouver in.../reply (8)

2011-02-06T00:28:58Z

I am a big fan of the question raised by Jes, the question allows us to interpret how pressing of a matter climate change is. The other questions would be much harder to interpret. Perhaps instead of listing 0 years, we could list not applicable, as climate change will not have a lasting effect on Canada's environment.

Thread:Course talk:POLI380JAN2011Owen/Survey/Environment--Citizens' sustainability choices/3. Government vs Individuals?/reply (6)

2011-02-06T00:22:33Z

ShannonLee: Reply to 3. Government vs Individuals?

I think this would be a great question in regard to questioning if people believe green/sustainability education should be made part of the public school system. Eg.

How important do you believe governmental policies, such as making green education part of school curriculum, are for encouraging greater sustainability within Canada going forward.

or simply,

Should green education be integrated into the public school system's curriculum?

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

2011-02-04T14:28:22Z

ShannonLee:

==Homework 13: Team Problem==

==pH: The Power of Understanding==

The pH scale measures acidic and basic chemical properties. The scale ranges from 0 to 14, with a pH of 7 being neutral. Any pH less than 7 is acidic, whereas a pH of greater than 7 is basic. The pH scale is a logarithmic scale, which means that each lower numerical value is ten times more acidic than the next higher numerical value (eg. pH of 3 is ten times more acidic than a pH of 4, and 100 times (10 times 10) more acidic than a pH of 5). This property works for basic values as well, with each higher value being ten times more alkaline, or basic, than the value below it (eg. pH of 9 is ten times more alkaline than pH of 8).

pH is a measure of the concentration of hydrogen ions. The definition of pH is the negative logarithm of a hydrogen ion concentration. The less ions that are present, the more alkaline the solution. If more hydrogen ions are found the more acidic. To find the pH, we need to find the logarithm of the H+ (hydrogen) concentration, and then reverse the sign - the only case in which the sign is not reversed is if the H+ concentration is greater than 1 M (which is highly unusual).

The fact that the pH scale is a logarithmic scale turns out to be very useful when scientists need to calculate very high values of hydrogen ions in solutions. In fact, strongly basic solutions might have, let’s say, 100.000.000.000.000 times less hydrogen ions than strongly basic solutions. It is then obvious that dealing with such large numbers is a source of complications for the scientists that need to analyze this data. A logarithmic scale helps scientists deal with extremely large numbers for pH in a more efficient way, with applications in various fields such as marine biology and oceanography (sea water), medicine (body fluids) and certainly chemistry.

For comparison purposes, the pH scale uses logs because logs can expand the values between 0 and 1 on a compact scale to make it into a large scale, this also means that the log scale changes. pH scale uses logs when the answer or the change in the values depends on the size of the the change in proportion to the value and not on the absolute size of change.
For example, when you graph two acids, one with a concentration of 0.001 and one with a weaker acid say with pH4 as opposed to the first, pH3, it would be 0.0001. When you graph this, the change barely shows and it comes up as the same value, implying that all neutral and basic values are indistinguishable. Since it is important to show the difference between the two, a weak and a weaker acid, then these values are needed, and therefore we use logs to find these values to plot on our graph.

The equation of pH is:

'''pH = -log [H+]'''

eg. what is the pH of a substance with an H+ concentration of 0.001?

''to find the pH, we first should convert the concentration value into exponential notation:''

0.001M = 10^-3

''we than plug this number into our formula of pH = -log [H+]''

pH = -log [10^-3]

pH = 3

This means that since pure water is pH 7

The concentration of hydrogen ions would be 0.0000001, plugged into the pH formula, it would look like pH = -log [10^-7]

[[Image:184phscale.gif]]

'''FEEL FREE TO CHECK OUT THE YOUTUBE VIDEO FOR A VISUAL EXPLANATION!'''

{{#ev:youtube|u837KYKyr9c|400}}

sources:
http://www.elmhurst.edu/~chm/vchembook/184ph.html
http://www.science.uwaterloo.ca/~cchieh/cact/c123/ph.html
http://mathforum.org/library/drmath/view/55574.html

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

2011-02-04T05:27:27Z

ShannonLee: /* pH: The Power of Understanding */

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

2011-02-04T05:27:03Z

ShannonLee: /* pH: The Power of Understanding */

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

2011-02-04T05:23:53Z

ShannonLee: /* pH: The Power of Understanding */

Thread:Course talk:POLI380JAN2011Owen/Survey/Environment--Citizens' sustainability choices/Opinion of the possible affects of global climate change/reply (15)

2011-02-02T05:59:53Z

ShannonLee: Reply to Opinion of the possible affects of global climate change

Yes thank-you so much for summarizing the discussion, it was getting to be all over the place. @NielChah, I really like the concept of the question. Maybe instead of phrasing it as a should/should not question we can ask if it actually is a factor:

"Is cost a factor in your decision to make sustainable choices?" (1)Yes (2)No

I guess I will go ahead and ask a question as well, please critique and edit, the basic premises is whether or not people truly believe that individual efforts make a difference in regards to sustainability and environmental issues.

Q2: "Individual efforts (eg. recycling, composting, biking, using public transit etc.)make substantial differences in environmental issues such as climate change?"
(1)strongly agree (2)somewhat agree (3)neutral (4)somewhat disagree (5)strongly disagree

Thread:Course talk:POLI380JAN2011Owen/Survey/Environment--Citizens' sustainability choices/Opinion of the possible affects of global climate change/reply (12)

2011-01-31T20:43:13Z

ShannonLee: Reply to Opinion of the possible affects of global climate change

Perhaps we can ask a question about how important environmental issues are when voting.

For example: To what extend does a candidate's environmental platform influence your vote?
and then have some sort of scale such as very applicable, not applicable etc. to measure it.

Thread:Course talk:POLI380JAN2011Owen/Survey/Environment--Citizens' sustainability choices/Opinion of the possible affects of global climate change/reply (10)

2011-01-30T22:55:40Z

ShannonLee: Reply to Opinion of the possible affects of global climate change

Hey guys, great discussion so far. I agree that we should focus on actions instead of opinions. When I asked the TA about the survey questions, he said that each question could only have one part, as in "Do you recycle?" But we should avoid multi-part question ie. "Do you recycle? If yes, how many times a week?"

As for defining sustainability

Merriam-webster definition:

Definition of SUSTAINABLE

1 capable of being sustained

2a : of, relating to, or being a method of harvesting or using a resource so that the resource is not depleted or permanently damaged <sustainable techniques> <sustainable agriculture> b : of or relating to a lifestyle involving the use of sustainable methods <sustainable society>

http://www.sustainabilitydictionary.com/

Development that meets the needs of the present without compromising the ability of future generations to meet their own needs.

This definition was created in 1987 at the World Commission on Environment and Development (the Brundtland Commission).

User:ShannonLee/Politics & math

2011-01-28T08:39:48Z

ShannonLee:

'''On your profile page within the wiki, write an essay describing a particular use of calculus in your ﬁeld of study, or in a ﬁeld of interest to you. Feel free of course to add pictures, links and/or videos on the page (that's the advantage of writing online versus on paper). Your essay should be at least 500 words long (think a nice and interesting full page of text).'''

http://s3.hubimg.com/u/270622_f520.jpg

The use of calculus at first does not seem relevant to Political Science. However, as one evaluates the relationship between these two subject areas more closely it becomes clear that even in the realm of politics, math plays a huge part. The study of statistics is a huge component of political science. Evaluating numbers, percentages and averages are necessary to help understand what the population wants/needs as well as for developing political platforms.

In the statistical politics class I am currently taking, learning how to critically engage numbers as a representative of popular opinion is crucial. Many important decisions are decided upon based on polls and surveys. Politicians decide upon their platform, and strategy as a direct result of the outcome of particular census data. Calculus plays a huge part in increasing one’s knowledge on how to manipulate models or formulas as well as how to interpret said models. Ultimately, statistics interprets people’s political ideas or opinions and applies them to polls. Through using numbers as representatives of opinions political scientist are able to track changing opinions and preferences. Statistics are extremely useful because they summarize vast amounts of information into a succinct fashion, represented by numbers and graphs. As a direct result of using numbers to represent data, political experts are moving closer and closer to being able to predict future wars or voter tendencies.

A key component to politics is number crunching. Budgets, population, cost/benefit analysis’ are all things in politics that incorporate the basic skills of calculus. The key component to almost any political movement, party, candidate etc. is budget. Calculus allows the experts to crunch numbers in order to find out things such as “will spending x amount on y bring more revenue or benefit, than spending x amount on z.” Many questions regarding budgetary concerns can be answered through creating particular models to evaluate potential expenditures and cost-benefits. Economics uses many mathematical devices, in turn, political science is heavily influenced and heavily influences economics; therefore, political science also employs the same economic mathematical devices.

[[Image:Mathpoli.png]]

Predicting population growth in certain regions as well as economic tendencies, environmental patterns, or trends in voting are all key concerns of political science. Through evaluating and applying previous patterns through the use of graphs, data and modeling, the probable outcome of a situation can be achieved. Despite the fact that it is very difficult, and some say impossible, to quantify beliefs, ideas and opinions, as math becomes increasing integrated into this field, political science tentatively crosses the border from social sciences to a “hard” sciences. Political scientists in the United States have been able to predict federal election outcomes in recent years by analyzing the country’s GDP and unemployment rates. Through using mathematical devices and statistics these outcomes could be obtained.

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

2011-01-28T08:38:49Z

ShannonLee:

'''Homework 12: Team Problem'''

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

*For this function, the height of the horizontal asymptote on the right is equal to the value of the numerator. If the whole function is multiplied by a value, this value will be where the horizontal asymptote on the right hand side is located.

eg. <math>P(t) = \frac{K}{1+e^{-t}}</math> <math>K=</math>height of right side horizontal asymptote.

*The Y-intercept will always be a value between the left side horizontal asymptote and the right side horizontal asymptote because these are the floor and the ceiling of the function. Since the left side horizontal asymptote is 0 and the right side horizontal asymptote is K, it is a matter of choosing which value between 0 and K is desired for the Y-intercept. The Y-intercept can be changed by shifting the function to the left and to the right, therefore, it will take different Y-values at t=0. Shifting the function left and right is done by adding or subtracting directly to the t.

eg.<math>P(t) = \frac{1}{1+e^{-t+j}}</math>

In order to select the specific number for the Y-intercept, you can use the following formula in place of j:

<math>ln( \frac{K}{J}-1)</math>

Where K= the horizontal asymptote on the right, and where J= the Y-value you have chosen for your Y-intercept. So when you combine these manipulations, you get:

<math>P(t) = \frac{K}{1+e^{-t+ln(\frac{K}{J}-1)}}</math>

*'''Bonus''' If you want to alter the slope of the function you need to add a coefficient to t. If you multiply t by an integer then the function will get from the horizontal asymptote on the left to the horizontal asymptote on the right sooner. If you divide t then the function will move from it's floor to it's ceiling more slowly.

eg.<math>P(t) = \frac{K}{1+e^{-Rt+ln(\frac{K}{J}-1)}}</math>

Here the value you plug in for R will control the slope of the function as it spans from one asymptote to the other.

In summary:
K = horizontal asymptote on the right
J = value of Y-intercept
R > 1 = steeper slope between asymptotes
0 < R < 1 = more gradual slope between asymptotes

'''Application of the graph to model squirrels population in Canada'''

[[Image:Mathgraph.jpg]]

<math>P(t) = \frac{1}{3+e^{-15t+6}}</math>

A projection of the population in 20 years is P(20)=0.3 periodic.

<math>P(t) = \frac{1}{3+e^{-15{20}+6}}</math> = 0.3 periodic.

We compute the derivative P'(t):

[[Image:Mathformuladeriv.jpg]]

In 20 years the instantaneous rate of growth P'(20) of the squirrel population will be 9.2 X 10^-8. This number tells us that the population is nearing carrying capacity. As time progresses and resources and space are abundant the growth of the squirrel population greatly increases; however, when space and resources become more limited the population of squirrels stops increasing and levels off at carrying capacity.

User:ShannonLee

2011-01-28T08:17:00Z

ShannonLee:

Hello, I'm a third year Political Science student, who is planning on doing a secondary degree in Environmental Sciences. I haven't done math in quite awhile so I'm quite rusty at it, hopefully this course will enable me to catch up and improve upon my mathematical skills!

http://wiki.ubc.ca/Politics%26math

The Pythagorean Theorem

<math> a^2+b^2=c^2 </math>

Image obtained from [http://www.mathopenref.com/pythagorastheorem.html]

The Pythagorean Theorem is used to solve for the hypotenuse or length of a side of a right angle triangle. The Pythagorean equation, which is <math>a^2 + b^2 = c^2 </math>, was created by Pythagoras, an ancient Greek mathematician; however, some argue that the theorem was realized centuries prior to Pythagoras by the Babylonians.

In the equation <math>a^2 + b^2 = c^2</math>, a and b refer to the lengths of the triangle, and c refers to the hypotenuse. Rearranging the equation to form <math>b= \sqrt{(c^2 - a^2)} </math>or <math>a= \sqrt{(c^2 +b^2)} </math> solves for the length of a side of the triangle given the length of one other side and the hypotenuse. Because the theorem is a statement of both areas and lengths, it is said to be an expression of both algebra and geometry.

There are many proofs which justify the Pythagorean Theorem, the following is a proof that was constructed by U.S. president James Garfield. The proof uses a trapezoid to prove that in regards to a right triangle, the sum of the squares of the sides is equal to the square of the hypotenuse. Ultimately this proof uses the formula <math> \frac{(a+b)(a+b)}{2} = ab+\frac{(c^2)}{2} </math>, with the left side expanded to <math>a^2+2ab+b^2 </math> to prove the equality found thousands of years previous by Pythagoras.

Image obtained from: [http://www.mathopenref.com/pythagorastheorem.html]

An example using the Pythagorean Theorem to find the hypotenuse of a right triangle:
Joan lives 8.0m east of her high school, Bill lives 10.0m north from the same school. How far do Joan and Bill live from one another?

To solve this question, let a = Joan’s distance from the school (8.0m) and let b= Bill’s distance from the school (10.0m). The Pythagorean Theorem states that <math>a^2+b^2=c^2</math>,
Therefore, <math>8.0m^2+10.0m^2 = c^2</math>,

<math>164.0m = c^2</math>
<math>c=\sqrt{164.0}</math>
<math>c=12.8m</math>

Sources:

Oracle Education Foundation. (August 25, 2010). The Pythagorean Theorem. Retrieved from: http://library.thinkquest.org/25672/newpage1.htm

Page, John. (2009). Pythagorean Theorem. Retrieved from:
http://www.mathopenref.com/pythagorastheorem.html

User:ShannonLee/Politics & math

2011-01-28T08:16:31Z

ShannonLee: Created page with "'''On your profile page within the wiki, write an essay describing a particular use of calculus in your ﬁeld of study, or in a ﬁeld of interest to you. Feel free of course to..."

'''On your profile page within the wiki, write an essay describing a particular use of calculus in your ﬁeld of study, or in a ﬁeld of interest to you. Feel free of course to add pictures, links and/or videos on the page (that's the advantage of writing online versus on paper). Your essay should be at least 500 words long (think a nice and interesting full page of text).'''

http://s3.hubimg.com/u/270622_f520.jpg

The use of calculus at first does not seem relevant to Political Science. However, as one evaluates the relationship between these two subject areas more closely it becomes clear that even in the realm of politics, math plays a huge part. The study of statistics is a huge component of political science. Evaluating numbers, percentages and averages are necessary to help understand what the population wants/needs as well as for developing political platforms.

In the statistical politics class I am currently taking, learning how to critically engage numbers as a representative of popular opinion is crucial. Many important decisions are decided upon based on polls and surveys. Politicians decide upon their platform, and strategy as a direct result of the outcome of particular census data. Calculus plays a huge part in increasing one’s knowledge on how to manipulate models or formulas as well as how to interpret said models. Ultimately, statistics interprets people’s political ideas or opinions and applies them to polls. Through using numbers as representatives of opinions political scientist are able to track changing opinions and preferences. Statistics are extremely useful because they summarize vast amounts of information into a succinct fashion, represented by numbers and graphs. As a direct result of using numbers to represent data, political experts are moving closer and closer to being able to predict future wars or voter tendencies.

A key component to politics is number crunching. Budgets, population, cost/benefit analysis’ are all things in politics that incorporate the basic skills of calculus. The key component to almost any political movement, party, candidate etc. is budget. Calculus allows the experts to crunch numbers in order to find out things such as “will spending x amount on y bring more revenue or benefit, than spending x amount on z.” Many questions regarding budgetary concerns can be answered through creating particular models to evaluate potential expenditures and cost-benefits. Economics uses many mathematical devices, in turn, political science is heavily influenced and heavily influences economics; therefore, political science also employs the same economic mathematical devices.

[[Image:Mathpoli.png]]

Predicting population growth in certain regions as well as economic tendencies, environmental patterns, or trends in voting are all key concerns of political science. Through evaluating and applying previous patterns through the use of graphs, data and modelling, the probable outcome of a situation can be achieved. Despite the fact that it is very difficult, and some say impossible, to quantify beliefs, ideas and opinions, as math becomes increasing integrated into this field, political science tentatively crosses the border from social sciences to a “hard” sciences. Political scientists in the United States have been able to predict federal election outcomes in recent years by analyzing the country’s GDP and unemployment rates. Through using mathematical devices and statistics these outcomes could be deduced.

File:Mathpoli.png

2011-01-28T08:02:17Z

ShannonLee:

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

2011-01-28T07:51:30Z

ShannonLee:

'''Homework 12: Team Problem'''

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

*For this function, the height of the horizontal asymptote on the right is equal to the value of the numerator. If the whole function is multiplied by a value, this value will be where the horizontal asymptote on the right hand side is located.

eg. <math>P(t) = \frac{K}{1+e^{-t}}</math> <math>K=</math>height of right side horizontal asymptote.

*The Y-intercept will always be a value between the left side horizontal asymptote and the right side horizontal asymptote because these are the floor and the ceiling of the function. Since the left side horizontal asymptote is 0 and the right side horizontal asymptote is K, it is a matter of choosing which value between 0 and K is desired for the Y-intercept. The Y-intercept can be changed by shifting the function to the left and to the right, therefore, it will take different Y-values at t=0. Shifting the function left and right is done by adding or subtracting directly to the t.

eg.<math>P(t) = \frac{1}{1+e^{-t+j}}</math>

In order to select the specific number for the Y-intercept, you can use the following formula in place of j:

<math>ln( \frac{K}{J}-1)</math>

Where K= the horizontal asymptote on the right, and where J= the Y-value you have chosen for your Y-intercept. So when you combine these manipulations, you get:

<math>P(t) = \frac{K}{1+e^{-t+ln(\frac{K}{J}-1)}}</math>

*'''Bonus''' If you want to alter the slope of the function you need to add a coefficient to t. If you multiply t by an integer then the function will get from the horizontal asymptote on the left to the horizontal asymptote on the right sooner. If you divide t then the function will move from it's floor to it's ceiling more slowly.

eg.<math>P(t) = \frac{K}{1+e^{-Rt+ln(\frac{K}{J}-1)}}</math>

Here the value you plug in for R will control the slope of the function as it spans from one asymptote to the other.

In summary:
K = horizontal asymptote on the right
J = value of Y-intercept
R > 1 = steeper slope between asymptotes
0 < R < 1 = more gradual slope between asymptotes

[[Image:Mathgraph.jpg]]

Application of the graph to the model

example 1: Population growth of squirrels in Vancouver (information not real)

[[Image:Mathformuladeriv.jpg]]

In 20 years the instantaneous rate of growth of the squirrel population is 9.2 X 10^-8. This number tells us that the population is nearing carrying capacity. As time progresses and resources and space are abundant the growth of the squirrel population greatly increases; however, when space and resources become more limited the population of squirrels stops increasing and levels off at carrying capacity.

File:Mathformuladeriv.jpg

2011-01-28T07:23:48Z

ShannonLee:

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

2011-01-28T07:21:56Z

ShannonLee:

'''Homework 12: Team Problem'''

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

*For this function, the height of the horizontal asymptote on the right is equal to the value of the numerator. If the whole function is multiplied by a value, this value will be where the horizontal asymptote on the right hand side is located.

eg. <math>P(t) = \frac{K}{1+e^{-t}}</math> <math>K=</math>height of right side horizontal asymptote.

*The Y-intercept will always be a value between the left side horizontal asymptote and the right side horizontal asymptote because these are the floor and the ceiling of the function. Since the left side horizontal asymptote is 0 and the right side horizontal asymptote is K, it is a matter of choosing which value between 0 and K is desired for the Y-intercept. The Y-intercept can be changed by shifting the function to the left and to the right, therefore, it will take different Y-values at t=0. Shifting the function left and right is done by adding or subtracting directly to the t.

eg.<math>P(t) = \frac{1}{1+e^{-t+j}}</math>

In order to select the specific number for the Y-intercept, you can use the following formula in place of j:

<math>ln( \frac{K}{J}-1)</math>

Where K= the horizontal asymptote on the right, and where J= the Y-value you have chosen for your Y-intercept. So when you combine these manipulations, you get:

<math>P(t) = \frac{K}{1+e^{-t+ln(\frac{K}{J}-1)}}</math>

*'''Bonus''' If you want to alter the slope of the function you need to add a coefficient to t. If you multiply t by an integer then the function will get from the horizontal asymptote on the left to the horizontal asymptote on the right sooner. If you divide t then the function will move from it's floor to it's ceiling more slowly.

eg.<math>P(t) = \frac{K}{1+e^{-Rt+ln(\frac{K}{J}-1)}}</math>

Here the value you plug in for R will control the slope of the function as it spans from one asymptote to the other.

In summary:
K = horizontal asymptote on the right
J = value of Y-intercept
R > 1 = steeper slope between asymptotes
0 < R < 1 = more gradual slope between asymptotes

[[Image:Mathgraph.jpg]]

Application of the graph to the model

example 1: Population growth of squirrels in Vancouver (information not real)

As time progresses and resources and space are abundant the growth of the squirrel population greatly increases; however, when space and resources become more limited the population of squirrels stops increasing and levels off at carrying capacity.

File:Mathgraph.jpg

2011-01-28T07:20:12Z

ShannonLee:

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

2011-01-28T06:28:03Z

ShannonLee:

'''Homework 12: Team Problem'''

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

*For this function, the height of the horizontal asymptote on the right is equal to the value of the numerator. If the whole function is multiplied by a value, this value will be where the horizontal asymptote on the right hand side is located.

eg. <math>P(t) = \frac{K}{1+e^{-t}}</math> <math>K=</math>height of right side horizontal asymptote.

*The Y-intercept will always be a value between the left side horizontal asymptote and the right side horizontal asymptote because these are the floor and the ceiling of the function. Since the left side horizontal asymptote is 0 and the right side horizontal asymptote is K, it is a matter of choosing which value between 0 and K is desired for the Y-intercept. The Y-intercept can be changed by shifting the function to the left and to the right, therefore, it will take different Y-values at t=0. Shifting the function left and right is done by adding or subtracting directly to the t.

eg.<math>P(t) = \frac{1}{1+e^{-t+j}}</math>

In order to select the specific number for the Y-intercept, you can use the following formula in place of j:

<math>ln( \frac{K}{J}-1)</math>

Where K= the horizontal asymptote on the right, and where J= the Y-value you have chosen for your Y-intercept. So when you combine these manipulations, you get:

<math>P(t) = \frac{K}{1+e^{-t+ln(\frac{K}{J}-1)}}</math>

*'''Bonus''' If you want to alter the slope of the function you need to add a coefficient to t. If you multiply t by an integer then the function will get from the horizontal asymptote on the left to the horizontal asymptote on the right sooner. If you divide t then the function will move from it's floor to it's ceiling more slowly.

eg.<math>P(t) = \frac{K}{1+e^{-Rt+ln(\frac{K}{J}-1)}}</math>

Here the value you plug in for R will control the slope of the function as it spans from one asymptote to the other.

In summary:
K = horizontal asymptote on the right
J = value of Y-intercept
R > 1 = steeper slope between asymptotes
0 < R < 1 = more gradual slope between asymptotes

[[Image:math2.gif]]

Application of the graph to the model

example 1: Population growth of squirrels in Vancouver (information not real)

As time progresses and resources and space are abundant the growth of the squirrel population greatly increases; however, when space and resources become more limited the population of squirrels stops increasing and levels off at carrying capacity.

File:Math2.gif

2011-01-28T06:27:48Z

ShannonLee:

File:Math.gif

2011-01-28T06:21:16Z

ShannonLee:

Course:POLI380JAN2011Owen/Survey/Environment--Citizens' sustainability choices

2011-01-24T21:57:25Z

ShannonLee:

Topic: == Environment--Citizens' sustainability choices (i.e. the actions individual citizens take (or do not take) in the interest of sustainability (such as personal carbon reduction efforts, sustainable food choices, etc) ==

Reminder: Write properly structured questions, including simple text for response options. Use the discussion tab above for discussion of advantages and disadvantages of particular questions and their wording.

'''Contributors''':
1. Caroline Jankech
2. Aaron Bains
3. Asena Cansu Yildiz
4. Niel Chah
5. Jesica Simkin
6. Shannon Lee
7.
8.
9.
10.
11.
12.
13.
14.
15.

Course:MATH110/Archive/2010-2011/003/Teams/Solothurn/Homework 11

2011-01-19T07:52:55Z

ShannonLee:

'''Write a linear model to predict the cost of producing flags of your team's 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.'''

C= mx+b

where C = total cost
mx = variable cost
b = fixed cost
m = marginal cost
x = quantity produced

$100= $7(20units) + b

b = $100 - $140

b = -$40

'''Describe your model. What does your model predict for a production of 150 items? According to your model, what happens to the average cost per item as production levels increase?

'''

Our model predicts the total cost for producing x number of flags given the marginal cost ($7) and a fixed cost (-$40). In the model above, when producing 20 units, the cost of producing a single unit is $5. The reason that this number is lower than the marginal cost is because of the negative fixed cost. One reasoning for the negative fixed cost could be government subsidies for the production of flags.

For the production of 150 items our model predicts a total cost of $1010.

C = $7(150units) + (-$40)

C = $1010

The cost per unit in this example is $1010/300 = $6.73

According to our model, as the production level increases, the average cost per item increases

examples:

C = $7(300units) + (-$40)

C = $2060

The cost per unit in this example is $2060/300 = $6.87

C = $7(1000units) + (-$40)

C = $6960

The cost per unit in this example is $2060/1000 = $6.96

'''Finally, find some other models (not necessarily linear) for which you get other behaviours such as: The average cost remains constant as production increases. The average cost diminishes as production increases. The average cost increases as production increases. 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. Any other interesting properties that you can think of and create a model for. Bonus points can be obtained for very interesting ideas.'''

'''The average cost remains constant as production increases'''

An example would b: TC=3Q

Q would be quantity and 3 is the marginal cost

Because in economics, AVC = TC/Q ( average cost = total cost / quantity)

If you make the cost function 3Q, then that means the average cost will be 3, which is constant no matter what your Q value is.

* there is no fixed cost or 'b' in the y=mx+b formula *

'''The average cost increases as production increases'''

An example would be: TC=100+3Q

The average cost remains constant but now there is a fixed cost added or 'b' because as the quantity increase, the fixed cost is shared by an increasing number, therefore the average cost diminishes.

'''The average cost increases as production increases'''

An example would be: TC=Q^2

Because AVC=TC/Q, and TC is Q^2, then AVC would then equal Q^2/Q which would make AVC=Q. Therefore as quantity increases average cost increases.

Course:MATH110/Archive/2010-2011/003/Teams/Solothurn/Homework 11

2011-01-19T07:20:42Z

ShannonLee:

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

2011-01-19T07:10:58Z

ShannonLee: /* Contact */

{{
Infobox MATH110 Teams
| team name = Solothurn
| member 1 = [[User:CurtisDoucette|Curtis Doucette]]
| member 2 = Fiona Ma
| member 3 = Gaia Silvestri
| member 4 = Shannon Lee
}}
= Contact =

'''Curtis Doucette'''
cell: 778-855-2355 email: <curtismdoucette@gmail.com> skype: curtismdoucette

'''Fiona Ma''' cell: 778 688 8537 email: <fionama@live.com> skype: fionama-

'''Shannon Lee''' cell: 604-351-5816 email: s.e.lee16@hotmail.com skype: shannon.elizabeth.l

http://wiki.ubc.ca/Course:MATH110/003/Teams/Solothurn/homework11

Course:MATH110/Archive/2010-2011/003/Teams/Solothurn/Homework 11

2011-01-19T07:09:35Z

ShannonLee: Created page with "'''Write a linear model to predict the cost of producing flags of your team's Canton under the assumptions that the marginal cost is $7 per unit and that at the current productio..."

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

2011-01-16T03:01:59Z

ShannonLee: /* Contact */

Course:MATH110/Archive/2010-2011/003/Groups/Group 11/Basic Skills: Trigonometric Functions

2010-12-02T02:33:50Z

ShannonLee: /* Solving Trig Functions */

Course:MATH110/Archive/2010-2011/003/Groups/Group 11/Basic Skills: Trigonometric Functions

2010-12-02T02:07:35Z

ShannonLee: /* Basic Skills Project: Trigonometric Functions */

Course:MATH110/Archive/2010-2011/003/Groups/Group 11/Basic Skills: Trigonometric Functions

2010-12-02T02:02:53Z

ShannonLee: /* Right-angled Triangle Definition */

Course:MATH110/Archive/2010-2011/003/Groups/Group 11/Basic Skills: Trigonometric Functions

2010-12-01T17:27:38Z

ShannonLee: /* Solving Trig Functions */

Course:MATH110/Archive/2010-2011/003/Groups/Group 11/Basic Skills: Trigonometric Functions

2010-12-01T07:08:49Z

ShannonLee: /* Graphing Trigonometric Functions */

File:Graphing translations.gif

2010-12-01T06:54:53Z

ShannonLee:

File:Cos.gif

2010-12-01T06:54:36Z

ShannonLee:

File:Sin.gif

2010-12-01T06:50:50Z

ShannonLee:

File:Tan.gif

2010-12-01T06:50:22Z

ShannonLee:

Course:MATH110/Archive/2010-2011/003/Groups/Group 11/Basic Skills: Trigonometric Functions

2010-12-01T06:40:31Z

ShannonLee: /* Solving Trig Functions */

Course:MATH110/Archive/2010-2011/003/Groups/Group 11

2010-12-01T06:36:12Z

ShannonLee: /* Basic Skills Project: Trigonometric Functions */

{{Infobox MATH110 Groups
| group number = 11
| member 1 = [[User:AbdulSherifAbdulrahaman|Abdul-Sherif Abdul-Rahaman]]
| member 2 = [[User:ShannonLee|Shannon Lee]]
| member 3 = [[User:AmirKushairyMohamadNazri|Amir Mohamad Nazri]]
| member 4 = [[User:MattVetter|Matt Vetter]]
| member 5 =
| member 6 =
}}
==http://wiki.ubc.ca/Math_110_Basic_Skills:_Trigonometric_Functions#Solving_Trig_Functions==

==Basic Skills Project: Trigonometric Functions==
[[http://wiki.ubc.ca/Math_110_Basic_Skills:_Trigonometric_Functions|Basic Skills Project: Trigonometric Functions]]

This page will focus on trigonometric function. Group 11 hopes that through providing examples, and step-by-step solution guides we can eliminate the struggles many of us face in regards to solving problems involving trigonometry. Hopefully after reviewing this page you will be able to relate trigonometric functions to the study of everyday occurrences, some of which include: waves, daily temperature averages through the seasons, alternating current, electronic signals, as well as other cyclic phenomena. In the "Just-in-Time" text book, chapter 7, titled "Cyclic Phenomena: The Six Basic Trigonometric Functions," can provide further assistance when reviewing trig functions.

A trigonometric function, also known as a circular function, is a function that relates the lengths of the sides of a right triangle with the angles of the triangle. The basic trig functions are sine, cosine, tangent, cotangent, secant and cosecant.

f(x)=sin x and f(x)=cos x are functions defined for any real number x.
The other basic trigonometric functions we also have to remember are

1)tangent: tan x = sin x/ cos x
2)cotangent: cot x = cos x/sin x
3)secant: 1/cos x
4)cosecant: 1/sin x

==Right-angled Triangle Definition==
If a value of x falls between 0 and π/2, so 0<x<π/2 then the definition of sin(x) and cos(x) is called the right-angled triangle definition. By this definition, sin(x)=opposite/ hypotenuse and cos(x)=adjacent/hypotenuse.
A helpful pneumonic for remembering how to solve the sides of a right triangle is SOH CAH TOA!

[[File:Right_triangle_.png|frame|Caption: Right Triangle.]]

{| class=wikitable style="margin-left:1em"
! style="text-align:left" | '''Function'''
! style="text-align:left" | '''Abbreviation'''
! style="text-align:left" | '''Description'''
! style="text-align:left" | '''Trigonometric Identities using Radians'''
|- style="background-color:#FFFFFF"
| '''Sine'''
| sin
| <math>\frac {\text{opposite}} {\text{hypotenuse}} </math>
| <math>\sin \theta \equiv \cos \left(\frac{\pi}{2} - \theta \right) \equiv \frac{1}{\csc \theta}</math>
|- style="background-color:#FFFFFF"
| '''Cosine'''
| cos
| <math>\frac {\text{adjacent}} {\text{hypotenuse}} </math>
| <math>\cos \theta \equiv \sin \left(\frac{\pi}{2} - \theta \right) \equiv \frac{1}{\sec \theta}\,</math>
|- style="background-color:#FFFFFF"
| '''Tangent'''
| tan
|align=center| <math>\frac {\text{opposite}} {\text{adjacent}} </math>
| <math>\tan \theta \equiv \frac{\sin \theta}{\cos \theta} \equiv \cot \left(\frac{\pi}{2} - \theta \right) \equiv \frac{1}{\cot \theta} </math>
|- style="background-color:#FFFFFF"
| '''Cotangent'''
| cot
|align=center| <math>\frac {\text{adjacent}} {\text{opposite}} </math>
| <math>\cot \theta \equiv \frac{\cos \theta}{\sin \theta} \equiv \tan \left(\frac{\pi}{2} - \theta \right) \equiv \frac{1}{\tan \theta} </math>
|- style="background-color:#FFFFFF"
| '''Secant'''
| sec
| <math>\frac {\text{hypotenuse}} {\text{adjacent}} </math>
| <math>\sec \theta \equiv \csc \left(\frac{\pi}{2} - \theta \right) \equiv\frac{1}{\cos \theta} </math>
|- style="background-color:#FFFFFF"
| '''Cosecant'''
| csc
| <math>\frac {\text{hypotenuse}} {\text{opposite}} </math>
| <math>\csc \theta \equiv \sec \left(\frac{\pi}{2} - \theta \right) \equiv\frac{1}{\sin \theta} </math>
|}{{clearright}}

To Begin learning about trigonometric functions, lets review the unit circle:

[[File:UnitCircle2.gif|center|frame|Caption: The Unit Circle.]]

The x coordinate on the unit circle is cos(x), and the y coordinate is sin(x)
(x,y) = (cos(x), sin(x))

Because cos(x) and sin(x) is a point on the unit circle, its coordinate must satisfy the equation of the circle: x^2+y^2 = 1

Therefore, Sin^2(x)+cos^2(x) = 1

The unit circle has a radius = 1
The circumference of the unit circle = 2π radians

each complete rotation of the unit circle will add 2π or 360°, for instance 2 complete revolutions of the unit circle = 4π or 720°,

On the Unit Circle, x and x + 2π correspond to the same point, this fact means that the graphs of y=sin(x) and y=cos(x) repeat over and over again, and gives us sin(x) = sin(x + 2kπ), where k is any whole number and represents the number of revolutions. From this we can say that sin(x) has a period of 2π. This can also be said about cos(x).

Converting Degrees to Radians:
Angle *(2π/360°)
for example: Convert 27° into radians:
27°*(2π/360°)= (3/40)*2π = 3π/20

Converting Radians to Degrees:
Radian * n(π), where π is equal to 180°
for example: Convert 3π into degrees
3π = 3(180°) = 540°

The following is a helpful video for understanding the Unit Circle, and remembering the points

{{#ev:youtube | cIVpemcoAlY| 400}}

==Solving Trig Functions==
Examples:

1.)Evaluate cos(3π/2)and sin(3π/2):

To determine the sine and cosine values of angles using the unit circle is quite simple. If you refer back to the picture of the unit circle above, you can see that 3π/2 is three quarters of the way around the unit circle and located at the point (0,-1). If you remember that (x,y) is equivalent to (cos(x), sin(x)), then we can determine that:

cos(3π/2) = 0 and sin(3π/2) = -1

2.)Evaluate cos(9π/2):

To solve this problem, we first need to determine that 9π/2 is equivalent to 4π +π/2. A complete revolution of the unit circle is = 2π; therefore, 4π is just equal to two complete revolutions of the unit circle, which does not change the value of your answer. So cos 4π+π/2 is equal to two complete rotations and another quarter rotation, or 90°. If we refer to the unit circle above, we can see that π/2 is equal to (0,1). Because cosine is equal to the x value, we can determine that:

cos(9π/2) = 0

For some extra practice in evaluating some basic angles of sin and cos, refer to the exercises in chapter 7.2 of your Just-in-Time text!

Finding Special Angles

examples:

1.) Evaluate Sin(π/4)

Solution: π/4 = 45°

[[File:Sine,cos_Unit_circle.GIF‎|center|frame|Caption: Sin and Cos of 45°.]]

The coordinates of the point in the question are (cos(x) and sin(x), x=π/4. The equation of the circle is sin^2(x)+cos^2(x)=1,

therefore, sin^2(π/4)+cos^2(π/4)= 1

so if we let s represent the unknown side values of the triangle (cos(x),

Solving Trigonometric Functions:
{{#ev:youtube | IE0FxGegdMg| 400}}

'''Sine graphs'''

If sin(x) is plotted on a set of "x" and "y" axis, the graph of sin(x) would be drawn as shown below:

http://i56.tinypic.com/2mmxnq0.jpg

The demonstration above is to show how the values from the unit circle can be plotted onto a set of "x" and "y" axis. The graph is not limited, it depends on a given scope. In other words, as long as the range of the sin function is not given, sin(x) graph will repeat as a wave until infinite. The sin(x) graph below shows for x values from 0 to 3π(540°).In order to calculate the points where slope = 0, we must make equate the derivate of sin(x) = 0.

http://i54.tinypic.com/okbkpd.jpg

'''Cosine graphs'''

Cosine have the exact same characteristics to that of a sine graph. The only difference between the two is that cosine function shifts to the right by a factor of 90° or π/2 in radians:

y = cos(x)

http://i55.tinypic.com/34hh2t2.jpg

As mentioned previously, cosine functions have the exact same behavior as the sine graph, and the only difference between them is that cosine functions shifts to the right by a factor of 90° or π/2 in radians. Mathematically speaking, if we were to expressed cos(x) in terms of sin(x), it would be written as; cos(x) = (sin(x+π/2)). The comparison between a sine and a cosine function is demonstrated as below;

http://i52.tinypic.com/s5920j.jpg

'''Tangent Graphs'''

http://i53.tinypic.com/2vd058x.jpg

Graphing Trigonometric Functions
{{#ev:youtube | R0RM6FeiD8k | 400}}
{{#ev:youtube | 80c_F0-7ZxE | 400}}

==Basic Skills Project-Group Contribution==
http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_11/Basic_Skills

== Homeworks ==

HOMEWORK 4[[http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_11/Homework_4]]

[[Course:MATH110/003/Groups/Group 11/Information on Basic Skills|Information on Basic Skills]]

== Homework 3==

==Question 1==
Well in both instance the travel time were exactly the same. One was given in hour and minute and the second was given in minutes. We know that an hour is compose of 60 minutes so at 1hour and 20 minute we have 80 total minutes. The Second trip back was 80 minutes so both trips took equal time.

==Question 2==
The the reason the police did not stop the woman was because she was walking on the street, she was not driving. If the officer was to not stop her as she drove on to oncoming traffic he would not be doing his job. If the officer did not try stopping her it is obvious she was not breaking any laws.

==Question 3==
Well in the first two boxes you could pick a fruit up and figured it is either apples or oranges, it is a homogenous mixture in the first two boxes, but the third is a heterogenous mixture. I would personally grab into each box three times so i figure it is all the same or a mixture of both

==Question 4==
He is a brother to the fiddler because he too is blind.

==Question 5==
The moving quarter has made one full revolution around the fixed coin.

== Question 6==
If we let the 3 kinds of apples be X, Y and Z respectively, we know from the question that all these types of apples will sum up to a value; let that value be a.
Our goal is to have at least two of the same type of apple, which leads to our question-how many apples do we need to draw in order to get a matching pair of apples?

In our attempt to find a solution we found the probability of drawing any type of apple out of the basket, concluding that it is necessary to draw 3 times minimally to have any chance of a matching pair being taken out.

On the 1st draw:
1/a chances of getting any of the types of apples.

1 apple drawn from a basket filled with a apples.

On the 2nd draw:
1/(a-1) chances of getting any of the types of apples, however since you have drawn one out already, one of the 3 types is short an apple, reducing the total by 1.

Another apple drawn from the basket filled with (a-1) apples.

On the 3rd draw:
1/(a-2) chances of getting any type of the apples, with one group short of an apple, possible the same group that you drew an apple from in the 1st and/or 2nd group, resulting in a matching pair of apples.

A third apple drawn from the basket filled with (a-2) apples.

The necessary information that are lacking in this question are the total number of apples in the basket and the total number of each type of apple. Polya's method was only useful in sorting out our thinking. In order to apply probability and combinatoric concepts and further Step 3 of Polya's method, more information is required.--[[User:BernadetteHii|BernadetteHii]] 17:55, 16 October 2010 (UTC)
== Question 7==
We know that there are 80 socks in a drawer, 40 blue socks and 40 brown socks.

i.) Our goal in the first half of the question is to draw out a matching pair, how many socks need to be taken from the drawer to ensure a matching pair is found?

Similarly to the preceding question, we use probability of drawing any of the socks out of the drawer, concluding that it is necessary to draw 3 socks, minimally, to have a pair of matching socks.

On the 1st draw:
The probability drawing a blue or brown sock is 1/80.

1/40 chance of drawing a blue sock
1/40 chance of drawing a brown sock

On the 2nd draw:
1/79 chance of drawing a blue or brown sock, however the amount of one colour of socks is not reduced by 1.

If we take the two scenarios:
a.) 1st draw=blue sock, you now have a 1/39 chance of drawing a blue sock and a 1/40 chance of drawing a brown one on the 2nd draw.
b.) 1st draw=brown sock, you now have a 1/39 chance of drawing a brown sock and a 1/40 chance of drawing a blue one on the 2nd draw.

On the 3rd draw:
1/78 chance of drawing a blue or brown sock, however the amount of one of the colours has been reduced again by 1 sock.

a.) 2nd draw=blue sock, you now have a 1/38 chance of drawing a blue sock on the 3rd draw, assuming the 1st draw also was a blue sock
b.)2nd draw=brown sock, you now have a 1/38 chance of drawing a bluce sock on the 3rd draw assuming the 1st draw was also a brown sock.

As more socks are taken from the drawer, one by one, the amount of one colour of sock reduces. Since the picking is at random one colour will not necessarily decrease faster than the other, assuming that they are mixed in the drawer not nicely laid out. Therefore, to ensure a matching pair is picked, 3 socks minimally need to be taken from the drawer.

ii.) To draw an unmatched pair of socks from this drawer, simply draw two socks. Picking is at random and assuming the socks are mixed up and not laid out within the drawer, drawing two socks is sufficient enough to two different socks are taken out.

1st draw:
1/40 chance of getting a blue sock
1/40 chance of getting a brown sock

On the first draw there is an equal chance of drawing either a blue or brown sock.

2nd draw:
1/40 chance of getting a brown sock
1/40 chance of getting a blue sock

On the second draw there is also an equal chance of drawing either a blue or brown sock, assuming that the sock draw the first time, is the opposite colour of the sock drawn the second time.--[[User:BernadetteHii|BernadetteHii]] 17:55, 16 October 2010 (UTC)
== Question 8==
If we assume Reuben's birthday is December 31, then on December 31 he turned 21 years old.
If it is January 1, 2 days ago the would have been December 30. On that day he was still 20 years old. By the end of this year he will turn 22 years old. One year from the day January 1 he will be 23.

Polya's method seems to help organize the wording in this question making it easier to interpret the information the question gives us.--[[User:BernadetteHii|BernadetteHii]] 17:55, 16 October 2010 (UTC)
== Question 9==
From the question we know that the distance between each rung is 1 foot. If 10 rungs are showing we can assume that these 10 rungs are hanging off the side of the boat. Also, we assume that these 10 rungs are not submerged in water yet. If the tide rises 5 feet, this is equivalent to the distance between the bottom rung and the 5th rung of the ladder, therefore 5 rungs are now submerged in water, leaving the 5 upper rungs out of the water.

Similar to question 8, Polya's method is helpful to organize the wording in this question and to deduce from the question an appropriate scenario, in this case, 10 rungs of the ladder on the side of a boat, not covered by water.--[[User:BernadetteHii|BernadetteHii]] 17:55, 16 October 2010 (UTC)
== Question 10==
Polya's method helps start the question off by helping us interpret what the question is telling us.
i.) one-half the population of women can be written as
(1/2) (1/2)=(1/4)

If we know half the population are women, then it follows the other half of the population are men.
One-half the population also equals
(1/2)(1/2)=(1/4)

Knowing that half the population are chocolate eaters, we an take (1/4) of male chocolate eaters and (1/4) of female chocolate eaters, adding these together we get

(1/4)+(1/4)=(1/2), which is consistent with the information given in the question.

ii.) One-half of all men is
(1/2)(1/2)=(1/4)

therefore it follows that 1/4 the population of men are chocolate eaters.--[[User:BernadetteHii|BernadetteHii]] 17:55, 16 October 2010 (UTC)

== Question 11 ==
A woman, her older brother, her son, and her daughter are chess players. The worst player's twin, who is one of the four players, and the best player are of opposite sex. The worst player and best have the same age. If this is possible, who is the worst player?

The first step in solving this problem is understanding what the question is asking us and determining what information we are given. The question is asking us to solve out of four possible players, which on is the worst.

The information we are given is:
-The twin of the worst player is the opposite sex of the best player
-the worst and best player have the same age

Through this information we can determine that it is not possible to solve for the worst player, if one of the twins is the worst player and the best player is the same age and opposite gender of the other twin, there is no way of determining which twin is the worst player; also, no other player fits the description of the best player.

== Question 12 ==
A Manhattan fellow had a girlfriend in the Bronx and a girlfriend in Brooklyn. He decided which girlfriend to visit by arriving randomly at he train station and taking the first of the Bronx or Brooklyn trains that arrived. The trains to Brooklyn and the Bronx each arrived regularly every 10 minutes. Not long after he began his scheme the man's Bronx girlfriend left him because he rarely visited. Give a (logical) explanation.

This question is asking us to conclude why it was that the man rarely visited his Bronx girlfriend in order to determine why he was broken up with.

The man's daily routine probably meant he arrived at the train station at a similar point every time, the train's also run on a consistent schedule. Therefore, it is likely that the Brooklyn train schedule matched the Manhattan fellow's typical schedule more often; consequently, he rarely visited his Bronx girlfriend.

== Question 13 ==
If a clock takes 5 seconds to strike 5:00 (with 5 equally spaced chimes), how long does it take to strike 10:00 (with 10 equally spaced chimes)?

x= #chimes/second

x= 5 seconds/ 5chimes = 1 chime/ second

from this we can conclude that the clock chimes at a rate of 1 chime per second; therefore, if we want to solve for the time it would take to strike 10:00 (with 10 equally spaced chimes), we divide number of chimes by the rate of chimes:

10chimes/ 1chime/s = 10 seconds

== Question 14 ==
One day in the maternity ward, the name tags for four girl babies become mixed up.
i)How many different ways could two of the babies be tagged correctly and two incorrectly?

The question is asking us to find the number of ways two correct and two incorrect names can be arrangged. There are four names for the four babies. Let the correct order be ABCD. If there are two correct name orders and two incorrect these are our possible arrangements: ABDC, ADCB, ACBD, BACD, DBCA, CBAD. There are 6 possible arrangements given that 2 of the names are correct and the other two are not.

ii) In how many ways could three of the babies be tagged correctly and one baby be tagged incorrectly?

This question wants to know how many, if at all, ways can three of the baby names be correct and one incorrect. If the correct order of the names is ABCD, and ABC are correct the only possible name assignment for the fourth baby is D; therefore there are no possible arrangements for three correctly ordered names and one incorrect.

== Question 15 ==
Alex says to you, "I'll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck." Should you accept his bet?

This question is asking if Alex's assumption that the number of red cards in the first half of a split deck as there are black cards in the second half of a split deck is correct.

To solve, let
R= red cards
B= black cards

R(in first half deck) = B(in second half deck)

52 cards = R+B, half a deck = 26 cards

if,
1st half of deck has 16R + 10B, that leaves 16B +10R for the 2nd half of the deck

or,
1st half of deck has 2R+24B, that leaves 2B +24R for the 2nd half of the deck

No, you should not accept his bet because you would owe him a lot of money. Alex's statement that the number of red cards in the first half of the deck is equal to the number of black cards in the second half of the deck is true. Half a deck = 26 cards, so if you have 14 red cards in your first half of the deck that means the remaining 12 cards are black; therefore, in the second half of the deck, there are 14 black cards and 12 red cards remaining. So the number of red in the first half of the deck is equal to the number of black in the second half of the deck.

== Qustion 16 ==
Suppose that each daughter in your family has the same number of brothers as she has sisters, and each son in your family has twice as many sisters as he has brothers. How many sons and daughters are in the family?

Assume that total kids: 7
Boys: 3
Girls: 4
From this each son would have twice as many sisters as he has brothers.

== Question 17 ==
The zero point on a bathroom scale is set incorrectly, but otherwise the scale is accurate. It shows 60 kg when Dan stands on the scale, 50 kg when Sarah stands on the scale, but 105 kg when Dan and Sarah both stand on the scale. Does the scale read too high or too low? Explain.

Theoretically, if both Dan and Sarah stand on the scale, it should have read “110” kg. Since it reads 105kg instead of 110 kg, it shows that the scale reads too high since it should reduce about 2.5 kg for each individual for it to be 105kg. Therefore, the actual reading for Dan should be 57.5 kg and as for Sarah’s it should be 47.5 kg. Both of these weights make up to 105kg.

== Question 18 ==
Alice takes one-third of the pennies from a large jar. Then Bret takes one-third of the remaining pennies from the jar. Finally, Carla takes one-third of the remaining pennies from the jar, leaving 40 pennies in the jar. How many pennies were in the jar at the start?

In order to calculate the total pennies in the jar, one has to work backwards. Starting from the time Carla takes 1/3 of the pennies from the jar. In order to find the previous amount of pennies in the jar before Carla took 1/3 of it, the amount left must be divided by a factor of 1/3. This can be written below as “x” represents the amount of pennies before Carla took 1/3 of it;
1/3 of pennies = 40
Pennies = 120
This number would then be divided by a factor of 1/3 to calculate the amount before Bret took 1/3 of it;
1/3 of pennies = 120
Pennies = 360
Then 360 would be divided again by the same factor of 1/3 to find the original amount of pennies in the jar;
1/3 pennies = 360
Pennies = 1080
In conclusion, there were 1080 pennies originally in the jar.

== Question 19 ==
One morning each member of Angela's family drank an eight-ounce cup of coffee and milk, with the (nonzero) amounts of coffee and milk varying from cup to cup. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. What is the least number of people in the family?

If Angela consumes a quarter of the total amount of milk, that means there could only be a minimum of 4 number of people in the family. Since (1/4) x 4 makes 1 carton of milk, it is reasonable to conclude that the least number of people in Angela’s family would be 4 members.

== Question 20 ==
Of two clocks next to each other, one runs 5 min per hour fast and the other runs 5 min per hour slow. At midnight the clocks show the same time. At what time are they are one hour apart?

At every hour, the first clock runs about 5 minutes faster, and the second clock runs 5 minutes per hour slow. Therefore, for each consecutive hour, the first clock would gain 5 minutes while the second clock would lose 5 minutes. As a result, the total of minutes gained by the first clock for every hour is 10 minutes compared to the second clock. Because there is 60 minutes in an hour, and the first clock gains a lead of 10 minutes every hour, the two clocks would be an hour apart once the first clock strikes 18:00.

== Question 21 ==
Sven was in ninth place among seventeen runners in the marathon. It states that Sven was in the exact middle of placement in the marathon, which automatically narrowed the number of racers down to an odd number. This is because any single odd number is surrounded by the same amount of even numbers, so in this case of 9th, there are 8 racers before him, and 8 racers after him, leading the number to seventeen racers. The insert about Dan finishing in 10th place ruled out 19 as another potential number of racers going with the same logic as before. And Lars finishing in 16th place made it obvious there was more than 16 racers. Now there could not have been more racers as with Sven finishing ahead of Dan in 10th place, and being in the exact middle, there could not possibly have been more than 17 racers.

== Question 22 ==
Given that the vacation can be no shorter than 13 days, as it states that the rain cannot endure for the entire day. In the morning if it rains, it is followed by a sunny afternoon, and if it rains in the afternoon it is sunny in the morning. There are two distinct time frames, morning and afternoon, per day. Since there are only 2 slots per day, it makes sense to add the numbers of each type of day; 13 periods of rain, 12 sunny afternoons, and 11 bright mornings. After the addition you come out to the number of 36, and since there are 2 time slots per day, 36/2 is equal to 18 days of vacation.

== Question 23 ==
Yes you would be able to find the ages of the children. It would be necessary however to know the date at least. The number set for dates alone is much to factor in, from the 1 - 31 at most. However definitely not as high as 31 since the product of the three children's ages is 36. The date narrows it down by quite a bit. It would be simple and very possible to come up with a list of viable ages through simple calculations, but to narrow that list down the date would help. For example; (9,2,2), (2,3,6), (18,2,1), (4,3,3), (6,6,1), (12,3,1) etc ...

== Question 24 ==
Considering the second candle took 3 hours to burn out, half the time it took the first candle to burn out (6 hours). I made each candle into a standard and equal fraction of 6/6. Then I began by reducing the first candle to 5/6. Since the first candle lasted twice as long as the second candle, it is safe to assume that the second candle would go down 2 fractional points compared to the first candle, making it 4/6. You do this for one more interval and you find that when the first candle is equal to 4/6 the second candle is down to 2/6, half the value of the first. So doing some simple adjustments, the first candle lasted for 6 hours, and it is double the height of the second candle at fractions 4/6, therefore it took 2 hours before the first candle was twice as big as the second candle.

== Question 25 ==
For this question, I started out by figuring out how long each candle had lasted. The L + 1 candle lasted for 6 hours, whereas the L candle lasted for 4 hours. It could then be concluded that the L + 1 candle lasted 3/2 longer than the L candle. And since (L + 1)/L = 6/4, by cross-multiplying this equation, you see that 6L = 4(L + 1), which in turn becomes 6L = 4L + 4. By subtracting 4L from both sides to get rid of it on the right side, the equation then becomes 2L = 4. With some simple division, you discover that L = 2

Course:MATH110/Archive/2010-2011/003/Groups/Group 11/Basic Skills: Trigonometric Functions

2010-12-01T06:32:37Z

ShannonLee: /* Solving Trig Functions */

Course:MATH110/Archive/2010-2011/003/Groups/Group 11

2010-12-01T06:23:43Z

ShannonLee: /* Solving Trig Functions */

{{Infobox MATH110 Groups
| group number = 11
| member 1 = [[User:AbdulSherifAbdulrahaman|Abdul-Sherif Abdul-Rahaman]]
| member 2 = [[User:ShannonLee|Shannon Lee]]
| member 3 = [[User:AmirKushairyMohamadNazri|Amir Mohamad Nazri]]
| member 4 = [[User:MattVetter|Matt Vetter]]
| member 5 =
| member 6 =
}}
==Basic Skills Project: Trigonometric Functions==
This page will focus on trigonometric function. Group 11 hopes that through providing examples, and step-by-step solution guides we can eliminate the struggles many of us face in regards to solving problems involving trigonometry. Hopefully after reviewing this page you will be able to relate trigonometric functions to the study of everyday occurrences, some of which include: waves, daily temperature averages through the seasons, alternating current, electronic signals, as well as other cyclic phenomena. In the "Just-in-Time" text book, chapter 7, titled "Cyclic Phenomena: The Six Basic Trigonometric Functions," can provide further assistance when reviewing trig functions.

A trigonometric function, also known as a circular function, is a function that relates the lengths of the sides of a right triangle with the angles of the triangle. The basic trig functions are sine, cosine, tangent, cotangent, secant and cosecant.

f(x)=sin x and f(x)=cos x are functions defined for any real number x.
The other basic trigonometric functions we also have to remember are

1)tangent: tan x = sin x/ cos x
2)cotangent: cot x = cos x/sin x
3)secant: 1/cos x
4)cosecant: 1/sin x

==Right-angled Triangle Definition==
If a value of x falls between 0 and π/2, so 0<x<π/2 then the definition of sin(x) and cos(x) is called the right-angled triangle definition. By this definition, sin(x)=opposite/ hypotenuse and cos(x)=adjacent/hypotenuse.
A helpful pneumonic for remembering how to solve the sides of a right triangle is SOH CAH TOA!

[[File:Right_triangle_.png|frame|Caption: Right Triangle.]]

{| class=wikitable style="margin-left:1em"
! style="text-align:left" | '''Function'''
! style="text-align:left" | '''Abbreviation'''
! style="text-align:left" | '''Description'''
! style="text-align:left" | '''Trigonometric Identities using Radians'''
|- style="background-color:#FFFFFF"
| '''Sine'''
| sin
| <math>\frac {\text{opposite}} {\text{hypotenuse}} </math>
| <math>\sin \theta \equiv \cos \left(\frac{\pi}{2} - \theta \right) \equiv \frac{1}{\csc \theta}</math>
|- style="background-color:#FFFFFF"
| '''Cosine'''
| cos
| <math>\frac {\text{adjacent}} {\text{hypotenuse}} </math>
| <math>\cos \theta \equiv \sin \left(\frac{\pi}{2} - \theta \right) \equiv \frac{1}{\sec \theta}\,</math>
|- style="background-color:#FFFFFF"
| '''Tangent'''
| tan
|align=center| <math>\frac {\text{opposite}} {\text{adjacent}} </math>
| <math>\tan \theta \equiv \frac{\sin \theta}{\cos \theta} \equiv \cot \left(\frac{\pi}{2} - \theta \right) \equiv \frac{1}{\cot \theta} </math>
|- style="background-color:#FFFFFF"
| '''Cotangent'''
| cot
|align=center| <math>\frac {\text{adjacent}} {\text{opposite}} </math>
| <math>\cot \theta \equiv \frac{\cos \theta}{\sin \theta} \equiv \tan \left(\frac{\pi}{2} - \theta \right) \equiv \frac{1}{\tan \theta} </math>
|- style="background-color:#FFFFFF"
| '''Secant'''
| sec
| <math>\frac {\text{hypotenuse}} {\text{adjacent}} </math>
| <math>\sec \theta \equiv \csc \left(\frac{\pi}{2} - \theta \right) \equiv\frac{1}{\cos \theta} </math>
|- style="background-color:#FFFFFF"
| '''Cosecant'''
| csc
| <math>\frac {\text{hypotenuse}} {\text{opposite}} </math>
| <math>\csc \theta \equiv \sec \left(\frac{\pi}{2} - \theta \right) \equiv\frac{1}{\sin \theta} </math>
|}{{clearright}}

To Begin learning about trigonometric functions, lets review the unit circle:

[[File:UnitCircle2.gif|center|frame|Caption: The Unit Circle.]]

The x coordinate on the unit circle is cos(x), and the y coordinate is sin(x)
(x,y) = (cos(x), sin(x))

Because cos(x) and sin(x) is a point on the unit circle, its coordinate must satisfy the equation of the circle: x^2+y^2 = 1

Therefore, Sin^2(x)+cos^2(x) = 1

The unit circle has a radius = 1
The circumference of the unit circle = 2π radians

each complete rotation of the unit circle will add 2π or 360°, for instance 2 complete revolutions of the unit circle = 4π or 720°,

On the Unit Circle, x and x + 2π correspond to the same point, this fact means that the graphs of y=sin(x) and y=cos(x) repeat over and over again, and gives us sin(x) = sin(x + 2kπ), where k is any whole number and represents the number of revolutions. From this we can say that sin(x) has a period of 2π. This can also be said about cos(x).

Converting Degrees to Radians:
Angle *(2π/360°)
for example: Convert 27° into radians:
27°*(2π/360°)= (3/40)*2π = 3π/20

Converting Radians to Degrees:
Radian * n(π), where π is equal to 180°
for example: Convert 3π into degrees
3π = 3(180°) = 540°

The following is a helpful video for understanding the Unit Circle, and remembering the points

{{#ev:youtube | cIVpemcoAlY| 400}}

==Solving Trig Functions==
Examples:

1.)Evaluate cos(3π/2)and sin(3π/2):

To determine the sine and cosine values of angles using the unit circle is quite simple. If you refer back to the picture of the unit circle above, you can see that 3π/2 is three quarters of the way around the unit circle and located at the point (0,-1). If you remember that (x,y) is equivalent to (cos(x), sin(x)), then we can determine that:

cos(3π/2) = 0 and sin(3π/2) = -1

2.)Evaluate cos(9π/2):

To solve this problem, we first need to determine that 9π/2 is equivalent to 4π +π/2. A complete revolution of the unit circle is = 2π; therefore, 4π is just equal to two complete revolutions of the unit circle, which does not change the value of your answer. So cos 4π+π/2 is equal to two complete rotations and another quarter rotation, or 90°. If we refer to the unit circle above, we can see that π/2 is equal to (0,1). Because cosine is equal to the x value, we can determine that:

cos(9π/2) = 0

For some extra practice in evaluating some basic angles of sin and cos, refer to the exercises in chapter 7.2 of your Just-in-Time text!

Finding Special Angles

examples:

1.) Evaluate Sin(π/4)

Solution: π/4 = 45°

[[File:Sine,cos_Unit_circle.GIF‎|center|frame|Caption: Sin and Cos of 45°.]]

The coordinates of the point in the question are (cos(x) and sin(x), x=π/4. The equation of the circle is sin^2(x)+cos^2(x)=1,

therefore, sin^2(π/4)+cos^2(π/4)= 1

so if we let s represent the unknown side values of the triangle (cos(x),

Solving Trigonometric Functions:
{{#ev:youtube | IE0FxGegdMg| 400}}

'''Sine graphs'''

If sin(x) is plotted on a set of "x" and "y" axis, the graph of sin(x) would be drawn as shown below:

http://i56.tinypic.com/2mmxnq0.jpg

The demonstration above is to show how the values from the unit circle can be plotted onto a set of "x" and "y" axis. The graph is not limited, it depends on a given scope. In other words, as long as the range of the sin function is not given, sin(x) graph will repeat as a wave until infinite. The sin(x) graph below shows for x values from 0 to 3π(540°).In order to calculate the points where slope = 0, we must make equate the derivate of sin(x) = 0.

http://i54.tinypic.com/okbkpd.jpg

'''Cosine graphs'''

Cosine have the exact same characteristics to that of a sine graph. The only difference between the two is that cosine function shifts to the right by a factor of 90° or π/2 in radians:

y = cos(x)

http://i55.tinypic.com/34hh2t2.jpg

As mentioned previously, cosine functions have the exact same behavior as the sine graph, and the only difference between them is that cosine functions shifts to the right by a factor of 90° or π/2 in radians. Mathematically speaking, if we were to expressed cos(x) in terms of sin(x), it would be written as; cos(x) = (sin(x+π/2)). The comparison between a sine and a cosine function is demonstrated as below;

http://i52.tinypic.com/s5920j.jpg

'''Tangent Graphs'''

http://i53.tinypic.com/2vd058x.jpg

Graphing Trigonometric Functions
{{#ev:youtube | R0RM6FeiD8k | 400}}
{{#ev:youtube | 80c_F0-7ZxE | 400}}

==Basic Skills Project-Group Contribution==
http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_11/Basic_Skills

== Homeworks ==

HOMEWORK 4[[http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_11/Homework_4]]

[[Course:MATH110/003/Groups/Group 11/Information on Basic Skills|Information on Basic Skills]]

== Homework 3==

==Question 1==
Well in both instance the travel time were exactly the same. One was given in hour and minute and the second was given in minutes. We know that an hour is compose of 60 minutes so at 1hour and 20 minute we have 80 total minutes. The Second trip back was 80 minutes so both trips took equal time.

==Question 2==
The the reason the police did not stop the woman was because she was walking on the street, she was not driving. If the officer was to not stop her as she drove on to oncoming traffic he would not be doing his job. If the officer did not try stopping her it is obvious she was not breaking any laws.

==Question 3==
Well in the first two boxes you could pick a fruit up and figured it is either apples or oranges, it is a homogenous mixture in the first two boxes, but the third is a heterogenous mixture. I would personally grab into each box three times so i figure it is all the same or a mixture of both

==Question 4==
He is a brother to the fiddler because he too is blind.

==Question 5==
The moving quarter has made one full revolution around the fixed coin.

== Question 6==
If we let the 3 kinds of apples be X, Y and Z respectively, we know from the question that all these types of apples will sum up to a value; let that value be a.
Our goal is to have at least two of the same type of apple, which leads to our question-how many apples do we need to draw in order to get a matching pair of apples?

In our attempt to find a solution we found the probability of drawing any type of apple out of the basket, concluding that it is necessary to draw 3 times minimally to have any chance of a matching pair being taken out.

On the 1st draw:
1/a chances of getting any of the types of apples.

1 apple drawn from a basket filled with a apples.

On the 2nd draw:
1/(a-1) chances of getting any of the types of apples, however since you have drawn one out already, one of the 3 types is short an apple, reducing the total by 1.

Another apple drawn from the basket filled with (a-1) apples.

On the 3rd draw:
1/(a-2) chances of getting any type of the apples, with one group short of an apple, possible the same group that you drew an apple from in the 1st and/or 2nd group, resulting in a matching pair of apples.

A third apple drawn from the basket filled with (a-2) apples.

The necessary information that are lacking in this question are the total number of apples in the basket and the total number of each type of apple. Polya's method was only useful in sorting out our thinking. In order to apply probability and combinatoric concepts and further Step 3 of Polya's method, more information is required.--[[User:BernadetteHii|BernadetteHii]] 17:55, 16 October 2010 (UTC)
== Question 7==
We know that there are 80 socks in a drawer, 40 blue socks and 40 brown socks.

i.) Our goal in the first half of the question is to draw out a matching pair, how many socks need to be taken from the drawer to ensure a matching pair is found?

Similarly to the preceding question, we use probability of drawing any of the socks out of the drawer, concluding that it is necessary to draw 3 socks, minimally, to have a pair of matching socks.

On the 1st draw:
The probability drawing a blue or brown sock is 1/80.

1/40 chance of drawing a blue sock
1/40 chance of drawing a brown sock

On the 2nd draw:
1/79 chance of drawing a blue or brown sock, however the amount of one colour of socks is not reduced by 1.

If we take the two scenarios:
a.) 1st draw=blue sock, you now have a 1/39 chance of drawing a blue sock and a 1/40 chance of drawing a brown one on the 2nd draw.
b.) 1st draw=brown sock, you now have a 1/39 chance of drawing a brown sock and a 1/40 chance of drawing a blue one on the 2nd draw.

On the 3rd draw:
1/78 chance of drawing a blue or brown sock, however the amount of one of the colours has been reduced again by 1 sock.

a.) 2nd draw=blue sock, you now have a 1/38 chance of drawing a blue sock on the 3rd draw, assuming the 1st draw also was a blue sock
b.)2nd draw=brown sock, you now have a 1/38 chance of drawing a bluce sock on the 3rd draw assuming the 1st draw was also a brown sock.

As more socks are taken from the drawer, one by one, the amount of one colour of sock reduces. Since the picking is at random one colour will not necessarily decrease faster than the other, assuming that they are mixed in the drawer not nicely laid out. Therefore, to ensure a matching pair is picked, 3 socks minimally need to be taken from the drawer.

ii.) To draw an unmatched pair of socks from this drawer, simply draw two socks. Picking is at random and assuming the socks are mixed up and not laid out within the drawer, drawing two socks is sufficient enough to two different socks are taken out.

1st draw:
1/40 chance of getting a blue sock
1/40 chance of getting a brown sock

On the first draw there is an equal chance of drawing either a blue or brown sock.

2nd draw:
1/40 chance of getting a brown sock
1/40 chance of getting a blue sock

On the second draw there is also an equal chance of drawing either a blue or brown sock, assuming that the sock draw the first time, is the opposite colour of the sock drawn the second time.--[[User:BernadetteHii|BernadetteHii]] 17:55, 16 October 2010 (UTC)
== Question 8==
If we assume Reuben's birthday is December 31, then on December 31 he turned 21 years old.
If it is January 1, 2 days ago the would have been December 30. On that day he was still 20 years old. By the end of this year he will turn 22 years old. One year from the day January 1 he will be 23.

Polya's method seems to help organize the wording in this question making it easier to interpret the information the question gives us.--[[User:BernadetteHii|BernadetteHii]] 17:55, 16 October 2010 (UTC)
== Question 9==
From the question we know that the distance between each rung is 1 foot. If 10 rungs are showing we can assume that these 10 rungs are hanging off the side of the boat. Also, we assume that these 10 rungs are not submerged in water yet. If the tide rises 5 feet, this is equivalent to the distance between the bottom rung and the 5th rung of the ladder, therefore 5 rungs are now submerged in water, leaving the 5 upper rungs out of the water.

Similar to question 8, Polya's method is helpful to organize the wording in this question and to deduce from the question an appropriate scenario, in this case, 10 rungs of the ladder on the side of a boat, not covered by water.--[[User:BernadetteHii|BernadetteHii]] 17:55, 16 October 2010 (UTC)
== Question 10==
Polya's method helps start the question off by helping us interpret what the question is telling us.
i.) one-half the population of women can be written as
(1/2) (1/2)=(1/4)

If we know half the population are women, then it follows the other half of the population are men.
One-half the population also equals
(1/2)(1/2)=(1/4)

Knowing that half the population are chocolate eaters, we an take (1/4) of male chocolate eaters and (1/4) of female chocolate eaters, adding these together we get

(1/4)+(1/4)=(1/2), which is consistent with the information given in the question.

ii.) One-half of all men is
(1/2)(1/2)=(1/4)

therefore it follows that 1/4 the population of men are chocolate eaters.--[[User:BernadetteHii|BernadetteHii]] 17:55, 16 October 2010 (UTC)

== Question 11 ==
A woman, her older brother, her son, and her daughter are chess players. The worst player's twin, who is one of the four players, and the best player are of opposite sex. The worst player and best have the same age. If this is possible, who is the worst player?

The first step in solving this problem is understanding what the question is asking us and determining what information we are given. The question is asking us to solve out of four possible players, which on is the worst.

The information we are given is:
-The twin of the worst player is the opposite sex of the best player
-the worst and best player have the same age

Through this information we can determine that it is not possible to solve for the worst player, if one of the twins is the worst player and the best player is the same age and opposite gender of the other twin, there is no way of determining which twin is the worst player; also, no other player fits the description of the best player.

== Question 12 ==
A Manhattan fellow had a girlfriend in the Bronx and a girlfriend in Brooklyn. He decided which girlfriend to visit by arriving randomly at he train station and taking the first of the Bronx or Brooklyn trains that arrived. The trains to Brooklyn and the Bronx each arrived regularly every 10 minutes. Not long after he began his scheme the man's Bronx girlfriend left him because he rarely visited. Give a (logical) explanation.

This question is asking us to conclude why it was that the man rarely visited his Bronx girlfriend in order to determine why he was broken up with.

The man's daily routine probably meant he arrived at the train station at a similar point every time, the train's also run on a consistent schedule. Therefore, it is likely that the Brooklyn train schedule matched the Manhattan fellow's typical schedule more often; consequently, he rarely visited his Bronx girlfriend.

== Question 13 ==
If a clock takes 5 seconds to strike 5:00 (with 5 equally spaced chimes), how long does it take to strike 10:00 (with 10 equally spaced chimes)?

x= #chimes/second

x= 5 seconds/ 5chimes = 1 chime/ second

from this we can conclude that the clock chimes at a rate of 1 chime per second; therefore, if we want to solve for the time it would take to strike 10:00 (with 10 equally spaced chimes), we divide number of chimes by the rate of chimes:

10chimes/ 1chime/s = 10 seconds

== Question 14 ==
One day in the maternity ward, the name tags for four girl babies become mixed up.
i)How many different ways could two of the babies be tagged correctly and two incorrectly?

The question is asking us to find the number of ways two correct and two incorrect names can be arrangged. There are four names for the four babies. Let the correct order be ABCD. If there are two correct name orders and two incorrect these are our possible arrangements: ABDC, ADCB, ACBD, BACD, DBCA, CBAD. There are 6 possible arrangements given that 2 of the names are correct and the other two are not.

ii) In how many ways could three of the babies be tagged correctly and one baby be tagged incorrectly?

This question wants to know how many, if at all, ways can three of the baby names be correct and one incorrect. If the correct order of the names is ABCD, and ABC are correct the only possible name assignment for the fourth baby is D; therefore there are no possible arrangements for three correctly ordered names and one incorrect.

== Question 15 ==
Alex says to you, "I'll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck." Should you accept his bet?

This question is asking if Alex's assumption that the number of red cards in the first half of a split deck as there are black cards in the second half of a split deck is correct.

To solve, let
R= red cards
B= black cards

R(in first half deck) = B(in second half deck)

52 cards = R+B, half a deck = 26 cards

if,
1st half of deck has 16R + 10B, that leaves 16B +10R for the 2nd half of the deck

or,
1st half of deck has 2R+24B, that leaves 2B +24R for the 2nd half of the deck

No, you should not accept his bet because you would owe him a lot of money. Alex's statement that the number of red cards in the first half of the deck is equal to the number of black cards in the second half of the deck is true. Half a deck = 26 cards, so if you have 14 red cards in your first half of the deck that means the remaining 12 cards are black; therefore, in the second half of the deck, there are 14 black cards and 12 red cards remaining. So the number of red in the first half of the deck is equal to the number of black in the second half of the deck.

== Qustion 16 ==
Suppose that each daughter in your family has the same number of brothers as she has sisters, and each son in your family has twice as many sisters as he has brothers. How many sons and daughters are in the family?

Assume that total kids: 7
Boys: 3
Girls: 4
From this each son would have twice as many sisters as he has brothers.

== Question 17 ==
The zero point on a bathroom scale is set incorrectly, but otherwise the scale is accurate. It shows 60 kg when Dan stands on the scale, 50 kg when Sarah stands on the scale, but 105 kg when Dan and Sarah both stand on the scale. Does the scale read too high or too low? Explain.

Theoretically, if both Dan and Sarah stand on the scale, it should have read “110” kg. Since it reads 105kg instead of 110 kg, it shows that the scale reads too high since it should reduce about 2.5 kg for each individual for it to be 105kg. Therefore, the actual reading for Dan should be 57.5 kg and as for Sarah’s it should be 47.5 kg. Both of these weights make up to 105kg.

== Question 18 ==
Alice takes one-third of the pennies from a large jar. Then Bret takes one-third of the remaining pennies from the jar. Finally, Carla takes one-third of the remaining pennies from the jar, leaving 40 pennies in the jar. How many pennies were in the jar at the start?

In order to calculate the total pennies in the jar, one has to work backwards. Starting from the time Carla takes 1/3 of the pennies from the jar. In order to find the previous amount of pennies in the jar before Carla took 1/3 of it, the amount left must be divided by a factor of 1/3. This can be written below as “x” represents the amount of pennies before Carla took 1/3 of it;
1/3 of pennies = 40
Pennies = 120
This number would then be divided by a factor of 1/3 to calculate the amount before Bret took 1/3 of it;
1/3 of pennies = 120
Pennies = 360
Then 360 would be divided again by the same factor of 1/3 to find the original amount of pennies in the jar;
1/3 pennies = 360
Pennies = 1080
In conclusion, there were 1080 pennies originally in the jar.

== Question 19 ==
One morning each member of Angela's family drank an eight-ounce cup of coffee and milk, with the (nonzero) amounts of coffee and milk varying from cup to cup. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. What is the least number of people in the family?

If Angela consumes a quarter of the total amount of milk, that means there could only be a minimum of 4 number of people in the family. Since (1/4) x 4 makes 1 carton of milk, it is reasonable to conclude that the least number of people in Angela’s family would be 4 members.

== Question 20 ==
Of two clocks next to each other, one runs 5 min per hour fast and the other runs 5 min per hour slow. At midnight the clocks show the same time. At what time are they are one hour apart?

At every hour, the first clock runs about 5 minutes faster, and the second clock runs 5 minutes per hour slow. Therefore, for each consecutive hour, the first clock would gain 5 minutes while the second clock would lose 5 minutes. As a result, the total of minutes gained by the first clock for every hour is 10 minutes compared to the second clock. Because there is 60 minutes in an hour, and the first clock gains a lead of 10 minutes every hour, the two clocks would be an hour apart once the first clock strikes 18:00.

== Question 21 ==
Sven was in ninth place among seventeen runners in the marathon. It states that Sven was in the exact middle of placement in the marathon, which automatically narrowed the number of racers down to an odd number. This is because any single odd number is surrounded by the same amount of even numbers, so in this case of 9th, there are 8 racers before him, and 8 racers after him, leading the number to seventeen racers. The insert about Dan finishing in 10th place ruled out 19 as another potential number of racers going with the same logic as before. And Lars finishing in 16th place made it obvious there was more than 16 racers. Now there could not have been more racers as with Sven finishing ahead of Dan in 10th place, and being in the exact middle, there could not possibly have been more than 17 racers.

== Question 22 ==
Given that the vacation can be no shorter than 13 days, as it states that the rain cannot endure for the entire day. In the morning if it rains, it is followed by a sunny afternoon, and if it rains in the afternoon it is sunny in the morning. There are two distinct time frames, morning and afternoon, per day. Since there are only 2 slots per day, it makes sense to add the numbers of each type of day; 13 periods of rain, 12 sunny afternoons, and 11 bright mornings. After the addition you come out to the number of 36, and since there are 2 time slots per day, 36/2 is equal to 18 days of vacation.

== Question 23 ==
Yes you would be able to find the ages of the children. It would be necessary however to know the date at least. The number set for dates alone is much to factor in, from the 1 - 31 at most. However definitely not as high as 31 since the product of the three children's ages is 36. The date narrows it down by quite a bit. It would be simple and very possible to come up with a list of viable ages through simple calculations, but to narrow that list down the date would help. For example; (9,2,2), (2,3,6), (18,2,1), (4,3,3), (6,6,1), (12,3,1) etc ...

== Question 24 ==
Considering the second candle took 3 hours to burn out, half the time it took the first candle to burn out (6 hours). I made each candle into a standard and equal fraction of 6/6. Then I began by reducing the first candle to 5/6. Since the first candle lasted twice as long as the second candle, it is safe to assume that the second candle would go down 2 fractional points compared to the first candle, making it 4/6. You do this for one more interval and you find that when the first candle is equal to 4/6 the second candle is down to 2/6, half the value of the first. So doing some simple adjustments, the first candle lasted for 6 hours, and it is double the height of the second candle at fractions 4/6, therefore it took 2 hours before the first candle was twice as big as the second candle.

== Question 25 ==
For this question, I started out by figuring out how long each candle had lasted. The L + 1 candle lasted for 6 hours, whereas the L candle lasted for 4 hours. It could then be concluded that the L + 1 candle lasted 3/2 longer than the L candle. And since (L + 1)/L = 6/4, by cross-multiplying this equation, you see that 6L = 4(L + 1), which in turn becomes 6L = 4L + 4. By subtracting 4L from both sides to get rid of it on the right side, the equation then becomes 2L = 4. With some simple division, you discover that L = 2

File:Sine,cos Unit circle.GIF

2010-12-01T06:05:16Z

ShannonLee:

Course:MATH110/Archive/2010-2011/003/Groups/Group 11

2010-12-01T06:01:31Z

ShannonLee: /* Solving Trig Functions */

{{Infobox MATH110 Groups
| group number = 11
| member 1 = [[User:AbdulSherifAbdulrahaman|Abdul-Sherif Abdul-Rahaman]]
| member 2 = [[User:ShannonLee|Shannon Lee]]
| member 3 = [[User:AmirKushairyMohamadNazri|Amir Mohamad Nazri]]
| member 4 = [[User:MattVetter|Matt Vetter]]
| member 5 =
| member 6 =
}}
==Basic Skills Project: Trigonometric Functions==
This page will focus on trigonometric function. Group 11 hopes that through providing examples, and step-by-step solution guides we can eliminate the struggles many of us face in regards to solving problems involving trigonometry. Hopefully after reviewing this page you will be able to relate trigonometric functions to the study of everyday occurrences, some of which include: waves, daily temperature averages through the seasons, alternating current, electronic signals, as well as other cyclic phenomena. In the "Just-in-Time" text book, chapter 7, titled "Cyclic Phenomena: The Six Basic Trigonometric Functions," can provide further assistance when reviewing trig functions.

A trigonometric function, also known as a circular function, is a function that relates the lengths of the sides of a right triangle with the angles of the triangle. The basic trig functions are sine, cosine, tangent, cotangent, secant and cosecant.

f(x)=sin x and f(x)=cos x are functions defined for any real number x.
The other basic trigonometric functions we also have to remember are

1)tangent: tan x = sin x/ cos x
2)cotangent: cot x = cos x/sin x
3)secant: 1/cos x
4)cosecant: 1/sin x

==Right-angled Triangle Definition==
If a value of x falls between 0 and π/2, so 0<x<π/2 then the definition of sin(x) and cos(x) is called the right-angled triangle definition. By this definition, sin(x)=opposite/ hypotenuse and cos(x)=adjacent/hypotenuse.
A helpful pneumonic for remembering how to solve the sides of a right triangle is SOH CAH TOA!

[[File:Right_triangle_.png|frame|Caption: Right Triangle.]]

{| class=wikitable style="margin-left:1em"
! style="text-align:left" | '''Function'''
! style="text-align:left" | '''Abbreviation'''
! style="text-align:left" | '''Description'''
! style="text-align:left" | '''Trigonometric Identities using Radians'''
|- style="background-color:#FFFFFF"
| '''Sine'''
| sin
| <math>\frac {\text{opposite}} {\text{hypotenuse}} </math>
| <math>\sin \theta \equiv \cos \left(\frac{\pi}{2} - \theta \right) \equiv \frac{1}{\csc \theta}</math>
|- style="background-color:#FFFFFF"
| '''Cosine'''
| cos
| <math>\frac {\text{adjacent}} {\text{hypotenuse}} </math>
| <math>\cos \theta \equiv \sin \left(\frac{\pi}{2} - \theta \right) \equiv \frac{1}{\sec \theta}\,</math>
|- style="background-color:#FFFFFF"
| '''Tangent'''
| tan
|align=center| <math>\frac {\text{opposite}} {\text{adjacent}} </math>
| <math>\tan \theta \equiv \frac{\sin \theta}{\cos \theta} \equiv \cot \left(\frac{\pi}{2} - \theta \right) \equiv \frac{1}{\cot \theta} </math>
|- style="background-color:#FFFFFF"
| '''Cotangent'''
| cot
|align=center| <math>\frac {\text{adjacent}} {\text{opposite}} </math>
| <math>\cot \theta \equiv \frac{\cos \theta}{\sin \theta} \equiv \tan \left(\frac{\pi}{2} - \theta \right) \equiv \frac{1}{\tan \theta} </math>
|- style="background-color:#FFFFFF"
| '''Secant'''
| sec
| <math>\frac {\text{hypotenuse}} {\text{adjacent}} </math>
| <math>\sec \theta \equiv \csc \left(\frac{\pi}{2} - \theta \right) \equiv\frac{1}{\cos \theta} </math>
|- style="background-color:#FFFFFF"
| '''Cosecant'''
| csc
| <math>\frac {\text{hypotenuse}} {\text{opposite}} </math>
| <math>\csc \theta \equiv \sec \left(\frac{\pi}{2} - \theta \right) \equiv\frac{1}{\sin \theta} </math>
|}{{clearright}}

To Begin learning about trigonometric functions, lets review the unit circle:

[[File:UnitCircle2.gif|center|frame|Caption: The Unit Circle.]]

The x coordinate on the unit circle is cos(x), and the y coordinate is sin(x)
(x,y) = (cos(x), sin(x))

Because cos(x) and sin(x) is a point on the unit circle, its coordinate must satisfy the equation of the circle: x^2+y^2 = 1

Therefore, Sin^2(x)+cos^2(x) = 1

The unit circle has a radius = 1
The circumference of the unit circle = 2π radians

each complete rotation of the unit circle will add 2π or 360°, for instance 2 complete revolutions of the unit circle = 4π or 720°,

On the Unit Circle, x and x + 2π correspond to the same point, this fact means that the graphs of y=sin(x) and y=cos(x) repeat over and over again, and gives us sin(x) = sin(x + 2kπ), where k is any whole number and represents the number of revolutions. From this we can say that sin(x) has a period of 2π. This can also be said about cos(x).

Converting Degrees to Radians:
Angle *(2π/360°)
for example: Convert 27° into radians:
27°*(2π/360°)= (3/40)*2π = 3π/20

Converting Radians to Degrees:
Radian * n(π), where π is equal to 180°
for example: Convert 3π into degrees
3π = 3(180°) = 540°

The following is a helpful video for understanding the Unit Circle, and remembering the points

{{#ev:youtube | cIVpemcoAlY| 400}}

==Solving Trig Functions==
Examples:

1.)Evaluate cos(3π/2)and sin(3π/2):

To determine the sine and cosine values of angles using the unit circle is quite simple. If you refer back to the picture of the unit circle above, you can see that 3π/2 is three quarters of the way around the unit circle and located at the point (0,-1). If you remember that (x,y) is equivalent to (cos(x), sin(x)), then we can determine that:

cos(3π/2) = 0 and sin(3π/2) = -1

2.)Evaluate cos(9π/2):

To solve this problem, we first need to determine that 9π/2 is equivalent to 4π +π/2. A complete revolution of the unit circle is = 2π; therefore, 4π is just equal to two complete revolutions of the unit circle, which does not change the value of your answer. So cos 4π+π/2 is equal to two complete rotations and another quarter rotation, or 90°. If we refer to the unit circle above, we can see that π/2 is equal to (0,1). Because cosine is equal to the x value, we can determine that:

cos(9π/2) = 0

For some extra practice in evaluating the angles of sin and cos, refer to the exercises in chapter 7.2 of your Just-in-Time text!

Solving Trigonometric Functions:
{{#ev:youtube | IE0FxGegdMg| 400}}

'''Sine graphs'''

If sin(x) is plotted on a set of "x" and "y" axis, the graph of sin(x) would be drawn as shown below:

http://i56.tinypic.com/2mmxnq0.jpg

The demonstration above is to show how the values from the unit circle can be plotted onto a set of "x" and "y" axis. The graph is not limited, it depends on a given scope. In other words, as long as the range of the sin function is not given, sin(x) graph will repeat as a wave until infinite. The sin(x) graph below shows for x values from 0 to 3π(540°).In order to calculate the points where slope = 0, we must make equate the derivate of sin(x) = 0.

http://i54.tinypic.com/okbkpd.jpg

'''Cosine graphs'''

Cosine have the exact same characteristics to that of a sine graph. The only difference between the two is that cosine function shifts to the right by a factor of 90° or π/2 in radians:

y = cos(x)

http://i55.tinypic.com/34hh2t2.jpg

As mentioned previously, cosine functions have the exact same behavior as the sine graph, and the only difference between them is that cosine functions shifts to the right by a factor of 90° or π/2 in radians. Mathematically speaking, if we were to expressed cos(x) in terms of sin(x), it would be written as; cos(x) = (sin(x+π/2)). The comparison between a sine and a cosine function is demonstrated as below;

http://i52.tinypic.com/s5920j.jpg

'''Tangent Graphs'''

http://i53.tinypic.com/2vd058x.jpg

Graphing Trigonometric Functions
{{#ev:youtube | R0RM6FeiD8k | 400}}
{{#ev:youtube | 80c_F0-7ZxE | 400}}

==Basic Skills Project-Group Contribution==
http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_11/Basic_Skills

== Homeworks ==

HOMEWORK 4[[http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_11/Homework_4]]

[[Course:MATH110/003/Groups/Group 11/Information on Basic Skills|Information on Basic Skills]]

== Homework 3==

==Question 1==
Well in both instance the travel time were exactly the same. One was given in hour and minute and the second was given in minutes. We know that an hour is compose of 60 minutes so at 1hour and 20 minute we have 80 total minutes. The Second trip back was 80 minutes so both trips took equal time.

==Question 2==
The the reason the police did not stop the woman was because she was walking on the street, she was not driving. If the officer was to not stop her as she drove on to oncoming traffic he would not be doing his job. If the officer did not try stopping her it is obvious she was not breaking any laws.

==Question 3==
Well in the first two boxes you could pick a fruit up and figured it is either apples or oranges, it is a homogenous mixture in the first two boxes, but the third is a heterogenous mixture. I would personally grab into each box three times so i figure it is all the same or a mixture of both

==Question 4==
He is a brother to the fiddler because he too is blind.

==Question 5==
The moving quarter has made one full revolution around the fixed coin.

== Question 6==
If we let the 3 kinds of apples be X, Y and Z respectively, we know from the question that all these types of apples will sum up to a value; let that value be a.
Our goal is to have at least two of the same type of apple, which leads to our question-how many apples do we need to draw in order to get a matching pair of apples?

In our attempt to find a solution we found the probability of drawing any type of apple out of the basket, concluding that it is necessary to draw 3 times minimally to have any chance of a matching pair being taken out.

On the 1st draw:
1/a chances of getting any of the types of apples.

1 apple drawn from a basket filled with a apples.

On the 2nd draw:
1/(a-1) chances of getting any of the types of apples, however since you have drawn one out already, one of the 3 types is short an apple, reducing the total by 1.

Another apple drawn from the basket filled with (a-1) apples.

On the 3rd draw:
1/(a-2) chances of getting any type of the apples, with one group short of an apple, possible the same group that you drew an apple from in the 1st and/or 2nd group, resulting in a matching pair of apples.

A third apple drawn from the basket filled with (a-2) apples.

The necessary information that are lacking in this question are the total number of apples in the basket and the total number of each type of apple. Polya's method was only useful in sorting out our thinking. In order to apply probability and combinatoric concepts and further Step 3 of Polya's method, more information is required.--[[User:BernadetteHii|BernadetteHii]] 17:55, 16 October 2010 (UTC)
== Question 7==
We know that there are 80 socks in a drawer, 40 blue socks and 40 brown socks.

i.) Our goal in the first half of the question is to draw out a matching pair, how many socks need to be taken from the drawer to ensure a matching pair is found?

Similarly to the preceding question, we use probability of drawing any of the socks out of the drawer, concluding that it is necessary to draw 3 socks, minimally, to have a pair of matching socks.

On the 1st draw:
The probability drawing a blue or brown sock is 1/80.

1/40 chance of drawing a blue sock
1/40 chance of drawing a brown sock

On the 2nd draw:
1/79 chance of drawing a blue or brown sock, however the amount of one colour of socks is not reduced by 1.

If we take the two scenarios:
a.) 1st draw=blue sock, you now have a 1/39 chance of drawing a blue sock and a 1/40 chance of drawing a brown one on the 2nd draw.
b.) 1st draw=brown sock, you now have a 1/39 chance of drawing a brown sock and a 1/40 chance of drawing a blue one on the 2nd draw.

On the 3rd draw:
1/78 chance of drawing a blue or brown sock, however the amount of one of the colours has been reduced again by 1 sock.

a.) 2nd draw=blue sock, you now have a 1/38 chance of drawing a blue sock on the 3rd draw, assuming the 1st draw also was a blue sock
b.)2nd draw=brown sock, you now have a 1/38 chance of drawing a bluce sock on the 3rd draw assuming the 1st draw was also a brown sock.

As more socks are taken from the drawer, one by one, the amount of one colour of sock reduces. Since the picking is at random one colour will not necessarily decrease faster than the other, assuming that they are mixed in the drawer not nicely laid out. Therefore, to ensure a matching pair is picked, 3 socks minimally need to be taken from the drawer.

ii.) To draw an unmatched pair of socks from this drawer, simply draw two socks. Picking is at random and assuming the socks are mixed up and not laid out within the drawer, drawing two socks is sufficient enough to two different socks are taken out.

1st draw:
1/40 chance of getting a blue sock
1/40 chance of getting a brown sock

On the first draw there is an equal chance of drawing either a blue or brown sock.

2nd draw:
1/40 chance of getting a brown sock
1/40 chance of getting a blue sock

On the second draw there is also an equal chance of drawing either a blue or brown sock, assuming that the sock draw the first time, is the opposite colour of the sock drawn the second time.--[[User:BernadetteHii|BernadetteHii]] 17:55, 16 October 2010 (UTC)
== Question 8==
If we assume Reuben's birthday is December 31, then on December 31 he turned 21 years old.
If it is January 1, 2 days ago the would have been December 30. On that day he was still 20 years old. By the end of this year he will turn 22 years old. One year from the day January 1 he will be 23.

Polya's method seems to help organize the wording in this question making it easier to interpret the information the question gives us.--[[User:BernadetteHii|BernadetteHii]] 17:55, 16 October 2010 (UTC)
== Question 9==
From the question we know that the distance between each rung is 1 foot. If 10 rungs are showing we can assume that these 10 rungs are hanging off the side of the boat. Also, we assume that these 10 rungs are not submerged in water yet. If the tide rises 5 feet, this is equivalent to the distance between the bottom rung and the 5th rung of the ladder, therefore 5 rungs are now submerged in water, leaving the 5 upper rungs out of the water.

Similar to question 8, Polya's method is helpful to organize the wording in this question and to deduce from the question an appropriate scenario, in this case, 10 rungs of the ladder on the side of a boat, not covered by water.--[[User:BernadetteHii|BernadetteHii]] 17:55, 16 October 2010 (UTC)
== Question 10==
Polya's method helps start the question off by helping us interpret what the question is telling us.
i.) one-half the population of women can be written as
(1/2) (1/2)=(1/4)

If we know half the population are women, then it follows the other half of the population are men.
One-half the population also equals
(1/2)(1/2)=(1/4)

Knowing that half the population are chocolate eaters, we an take (1/4) of male chocolate eaters and (1/4) of female chocolate eaters, adding these together we get

(1/4)+(1/4)=(1/2), which is consistent with the information given in the question.

ii.) One-half of all men is
(1/2)(1/2)=(1/4)

therefore it follows that 1/4 the population of men are chocolate eaters.--[[User:BernadetteHii|BernadetteHii]] 17:55, 16 October 2010 (UTC)

== Question 11 ==
A woman, her older brother, her son, and her daughter are chess players. The worst player's twin, who is one of the four players, and the best player are of opposite sex. The worst player and best have the same age. If this is possible, who is the worst player?

The first step in solving this problem is understanding what the question is asking us and determining what information we are given. The question is asking us to solve out of four possible players, which on is the worst.

The information we are given is:
-The twin of the worst player is the opposite sex of the best player
-the worst and best player have the same age

Through this information we can determine that it is not possible to solve for the worst player, if one of the twins is the worst player and the best player is the same age and opposite gender of the other twin, there is no way of determining which twin is the worst player; also, no other player fits the description of the best player.

== Question 12 ==
A Manhattan fellow had a girlfriend in the Bronx and a girlfriend in Brooklyn. He decided which girlfriend to visit by arriving randomly at he train station and taking the first of the Bronx or Brooklyn trains that arrived. The trains to Brooklyn and the Bronx each arrived regularly every 10 minutes. Not long after he began his scheme the man's Bronx girlfriend left him because he rarely visited. Give a (logical) explanation.

This question is asking us to conclude why it was that the man rarely visited his Bronx girlfriend in order to determine why he was broken up with.

The man's daily routine probably meant he arrived at the train station at a similar point every time, the train's also run on a consistent schedule. Therefore, it is likely that the Brooklyn train schedule matched the Manhattan fellow's typical schedule more often; consequently, he rarely visited his Bronx girlfriend.

== Question 13 ==
If a clock takes 5 seconds to strike 5:00 (with 5 equally spaced chimes), how long does it take to strike 10:00 (with 10 equally spaced chimes)?

x= #chimes/second

x= 5 seconds/ 5chimes = 1 chime/ second

from this we can conclude that the clock chimes at a rate of 1 chime per second; therefore, if we want to solve for the time it would take to strike 10:00 (with 10 equally spaced chimes), we divide number of chimes by the rate of chimes:

10chimes/ 1chime/s = 10 seconds

== Question 14 ==
One day in the maternity ward, the name tags for four girl babies become mixed up.
i)How many different ways could two of the babies be tagged correctly and two incorrectly?

The question is asking us to find the number of ways two correct and two incorrect names can be arrangged. There are four names for the four babies. Let the correct order be ABCD. If there are two correct name orders and two incorrect these are our possible arrangements: ABDC, ADCB, ACBD, BACD, DBCA, CBAD. There are 6 possible arrangements given that 2 of the names are correct and the other two are not.

ii) In how many ways could three of the babies be tagged correctly and one baby be tagged incorrectly?

This question wants to know how many, if at all, ways can three of the baby names be correct and one incorrect. If the correct order of the names is ABCD, and ABC are correct the only possible name assignment for the fourth baby is D; therefore there are no possible arrangements for three correctly ordered names and one incorrect.

== Question 15 ==
Alex says to you, "I'll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck." Should you accept his bet?

This question is asking if Alex's assumption that the number of red cards in the first half of a split deck as there are black cards in the second half of a split deck is correct.

To solve, let
R= red cards
B= black cards

R(in first half deck) = B(in second half deck)

52 cards = R+B, half a deck = 26 cards

if,
1st half of deck has 16R + 10B, that leaves 16B +10R for the 2nd half of the deck

or,
1st half of deck has 2R+24B, that leaves 2B +24R for the 2nd half of the deck

No, you should not accept his bet because you would owe him a lot of money. Alex's statement that the number of red cards in the first half of the deck is equal to the number of black cards in the second half of the deck is true. Half a deck = 26 cards, so if you have 14 red cards in your first half of the deck that means the remaining 12 cards are black; therefore, in the second half of the deck, there are 14 black cards and 12 red cards remaining. So the number of red in the first half of the deck is equal to the number of black in the second half of the deck.

== Qustion 16 ==
Suppose that each daughter in your family has the same number of brothers as she has sisters, and each son in your family has twice as many sisters as he has brothers. How many sons and daughters are in the family?

Assume that total kids: 7
Boys: 3
Girls: 4
From this each son would have twice as many sisters as he has brothers.

== Question 17 ==
The zero point on a bathroom scale is set incorrectly, but otherwise the scale is accurate. It shows 60 kg when Dan stands on the scale, 50 kg when Sarah stands on the scale, but 105 kg when Dan and Sarah both stand on the scale. Does the scale read too high or too low? Explain.

Theoretically, if both Dan and Sarah stand on the scale, it should have read “110” kg. Since it reads 105kg instead of 110 kg, it shows that the scale reads too high since it should reduce about 2.5 kg for each individual for it to be 105kg. Therefore, the actual reading for Dan should be 57.5 kg and as for Sarah’s it should be 47.5 kg. Both of these weights make up to 105kg.

== Question 18 ==
Alice takes one-third of the pennies from a large jar. Then Bret takes one-third of the remaining pennies from the jar. Finally, Carla takes one-third of the remaining pennies from the jar, leaving 40 pennies in the jar. How many pennies were in the jar at the start?

In order to calculate the total pennies in the jar, one has to work backwards. Starting from the time Carla takes 1/3 of the pennies from the jar. In order to find the previous amount of pennies in the jar before Carla took 1/3 of it, the amount left must be divided by a factor of 1/3. This can be written below as “x” represents the amount of pennies before Carla took 1/3 of it;
1/3 of pennies = 40
Pennies = 120
This number would then be divided by a factor of 1/3 to calculate the amount before Bret took 1/3 of it;
1/3 of pennies = 120
Pennies = 360
Then 360 would be divided again by the same factor of 1/3 to find the original amount of pennies in the jar;
1/3 pennies = 360
Pennies = 1080
In conclusion, there were 1080 pennies originally in the jar.

== Question 19 ==
One morning each member of Angela's family drank an eight-ounce cup of coffee and milk, with the (nonzero) amounts of coffee and milk varying from cup to cup. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. What is the least number of people in the family?

If Angela consumes a quarter of the total amount of milk, that means there could only be a minimum of 4 number of people in the family. Since (1/4) x 4 makes 1 carton of milk, it is reasonable to conclude that the least number of people in Angela’s family would be 4 members.

== Question 20 ==
Of two clocks next to each other, one runs 5 min per hour fast and the other runs 5 min per hour slow. At midnight the clocks show the same time. At what time are they are one hour apart?

At every hour, the first clock runs about 5 minutes faster, and the second clock runs 5 minutes per hour slow. Therefore, for each consecutive hour, the first clock would gain 5 minutes while the second clock would lose 5 minutes. As a result, the total of minutes gained by the first clock for every hour is 10 minutes compared to the second clock. Because there is 60 minutes in an hour, and the first clock gains a lead of 10 minutes every hour, the two clocks would be an hour apart once the first clock strikes 18:00.

== Question 21 ==
Sven was in ninth place among seventeen runners in the marathon. It states that Sven was in the exact middle of placement in the marathon, which automatically narrowed the number of racers down to an odd number. This is because any single odd number is surrounded by the same amount of even numbers, so in this case of 9th, there are 8 racers before him, and 8 racers after him, leading the number to seventeen racers. The insert about Dan finishing in 10th place ruled out 19 as another potential number of racers going with the same logic as before. And Lars finishing in 16th place made it obvious there was more than 16 racers. Now there could not have been more racers as with Sven finishing ahead of Dan in 10th place, and being in the exact middle, there could not possibly have been more than 17 racers.

== Question 22 ==
Given that the vacation can be no shorter than 13 days, as it states that the rain cannot endure for the entire day. In the morning if it rains, it is followed by a sunny afternoon, and if it rains in the afternoon it is sunny in the morning. There are two distinct time frames, morning and afternoon, per day. Since there are only 2 slots per day, it makes sense to add the numbers of each type of day; 13 periods of rain, 12 sunny afternoons, and 11 bright mornings. After the addition you come out to the number of 36, and since there are 2 time slots per day, 36/2 is equal to 18 days of vacation.

== Question 23 ==
Yes you would be able to find the ages of the children. It would be necessary however to know the date at least. The number set for dates alone is much to factor in, from the 1 - 31 at most. However definitely not as high as 31 since the product of the three children's ages is 36. The date narrows it down by quite a bit. It would be simple and very possible to come up with a list of viable ages through simple calculations, but to narrow that list down the date would help. For example; (9,2,2), (2,3,6), (18,2,1), (4,3,3), (6,6,1), (12,3,1) etc ...

== Question 24 ==
Considering the second candle took 3 hours to burn out, half the time it took the first candle to burn out (6 hours). I made each candle into a standard and equal fraction of 6/6. Then I began by reducing the first candle to 5/6. Since the first candle lasted twice as long as the second candle, it is safe to assume that the second candle would go down 2 fractional points compared to the first candle, making it 4/6. You do this for one more interval and you find that when the first candle is equal to 4/6 the second candle is down to 2/6, half the value of the first. So doing some simple adjustments, the first candle lasted for 6 hours, and it is double the height of the second candle at fractions 4/6, therefore it took 2 hours before the first candle was twice as big as the second candle.

== Question 25 ==
For this question, I started out by figuring out how long each candle had lasted. The L + 1 candle lasted for 6 hours, whereas the L candle lasted for 4 hours. It could then be concluded that the L + 1 candle lasted 3/2 longer than the L candle. And since (L + 1)/L = 6/4, by cross-multiplying this equation, you see that 6L = 4(L + 1), which in turn becomes 6L = 4L + 4. By subtracting 4L from both sides to get rid of it on the right side, the equation then becomes 2L = 4. With some simple division, you discover that L = 2

Course:MATH110/Archive/2010-2011/003/Groups/Group 11

2010-12-01T05:33:51Z

ShannonLee: /* Basic Skills Project: Trigonometric Functions */

{{Infobox MATH110 Groups
| group number = 11
| member 1 = [[User:AbdulSherifAbdulrahaman|Abdul-Sherif Abdul-Rahaman]]
| member 2 = [[User:ShannonLee|Shannon Lee]]
| member 3 = [[User:AmirKushairyMohamadNazri|Amir Mohamad Nazri]]
| member 4 = [[User:MattVetter|Matt Vetter]]
| member 5 =
| member 6 =
}}
==Basic Skills Project: Trigonometric Functions==
This page will focus on trigonometric function. Group 11 hopes that through providing examples, and step-by-step solution guides we can eliminate the struggles many of us face in regards to solving problems involving trigonometry. Hopefully after reviewing this page you will be able to relate trigonometric functions to the study of everyday occurrences, some of which include: waves, daily temperature averages through the seasons, alternating current, electronic signals, as well as other cyclic phenomena. In the "Just-in-Time" text book, chapter 7, titled "Cyclic Phenomena: The Six Basic Trigonometric Functions," can provide further assistance when reviewing trig functions.

A trigonometric function, also known as a circular function, is a function that relates the lengths of the sides of a right triangle with the angles of the triangle. The basic trig functions are sine, cosine, tangent, cotangent, secant and cosecant.

f(x)=sin x and f(x)=cos x are functions defined for any real number x.
The other basic trigonometric functions we also have to remember are

1)tangent: tan x = sin x/ cos x
2)cotangent: cot x = cos x/sin x
3)secant: 1/cos x
4)cosecant: 1/sin x

==Right-angled Triangle Definition==
If a value of x falls between 0 and π/2, so 0<x<π/2 then the definition of sin(x) and cos(x) is called the right-angled triangle definition. By this definition, sin(x)=opposite/ hypotenuse and cos(x)=adjacent/hypotenuse.
A helpful pneumonic for remembering how to solve the sides of a right triangle is SOH CAH TOA!

[[File:Right_triangle_.png|frame|Caption: Right Triangle.]]

{| class=wikitable style="margin-left:1em"
! style="text-align:left" | '''Function'''
! style="text-align:left" | '''Abbreviation'''
! style="text-align:left" | '''Description'''
! style="text-align:left" | '''Trigonometric Identities using Radians'''
|- style="background-color:#FFFFFF"
| '''Sine'''
| sin
| <math>\frac {\text{opposite}} {\text{hypotenuse}} </math>
| <math>\sin \theta \equiv \cos \left(\frac{\pi}{2} - \theta \right) \equiv \frac{1}{\csc \theta}</math>
|- style="background-color:#FFFFFF"
| '''Cosine'''
| cos
| <math>\frac {\text{adjacent}} {\text{hypotenuse}} </math>
| <math>\cos \theta \equiv \sin \left(\frac{\pi}{2} - \theta \right) \equiv \frac{1}{\sec \theta}\,</math>
|- style="background-color:#FFFFFF"
| '''Tangent'''
| tan
|align=center| <math>\frac {\text{opposite}} {\text{adjacent}} </math>
| <math>\tan \theta \equiv \frac{\sin \theta}{\cos \theta} \equiv \cot \left(\frac{\pi}{2} - \theta \right) \equiv \frac{1}{\cot \theta} </math>
|- style="background-color:#FFFFFF"
| '''Cotangent'''
| cot
|align=center| <math>\frac {\text{adjacent}} {\text{opposite}} </math>
| <math>\cot \theta \equiv \frac{\cos \theta}{\sin \theta} \equiv \tan \left(\frac{\pi}{2} - \theta \right) \equiv \frac{1}{\tan \theta} </math>
|- style="background-color:#FFFFFF"
| '''Secant'''
| sec
| <math>\frac {\text{hypotenuse}} {\text{adjacent}} </math>
| <math>\sec \theta \equiv \csc \left(\frac{\pi}{2} - \theta \right) \equiv\frac{1}{\cos \theta} </math>
|- style="background-color:#FFFFFF"
| '''Cosecant'''
| csc
| <math>\frac {\text{hypotenuse}} {\text{opposite}} </math>
| <math>\csc \theta \equiv \sec \left(\frac{\pi}{2} - \theta \right) \equiv\frac{1}{\sin \theta} </math>
|}{{clearright}}

To Begin learning about trigonometric functions, lets review the unit circle:

[[File:UnitCircle2.gif|center|frame|Caption: The Unit Circle.]]

The x coordinate on the unit circle is cos(x), and the y coordinate is sin(x)
(x,y) = (cos(x), sin(x))

Because cos(x) and sin(x) is a point on the unit circle, its coordinate must satisfy the equation of the circle: x^2+y^2 = 1

Therefore, Sin^2(x)+cos^2(x) = 1

The unit circle has a radius = 1
The circumference of the unit circle = 2π radians

each complete rotation of the unit circle will add 2π or 360°, for instance 2 complete revolutions of the unit circle = 4π or 720°,

On the Unit Circle, x and x + 2π correspond to the same point, this fact means that the graphs of y=sin(x) and y=cos(x) repeat over and over again, and gives us sin(x) = sin(x + 2kπ), where k is any whole number and represents the number of revolutions. From this we can say that sin(x) has a period of 2π. This can also be said about cos(x).

Converting Degrees to Radians:
Angle *(2π/360°)
for example: Convert 27° into radians:
27°*(2π/360°)= (3/40)*2π = 3π/20

Converting Radians to Degrees:
Radian * n(π), where π is equal to 180°
for example: Convert 3π into degrees
3π = 3(180°) = 540°

The following is a helpful video for understanding the Unit Circle, and remembering the points

{{#ev:youtube | cIVpemcoAlY| 400}}

==Solving Trig Functions==
Examples:

1.)Evaluate cos(3π/2)and sin(3π/2):

To determine the sine and cosine values of angles using the unit circle is quite simple. If you refer back to the picture of the unit circle above, you can see that 3π/2 is three quarters of the way around the unit circle and located at the point (0,-1). If you remember that (x,y) is equivalent to (cos(x), sin(x)), then we can determine that:

cos(3π/2) = 0 and sin(3π/2) = -1

2.)Evaluate cos(9π/2):

To solve this problem, we first need to determine that 9π/2 is equivalent to 4π +π/2. A complete revolution of the unit circle is = 2π; therefore, 4π is just equal to two complete revolutions of the unit circle, which does not change the value of your answer. So cos 4π+π/2 is equal to two complete rotations and another quarter rotation, or 90°. If we refer to the unit circle above, we can see that π/2 is equal to (0,1). Because cosine is equal to the x value, we can determine that:

cos(9π/2) = 0

For some extra practice in evaluating the angles of sin and cos, refer to the exercises in chapter 7.2 of your Just-in-Time text!

Solving Trigonometric Functions:
{{#ev:youtube | IE0FxGegdMg| 400}}

'''Sine graphs'''

If sin(x) is plotted on a set of "x" and "y" axis, the graph of sin(x) would be drawn as shown below:

http://i56.tinypic.com/2mmxnq0.jpg

The demonstration above is to show how the values from the unit circle can be plotted onto a set of "x" and "y" axis. The graph is not limited, it depends on a given scope. In other words, as long as the range of the sin function is not given, sin(x) graph will repeat as a wave until infinite. The sin(x) graph below shows for x values from 0 to 3π(540°).In order to calculate the points where slope = 0, we must make equate the derivate of sin(x) = 0.

http://i54.tinypic.com/okbkpd.jpg

'''Cosine graphs'''

Cosine have the exact same characteristics to that of a sine graph. The only difference between the two is that cosine function shifts to the right by a factor of 90° or π/2 in radians:

y = cos(x)

http://i55.tinypic.com/34hh2t2.jpg

As mentioned previously, cosine functions have the exact same behavior as the sine graph, and the only difference between them is that cosine functions shifts to the right by a factor of 90° or π/2 in radians. Mathematically speaking, if we were to expressed cos(x) in terms of sin(x), it would be written as; cos(x) = (sin(x+π/2)). The comparison between a sine and a cosine function is demonstrated as below;

http://i52.tinypic.com/s5920j.jpg

'''Tangent Graphs'''

http://i53.tinypic.com/2vd058x.jpg

Graphing Trigonometric Functions
{{#ev:youtube | R0RM6FeiD8k | 400}}

==Basic Skills Project-Group Contribution==
http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_11/Basic_Skills

== Homeworks ==

HOMEWORK 4[[http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_11/Homework_4]]

[[Course:MATH110/003/Groups/Group 11/Information on Basic Skills|Information on Basic Skills]]

== Homework 3==

==Question 1==
Well in both instance the travel time were exactly the same. One was given in hour and minute and the second was given in minutes. We know that an hour is compose of 60 minutes so at 1hour and 20 minute we have 80 total minutes. The Second trip back was 80 minutes so both trips took equal time.

==Question 2==
The the reason the police did not stop the woman was because she was walking on the street, she was not driving. If the officer was to not stop her as she drove on to oncoming traffic he would not be doing his job. If the officer did not try stopping her it is obvious she was not breaking any laws.

==Question 3==
Well in the first two boxes you could pick a fruit up and figured it is either apples or oranges, it is a homogenous mixture in the first two boxes, but the third is a heterogenous mixture. I would personally grab into each box three times so i figure it is all the same or a mixture of both

==Question 4==
He is a brother to the fiddler because he too is blind.

==Question 5==
The moving quarter has made one full revolution around the fixed coin.

== Question 6==
If we let the 3 kinds of apples be X, Y and Z respectively, we know from the question that all these types of apples will sum up to a value; let that value be a.
Our goal is to have at least two of the same type of apple, which leads to our question-how many apples do we need to draw in order to get a matching pair of apples?

In our attempt to find a solution we found the probability of drawing any type of apple out of the basket, concluding that it is necessary to draw 3 times minimally to have any chance of a matching pair being taken out.

On the 1st draw:
1/a chances of getting any of the types of apples.

1 apple drawn from a basket filled with a apples.

On the 2nd draw:
1/(a-1) chances of getting any of the types of apples, however since you have drawn one out already, one of the 3 types is short an apple, reducing the total by 1.

Another apple drawn from the basket filled with (a-1) apples.

On the 3rd draw:
1/(a-2) chances of getting any type of the apples, with one group short of an apple, possible the same group that you drew an apple from in the 1st and/or 2nd group, resulting in a matching pair of apples.

A third apple drawn from the basket filled with (a-2) apples.

The necessary information that are lacking in this question are the total number of apples in the basket and the total number of each type of apple. Polya's method was only useful in sorting out our thinking. In order to apply probability and combinatoric concepts and further Step 3 of Polya's method, more information is required.--[[User:BernadetteHii|BernadetteHii]] 17:55, 16 October 2010 (UTC)
== Question 7==
We know that there are 80 socks in a drawer, 40 blue socks and 40 brown socks.

i.) Our goal in the first half of the question is to draw out a matching pair, how many socks need to be taken from the drawer to ensure a matching pair is found?

Similarly to the preceding question, we use probability of drawing any of the socks out of the drawer, concluding that it is necessary to draw 3 socks, minimally, to have a pair of matching socks.

On the 1st draw:
The probability drawing a blue or brown sock is 1/80.

1/40 chance of drawing a blue sock
1/40 chance of drawing a brown sock

On the 2nd draw:
1/79 chance of drawing a blue or brown sock, however the amount of one colour of socks is not reduced by 1.

If we take the two scenarios:
a.) 1st draw=blue sock, you now have a 1/39 chance of drawing a blue sock and a 1/40 chance of drawing a brown one on the 2nd draw.
b.) 1st draw=brown sock, you now have a 1/39 chance of drawing a brown sock and a 1/40 chance of drawing a blue one on the 2nd draw.

On the 3rd draw:
1/78 chance of drawing a blue or brown sock, however the amount of one of the colours has been reduced again by 1 sock.

a.) 2nd draw=blue sock, you now have a 1/38 chance of drawing a blue sock on the 3rd draw, assuming the 1st draw also was a blue sock
b.)2nd draw=brown sock, you now have a 1/38 chance of drawing a bluce sock on the 3rd draw assuming the 1st draw was also a brown sock.

As more socks are taken from the drawer, one by one, the amount of one colour of sock reduces. Since the picking is at random one colour will not necessarily decrease faster than the other, assuming that they are mixed in the drawer not nicely laid out. Therefore, to ensure a matching pair is picked, 3 socks minimally need to be taken from the drawer.

ii.) To draw an unmatched pair of socks from this drawer, simply draw two socks. Picking is at random and assuming the socks are mixed up and not laid out within the drawer, drawing two socks is sufficient enough to two different socks are taken out.

1st draw:
1/40 chance of getting a blue sock
1/40 chance of getting a brown sock

On the first draw there is an equal chance of drawing either a blue or brown sock.

2nd draw:
1/40 chance of getting a brown sock
1/40 chance of getting a blue sock

On the second draw there is also an equal chance of drawing either a blue or brown sock, assuming that the sock draw the first time, is the opposite colour of the sock drawn the second time.--[[User:BernadetteHii|BernadetteHii]] 17:55, 16 October 2010 (UTC)
== Question 8==
If we assume Reuben's birthday is December 31, then on December 31 he turned 21 years old.
If it is January 1, 2 days ago the would have been December 30. On that day he was still 20 years old. By the end of this year he will turn 22 years old. One year from the day January 1 he will be 23.

Polya's method seems to help organize the wording in this question making it easier to interpret the information the question gives us.--[[User:BernadetteHii|BernadetteHii]] 17:55, 16 October 2010 (UTC)
== Question 9==
From the question we know that the distance between each rung is 1 foot. If 10 rungs are showing we can assume that these 10 rungs are hanging off the side of the boat. Also, we assume that these 10 rungs are not submerged in water yet. If the tide rises 5 feet, this is equivalent to the distance between the bottom rung and the 5th rung of the ladder, therefore 5 rungs are now submerged in water, leaving the 5 upper rungs out of the water.

Similar to question 8, Polya's method is helpful to organize the wording in this question and to deduce from the question an appropriate scenario, in this case, 10 rungs of the ladder on the side of a boat, not covered by water.--[[User:BernadetteHii|BernadetteHii]] 17:55, 16 October 2010 (UTC)
== Question 10==
Polya's method helps start the question off by helping us interpret what the question is telling us.
i.) one-half the population of women can be written as
(1/2) (1/2)=(1/4)

If we know half the population are women, then it follows the other half of the population are men.
One-half the population also equals
(1/2)(1/2)=(1/4)

Knowing that half the population are chocolate eaters, we an take (1/4) of male chocolate eaters and (1/4) of female chocolate eaters, adding these together we get

(1/4)+(1/4)=(1/2), which is consistent with the information given in the question.

ii.) One-half of all men is
(1/2)(1/2)=(1/4)

therefore it follows that 1/4 the population of men are chocolate eaters.--[[User:BernadetteHii|BernadetteHii]] 17:55, 16 October 2010 (UTC)

== Question 11 ==
A woman, her older brother, her son, and her daughter are chess players. The worst player's twin, who is one of the four players, and the best player are of opposite sex. The worst player and best have the same age. If this is possible, who is the worst player?

The first step in solving this problem is understanding what the question is asking us and determining what information we are given. The question is asking us to solve out of four possible players, which on is the worst.

The information we are given is:
-The twin of the worst player is the opposite sex of the best player
-the worst and best player have the same age

Through this information we can determine that it is not possible to solve for the worst player, if one of the twins is the worst player and the best player is the same age and opposite gender of the other twin, there is no way of determining which twin is the worst player; also, no other player fits the description of the best player.

== Question 12 ==
A Manhattan fellow had a girlfriend in the Bronx and a girlfriend in Brooklyn. He decided which girlfriend to visit by arriving randomly at he train station and taking the first of the Bronx or Brooklyn trains that arrived. The trains to Brooklyn and the Bronx each arrived regularly every 10 minutes. Not long after he began his scheme the man's Bronx girlfriend left him because he rarely visited. Give a (logical) explanation.

This question is asking us to conclude why it was that the man rarely visited his Bronx girlfriend in order to determine why he was broken up with.

The man's daily routine probably meant he arrived at the train station at a similar point every time, the train's also run on a consistent schedule. Therefore, it is likely that the Brooklyn train schedule matched the Manhattan fellow's typical schedule more often; consequently, he rarely visited his Bronx girlfriend.

== Question 13 ==
If a clock takes 5 seconds to strike 5:00 (with 5 equally spaced chimes), how long does it take to strike 10:00 (with 10 equally spaced chimes)?

x= #chimes/second

x= 5 seconds/ 5chimes = 1 chime/ second

from this we can conclude that the clock chimes at a rate of 1 chime per second; therefore, if we want to solve for the time it would take to strike 10:00 (with 10 equally spaced chimes), we divide number of chimes by the rate of chimes:

10chimes/ 1chime/s = 10 seconds

== Question 14 ==
One day in the maternity ward, the name tags for four girl babies become mixed up.
i)How many different ways could two of the babies be tagged correctly and two incorrectly?

The question is asking us to find the number of ways two correct and two incorrect names can be arrangged. There are four names for the four babies. Let the correct order be ABCD. If there are two correct name orders and two incorrect these are our possible arrangements: ABDC, ADCB, ACBD, BACD, DBCA, CBAD. There are 6 possible arrangements given that 2 of the names are correct and the other two are not.

ii) In how many ways could three of the babies be tagged correctly and one baby be tagged incorrectly?

This question wants to know how many, if at all, ways can three of the baby names be correct and one incorrect. If the correct order of the names is ABCD, and ABC are correct the only possible name assignment for the fourth baby is D; therefore there are no possible arrangements for three correctly ordered names and one incorrect.

== Question 15 ==
Alex says to you, "I'll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck." Should you accept his bet?

This question is asking if Alex's assumption that the number of red cards in the first half of a split deck as there are black cards in the second half of a split deck is correct.

To solve, let
R= red cards
B= black cards

R(in first half deck) = B(in second half deck)

52 cards = R+B, half a deck = 26 cards

if,
1st half of deck has 16R + 10B, that leaves 16B +10R for the 2nd half of the deck

or,
1st half of deck has 2R+24B, that leaves 2B +24R for the 2nd half of the deck

No, you should not accept his bet because you would owe him a lot of money. Alex's statement that the number of red cards in the first half of the deck is equal to the number of black cards in the second half of the deck is true. Half a deck = 26 cards, so if you have 14 red cards in your first half of the deck that means the remaining 12 cards are black; therefore, in the second half of the deck, there are 14 black cards and 12 red cards remaining. So the number of red in the first half of the deck is equal to the number of black in the second half of the deck.

== Qustion 16 ==
Suppose that each daughter in your family has the same number of brothers as she has sisters, and each son in your family has twice as many sisters as he has brothers. How many sons and daughters are in the family?

Assume that total kids: 7
Boys: 3
Girls: 4
From this each son would have twice as many sisters as he has brothers.

== Question 17 ==
The zero point on a bathroom scale is set incorrectly, but otherwise the scale is accurate. It shows 60 kg when Dan stands on the scale, 50 kg when Sarah stands on the scale, but 105 kg when Dan and Sarah both stand on the scale. Does the scale read too high or too low? Explain.

Theoretically, if both Dan and Sarah stand on the scale, it should have read “110” kg. Since it reads 105kg instead of 110 kg, it shows that the scale reads too high since it should reduce about 2.5 kg for each individual for it to be 105kg. Therefore, the actual reading for Dan should be 57.5 kg and as for Sarah’s it should be 47.5 kg. Both of these weights make up to 105kg.

== Question 18 ==
Alice takes one-third of the pennies from a large jar. Then Bret takes one-third of the remaining pennies from the jar. Finally, Carla takes one-third of the remaining pennies from the jar, leaving 40 pennies in the jar. How many pennies were in the jar at the start?

In order to calculate the total pennies in the jar, one has to work backwards. Starting from the time Carla takes 1/3 of the pennies from the jar. In order to find the previous amount of pennies in the jar before Carla took 1/3 of it, the amount left must be divided by a factor of 1/3. This can be written below as “x” represents the amount of pennies before Carla took 1/3 of it;
1/3 of pennies = 40
Pennies = 120
This number would then be divided by a factor of 1/3 to calculate the amount before Bret took 1/3 of it;
1/3 of pennies = 120
Pennies = 360
Then 360 would be divided again by the same factor of 1/3 to find the original amount of pennies in the jar;
1/3 pennies = 360
Pennies = 1080
In conclusion, there were 1080 pennies originally in the jar.

== Question 19 ==
One morning each member of Angela's family drank an eight-ounce cup of coffee and milk, with the (nonzero) amounts of coffee and milk varying from cup to cup. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. What is the least number of people in the family?

If Angela consumes a quarter of the total amount of milk, that means there could only be a minimum of 4 number of people in the family. Since (1/4) x 4 makes 1 carton of milk, it is reasonable to conclude that the least number of people in Angela’s family would be 4 members.

== Question 20 ==
Of two clocks next to each other, one runs 5 min per hour fast and the other runs 5 min per hour slow. At midnight the clocks show the same time. At what time are they are one hour apart?

At every hour, the first clock runs about 5 minutes faster, and the second clock runs 5 minutes per hour slow. Therefore, for each consecutive hour, the first clock would gain 5 minutes while the second clock would lose 5 minutes. As a result, the total of minutes gained by the first clock for every hour is 10 minutes compared to the second clock. Because there is 60 minutes in an hour, and the first clock gains a lead of 10 minutes every hour, the two clocks would be an hour apart once the first clock strikes 18:00.

== Question 21 ==
Sven was in ninth place among seventeen runners in the marathon. It states that Sven was in the exact middle of placement in the marathon, which automatically narrowed the number of racers down to an odd number. This is because any single odd number is surrounded by the same amount of even numbers, so in this case of 9th, there are 8 racers before him, and 8 racers after him, leading the number to seventeen racers. The insert about Dan finishing in 10th place ruled out 19 as another potential number of racers going with the same logic as before. And Lars finishing in 16th place made it obvious there was more than 16 racers. Now there could not have been more racers as with Sven finishing ahead of Dan in 10th place, and being in the exact middle, there could not possibly have been more than 17 racers.

== Question 22 ==
Given that the vacation can be no shorter than 13 days, as it states that the rain cannot endure for the entire day. In the morning if it rains, it is followed by a sunny afternoon, and if it rains in the afternoon it is sunny in the morning. There are two distinct time frames, morning and afternoon, per day. Since there are only 2 slots per day, it makes sense to add the numbers of each type of day; 13 periods of rain, 12 sunny afternoons, and 11 bright mornings. After the addition you come out to the number of 36, and since there are 2 time slots per day, 36/2 is equal to 18 days of vacation.

== Question 23 ==
Yes you would be able to find the ages of the children. It would be necessary however to know the date at least. The number set for dates alone is much to factor in, from the 1 - 31 at most. However definitely not as high as 31 since the product of the three children's ages is 36. The date narrows it down by quite a bit. It would be simple and very possible to come up with a list of viable ages through simple calculations, but to narrow that list down the date would help. For example; (9,2,2), (2,3,6), (18,2,1), (4,3,3), (6,6,1), (12,3,1) etc ...

== Question 24 ==
Considering the second candle took 3 hours to burn out, half the time it took the first candle to burn out (6 hours). I made each candle into a standard and equal fraction of 6/6. Then I began by reducing the first candle to 5/6. Since the first candle lasted twice as long as the second candle, it is safe to assume that the second candle would go down 2 fractional points compared to the first candle, making it 4/6. You do this for one more interval and you find that when the first candle is equal to 4/6 the second candle is down to 2/6, half the value of the first. So doing some simple adjustments, the first candle lasted for 6 hours, and it is double the height of the second candle at fractions 4/6, therefore it took 2 hours before the first candle was twice as big as the second candle.

== Question 25 ==
For this question, I started out by figuring out how long each candle had lasted. The L + 1 candle lasted for 6 hours, whereas the L candle lasted for 4 hours. It could then be concluded that the L + 1 candle lasted 3/2 longer than the L candle. And since (L + 1)/L = 6/4, by cross-multiplying this equation, you see that 6L = 4(L + 1), which in turn becomes 6L = 4L + 4. By subtracting 4L from both sides to get rid of it on the right side, the equation then becomes 2L = 4. With some simple division, you discover that L = 2

Course:MATH110/Archive/2010-2011/003/Groups/Group 11

2010-12-01T00:51:40Z

ShannonLee: /* Basic Skills Project: Trigonometric Functions */

{{Infobox MATH110 Groups
| group number = 11
| member 1 = [[User:AbdulSherifAbdulrahaman|Abdul-Sherif Abdul-Rahaman]]
| member 2 = [[User:ShannonLee|Shannon Lee]]
| member 3 = [[User:AmirKushairyMohamadNazri|Amir Mohamad Nazri]]
| member 4 = [[User:MattVetter|Matt Vetter]]
| member 5 =
| member 6 =
}}
==Basic Skills Project: Trigonometric Functions==
This page will focus on trigonometric function. Group 11 hopes that through providing examples, and step-by-step solution guides we can eliminate the struggles many of us face in regards to solving problems involving trigonometry. Hopefully after reviewing this page you will be able to relate trigonometric functions to the study of everyday occurrences, some of which include: waves, daily temperature averages through the seasons, alternating current, electronic signals, as well as other cyclic phenomena. In the "Just-in-Time" text book, chapter 7, titled "Cyclic Phenomena: The Six Basic Trigonometric Functions," can provide further assistance when reviewing trig functions.

A trigonometric function, also known as a circular function, is a function that relates the lengths of the sides of a right triangle with the angles of the triangle. The basic trig functions are sine, cosine, tangent, cotangent, secant and cosecant.

f(x)=sin x and f(x)=cos x are functions defined for any real number x.
The other basic trigonometric functions we also have to remember are

1)tangent: tan x = sin x/ cos x
2)cotangent: cot x = cos x/sin x
3)secant: 1/cos x
4)cosecant: 1/sin x

[[File:Right_triangle_.png|frame|Caption: Right Triangle.]]

{| class=wikitable style="margin-left:1em"
! style="text-align:left" | '''Function'''
! style="text-align:left" | '''Abbreviation'''
! style="text-align:left" | '''Description'''
! style="text-align:left" | '''Trigonometric Identities using Radians'''
|- style="background-color:#FFFFFF"
| '''Sine'''
| sin
| <math>\frac {\text{opposite}} {\text{hypotenuse}} </math>
| <math>\sin \theta \equiv \cos \left(\frac{\pi}{2} - \theta \right) \equiv \frac{1}{\csc \theta}</math>
|- style="background-color:#FFFFFF"
| '''Cosine'''
| cos
| <math>\frac {\text{adjacent}} {\text{hypotenuse}} </math>
| <math>\cos \theta \equiv \sin \left(\frac{\pi}{2} - \theta \right) \equiv \frac{1}{\sec \theta}\,</math>
|- style="background-color:#FFFFFF"
| '''Tangent'''
| tan
|align=center| <math>\frac {\text{opposite}} {\text{adjacent}} </math>
| <math>\tan \theta \equiv \frac{\sin \theta}{\cos \theta} \equiv \cot \left(\frac{\pi}{2} - \theta \right) \equiv \frac{1}{\cot \theta} </math>
|- style="background-color:#FFFFFF"
| '''Cotangent'''
| cot
|align=center| <math>\frac {\text{adjacent}} {\text{opposite}} </math>
| <math>\cot \theta \equiv \frac{\cos \theta}{\sin \theta} \equiv \tan \left(\frac{\pi}{2} - \theta \right) \equiv \frac{1}{\tan \theta} </math>
|- style="background-color:#FFFFFF"
| '''Secant'''
| sec
| <math>\frac {\text{hypotenuse}} {\text{adjacent}} </math>
| <math>\sec \theta \equiv \csc \left(\frac{\pi}{2} - \theta \right) \equiv\frac{1}{\cos \theta} </math>
|- style="background-color:#FFFFFF"
| '''Cosecant'''
| csc
| <math>\frac {\text{hypotenuse}} {\text{opposite}} </math>
| <math>\csc \theta \equiv \sec \left(\frac{\pi}{2} - \theta \right) \equiv\frac{1}{\sin \theta} </math>
|}{{clearright}}

A helpful pneumonic for remembering how to solve the sides of a right triangle is SOH CAH TOA!

To Begin learning about trigonometric functions, lets review the unit circle:

[[File:UnitCircle2.gif|center|frame|Caption: The Unit Circle.]]

The x coordinate on the unit circle is cos(x), and the y coordinate is sin(x)
(x,y) = (cos(x), sin(x))

Because cos(x) and sin(x) is a point on the unit circle, its coordinate must satisfy the equation of the circle: x^2+y^2 = 1

Therefore, Sin^2(x)+cos^2(x) = 1

The unit circle has a radius = 1
The circumference of the unit circle = 2π radians

each complete rotation of the unit circle will add 2π or 360°, for instance 2 complete revolutions of the unit circle = 4π or 720°,

On the Unit Circle, x and x + 2π correspond to the same point, this fact means that the graphs of y=sin(x) and y=cos(x) repeat over and over again, and gives us sin(x) = sin(x + 2kπ), where k is any whole number and represents the number of revolutions. From this we can say that sin(x) has a period of 2π. This can also be said about cos(x).

Converting Degrees to Radians:
Angle *(2π/360°)
for example: Convert 27° into radians:
27°*(2π/360°)= (3/40)*2π = 3π/20

Converting Radians to Degrees:
Radian * n(π), where π is equal to 180°
for example: Convert 3π into degrees
3π = 3(180°) = 540°

The following is a helpful video for understanding the Unit Circle, and remembering the points

{{#ev:youtube | cIVpemcoAlY| 400}}

Solving Trigonometric Functions:
{{#ev:youtube | IE0FxGegdMg| 400}}

'''Sine graphs'''

If sin(x) is plotted on a set of "x" and "y" axis, the graph of sin(x) would be drawn as shown below:

http://i56.tinypic.com/2mmxnq0.jpg

The demonstration above is to show how the values from the unit circle can be plotted onto a set of "x" and "y" axis. The graph is not limited, it depends on a given scope. In other words, as long as the range of the sin function is not given, sin(x) graph will repeat as a wave until infinite. The sin(x) graph below shows for x values from 0 to 3π(540°).In order to calculate the points where slope = 0, we must make equate the derivate of sin(x) = 0.

http://i54.tinypic.com/okbkpd.jpg

'''Cosine graphs'''

Cosine have the exact same characteristics to that of a sine graph. The only difference between the two is that cosine function shifts to the right by a factor of 90° or π/2 in radians:

y = cos(x)

http://i55.tinypic.com/34hh2t2.jpg

As mentioned previously, cosine functions have the exact same behavior as the sine graph, and the only difference between them is that cosine functions shifts to the right by a factor of 90° or π/2 in radians. Mathematically speaking, if we were to expressed cos(x) in terms of sin(x), it would be written as; cos(x) = (sin(x+π/2)). The comparison between a sine and a cosine function is demonstrated as below;

http://i52.tinypic.com/s5920j.jpg

'''Tangent Graphs'''

http://i53.tinypic.com/2vd058x.jpg

Graphing Trigonometric Functions
{{#ev:youtube | R0RM6FeiD8k | 400}}

==Basic Skills Project-Group Contribution==
http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_11/Basic_Skills

== Homeworks ==

HOMEWORK 4[[http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_11/Homework_4]]

[[Course:MATH110/003/Groups/Group 11/Information on Basic Skills|Information on Basic Skills]]

== Homework 3==

==Question 1==
Well in both instance the travel time were exactly the same. One was given in hour and minute and the second was given in minutes. We know that an hour is compose of 60 minutes so at 1hour and 20 minute we have 80 total minutes. The Second trip back was 80 minutes so both trips took equal time.

==Question 2==
The the reason the police did not stop the woman was because she was walking on the street, she was not driving. If the officer was to not stop her as she drove on to oncoming traffic he would not be doing his job. If the officer did not try stopping her it is obvious she was not breaking any laws.

==Question 3==
Well in the first two boxes you could pick a fruit up and figured it is either apples or oranges, it is a homogenous mixture in the first two boxes, but the third is a heterogenous mixture. I would personally grab into each box three times so i figure it is all the same or a mixture of both

==Question 4==
He is a brother to the fiddler because he too is blind.

==Question 5==
The moving quarter has made one full revolution around the fixed coin.

== Question 6==
If we let the 3 kinds of apples be X, Y and Z respectively, we know from the question that all these types of apples will sum up to a value; let that value be a.
Our goal is to have at least two of the same type of apple, which leads to our question-how many apples do we need to draw in order to get a matching pair of apples?

In our attempt to find a solution we found the probability of drawing any type of apple out of the basket, concluding that it is necessary to draw 3 times minimally to have any chance of a matching pair being taken out.

On the 1st draw:
1/a chances of getting any of the types of apples.

1 apple drawn from a basket filled with a apples.

On the 2nd draw:
1/(a-1) chances of getting any of the types of apples, however since you have drawn one out already, one of the 3 types is short an apple, reducing the total by 1.

Another apple drawn from the basket filled with (a-1) apples.

On the 3rd draw:
1/(a-2) chances of getting any type of the apples, with one group short of an apple, possible the same group that you drew an apple from in the 1st and/or 2nd group, resulting in a matching pair of apples.

A third apple drawn from the basket filled with (a-2) apples.

The necessary information that are lacking in this question are the total number of apples in the basket and the total number of each type of apple. Polya's method was only useful in sorting out our thinking. In order to apply probability and combinatoric concepts and further Step 3 of Polya's method, more information is required.--[[User:BernadetteHii|BernadetteHii]] 17:55, 16 October 2010 (UTC)
== Question 7==
We know that there are 80 socks in a drawer, 40 blue socks and 40 brown socks.

i.) Our goal in the first half of the question is to draw out a matching pair, how many socks need to be taken from the drawer to ensure a matching pair is found?

Similarly to the preceding question, we use probability of drawing any of the socks out of the drawer, concluding that it is necessary to draw 3 socks, minimally, to have a pair of matching socks.

On the 1st draw:
The probability drawing a blue or brown sock is 1/80.

1/40 chance of drawing a blue sock
1/40 chance of drawing a brown sock

On the 2nd draw:
1/79 chance of drawing a blue or brown sock, however the amount of one colour of socks is not reduced by 1.

If we take the two scenarios:
a.) 1st draw=blue sock, you now have a 1/39 chance of drawing a blue sock and a 1/40 chance of drawing a brown one on the 2nd draw.
b.) 1st draw=brown sock, you now have a 1/39 chance of drawing a brown sock and a 1/40 chance of drawing a blue one on the 2nd draw.

On the 3rd draw:
1/78 chance of drawing a blue or brown sock, however the amount of one of the colours has been reduced again by 1 sock.

a.) 2nd draw=blue sock, you now have a 1/38 chance of drawing a blue sock on the 3rd draw, assuming the 1st draw also was a blue sock
b.)2nd draw=brown sock, you now have a 1/38 chance of drawing a bluce sock on the 3rd draw assuming the 1st draw was also a brown sock.

As more socks are taken from the drawer, one by one, the amount of one colour of sock reduces. Since the picking is at random one colour will not necessarily decrease faster than the other, assuming that they are mixed in the drawer not nicely laid out. Therefore, to ensure a matching pair is picked, 3 socks minimally need to be taken from the drawer.

ii.) To draw an unmatched pair of socks from this drawer, simply draw two socks. Picking is at random and assuming the socks are mixed up and not laid out within the drawer, drawing two socks is sufficient enough to two different socks are taken out.

1st draw:
1/40 chance of getting a blue sock
1/40 chance of getting a brown sock

On the first draw there is an equal chance of drawing either a blue or brown sock.

2nd draw:
1/40 chance of getting a brown sock
1/40 chance of getting a blue sock

On the second draw there is also an equal chance of drawing either a blue or brown sock, assuming that the sock draw the first time, is the opposite colour of the sock drawn the second time.--[[User:BernadetteHii|BernadetteHii]] 17:55, 16 October 2010 (UTC)
== Question 8==
If we assume Reuben's birthday is December 31, then on December 31 he turned 21 years old.
If it is January 1, 2 days ago the would have been December 30. On that day he was still 20 years old. By the end of this year he will turn 22 years old. One year from the day January 1 he will be 23.

Polya's method seems to help organize the wording in this question making it easier to interpret the information the question gives us.--[[User:BernadetteHii|BernadetteHii]] 17:55, 16 October 2010 (UTC)
== Question 9==
From the question we know that the distance between each rung is 1 foot. If 10 rungs are showing we can assume that these 10 rungs are hanging off the side of the boat. Also, we assume that these 10 rungs are not submerged in water yet. If the tide rises 5 feet, this is equivalent to the distance between the bottom rung and the 5th rung of the ladder, therefore 5 rungs are now submerged in water, leaving the 5 upper rungs out of the water.

Similar to question 8, Polya's method is helpful to organize the wording in this question and to deduce from the question an appropriate scenario, in this case, 10 rungs of the ladder on the side of a boat, not covered by water.--[[User:BernadetteHii|BernadetteHii]] 17:55, 16 October 2010 (UTC)
== Question 10==
Polya's method helps start the question off by helping us interpret what the question is telling us.
i.) one-half the population of women can be written as
(1/2) (1/2)=(1/4)

If we know half the population are women, then it follows the other half of the population are men.
One-half the population also equals
(1/2)(1/2)=(1/4)

Knowing that half the population are chocolate eaters, we an take (1/4) of male chocolate eaters and (1/4) of female chocolate eaters, adding these together we get

(1/4)+(1/4)=(1/2), which is consistent with the information given in the question.

ii.) One-half of all men is
(1/2)(1/2)=(1/4)

therefore it follows that 1/4 the population of men are chocolate eaters.--[[User:BernadetteHii|BernadetteHii]] 17:55, 16 October 2010 (UTC)

== Question 11 ==
A woman, her older brother, her son, and her daughter are chess players. The worst player's twin, who is one of the four players, and the best player are of opposite sex. The worst player and best have the same age. If this is possible, who is the worst player?

The first step in solving this problem is understanding what the question is asking us and determining what information we are given. The question is asking us to solve out of four possible players, which on is the worst.

The information we are given is:
-The twin of the worst player is the opposite sex of the best player
-the worst and best player have the same age

Through this information we can determine that it is not possible to solve for the worst player, if one of the twins is the worst player and the best player is the same age and opposite gender of the other twin, there is no way of determining which twin is the worst player; also, no other player fits the description of the best player.

== Question 12 ==
A Manhattan fellow had a girlfriend in the Bronx and a girlfriend in Brooklyn. He decided which girlfriend to visit by arriving randomly at he train station and taking the first of the Bronx or Brooklyn trains that arrived. The trains to Brooklyn and the Bronx each arrived regularly every 10 minutes. Not long after he began his scheme the man's Bronx girlfriend left him because he rarely visited. Give a (logical) explanation.

This question is asking us to conclude why it was that the man rarely visited his Bronx girlfriend in order to determine why he was broken up with.

The man's daily routine probably meant he arrived at the train station at a similar point every time, the train's also run on a consistent schedule. Therefore, it is likely that the Brooklyn train schedule matched the Manhattan fellow's typical schedule more often; consequently, he rarely visited his Bronx girlfriend.

== Question 13 ==
If a clock takes 5 seconds to strike 5:00 (with 5 equally spaced chimes), how long does it take to strike 10:00 (with 10 equally spaced chimes)?

x= #chimes/second

x= 5 seconds/ 5chimes = 1 chime/ second

from this we can conclude that the clock chimes at a rate of 1 chime per second; therefore, if we want to solve for the time it would take to strike 10:00 (with 10 equally spaced chimes), we divide number of chimes by the rate of chimes:

10chimes/ 1chime/s = 10 seconds

== Question 14 ==
One day in the maternity ward, the name tags for four girl babies become mixed up.
i)How many different ways could two of the babies be tagged correctly and two incorrectly?

The question is asking us to find the number of ways two correct and two incorrect names can be arrangged. There are four names for the four babies. Let the correct order be ABCD. If there are two correct name orders and two incorrect these are our possible arrangements: ABDC, ADCB, ACBD, BACD, DBCA, CBAD. There are 6 possible arrangements given that 2 of the names are correct and the other two are not.

ii) In how many ways could three of the babies be tagged correctly and one baby be tagged incorrectly?

This question wants to know how many, if at all, ways can three of the baby names be correct and one incorrect. If the correct order of the names is ABCD, and ABC are correct the only possible name assignment for the fourth baby is D; therefore there are no possible arrangements for three correctly ordered names and one incorrect.

== Question 15 ==
Alex says to you, "I'll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck." Should you accept his bet?

This question is asking if Alex's assumption that the number of red cards in the first half of a split deck as there are black cards in the second half of a split deck is correct.

To solve, let
R= red cards
B= black cards

R(in first half deck) = B(in second half deck)

52 cards = R+B, half a deck = 26 cards

if,
1st half of deck has 16R + 10B, that leaves 16B +10R for the 2nd half of the deck

or,
1st half of deck has 2R+24B, that leaves 2B +24R for the 2nd half of the deck

No, you should not accept his bet because you would owe him a lot of money. Alex's statement that the number of red cards in the first half of the deck is equal to the number of black cards in the second half of the deck is true. Half a deck = 26 cards, so if you have 14 red cards in your first half of the deck that means the remaining 12 cards are black; therefore, in the second half of the deck, there are 14 black cards and 12 red cards remaining. So the number of red in the first half of the deck is equal to the number of black in the second half of the deck.

== Qustion 16 ==
Suppose that each daughter in your family has the same number of brothers as she has sisters, and each son in your family has twice as many sisters as he has brothers. How many sons and daughters are in the family?

Assume that total kids: 7
Boys: 3
Girls: 4
From this each son would have twice as many sisters as he has brothers.

== Question 17 ==
The zero point on a bathroom scale is set incorrectly, but otherwise the scale is accurate. It shows 60 kg when Dan stands on the scale, 50 kg when Sarah stands on the scale, but 105 kg when Dan and Sarah both stand on the scale. Does the scale read too high or too low? Explain.

Theoretically, if both Dan and Sarah stand on the scale, it should have read “110” kg. Since it reads 105kg instead of 110 kg, it shows that the scale reads too high since it should reduce about 2.5 kg for each individual for it to be 105kg. Therefore, the actual reading for Dan should be 57.5 kg and as for Sarah’s it should be 47.5 kg. Both of these weights make up to 105kg.

== Question 18 ==
Alice takes one-third of the pennies from a large jar. Then Bret takes one-third of the remaining pennies from the jar. Finally, Carla takes one-third of the remaining pennies from the jar, leaving 40 pennies in the jar. How many pennies were in the jar at the start?

In order to calculate the total pennies in the jar, one has to work backwards. Starting from the time Carla takes 1/3 of the pennies from the jar. In order to find the previous amount of pennies in the jar before Carla took 1/3 of it, the amount left must be divided by a factor of 1/3. This can be written below as “x” represents the amount of pennies before Carla took 1/3 of it;
1/3 of pennies = 40
Pennies = 120
This number would then be divided by a factor of 1/3 to calculate the amount before Bret took 1/3 of it;
1/3 of pennies = 120
Pennies = 360
Then 360 would be divided again by the same factor of 1/3 to find the original amount of pennies in the jar;
1/3 pennies = 360
Pennies = 1080
In conclusion, there were 1080 pennies originally in the jar.

== Question 19 ==
One morning each member of Angela's family drank an eight-ounce cup of coffee and milk, with the (nonzero) amounts of coffee and milk varying from cup to cup. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. What is the least number of people in the family?

If Angela consumes a quarter of the total amount of milk, that means there could only be a minimum of 4 number of people in the family. Since (1/4) x 4 makes 1 carton of milk, it is reasonable to conclude that the least number of people in Angela’s family would be 4 members.

== Question 20 ==
Of two clocks next to each other, one runs 5 min per hour fast and the other runs 5 min per hour slow. At midnight the clocks show the same time. At what time are they are one hour apart?

At every hour, the first clock runs about 5 minutes faster, and the second clock runs 5 minutes per hour slow. Therefore, for each consecutive hour, the first clock would gain 5 minutes while the second clock would lose 5 minutes. As a result, the total of minutes gained by the first clock for every hour is 10 minutes compared to the second clock. Because there is 60 minutes in an hour, and the first clock gains a lead of 10 minutes every hour, the two clocks would be an hour apart once the first clock strikes 18:00.

== Question 21 ==
Sven was in ninth place among seventeen runners in the marathon. It states that Sven was in the exact middle of placement in the marathon, which automatically narrowed the number of racers down to an odd number. This is because any single odd number is surrounded by the same amount of even numbers, so in this case of 9th, there are 8 racers before him, and 8 racers after him, leading the number to seventeen racers. The insert about Dan finishing in 10th place ruled out 19 as another potential number of racers going with the same logic as before. And Lars finishing in 16th place made it obvious there was more than 16 racers. Now there could not have been more racers as with Sven finishing ahead of Dan in 10th place, and being in the exact middle, there could not possibly have been more than 17 racers.

== Question 22 ==
Given that the vacation can be no shorter than 13 days, as it states that the rain cannot endure for the entire day. In the morning if it rains, it is followed by a sunny afternoon, and if it rains in the afternoon it is sunny in the morning. There are two distinct time frames, morning and afternoon, per day. Since there are only 2 slots per day, it makes sense to add the numbers of each type of day; 13 periods of rain, 12 sunny afternoons, and 11 bright mornings. After the addition you come out to the number of 36, and since there are 2 time slots per day, 36/2 is equal to 18 days of vacation.

== Question 23 ==
Yes you would be able to find the ages of the children. It would be necessary however to know the date at least. The number set for dates alone is much to factor in, from the 1 - 31 at most. However definitely not as high as 31 since the product of the three children's ages is 36. The date narrows it down by quite a bit. It would be simple and very possible to come up with a list of viable ages through simple calculations, but to narrow that list down the date would help. For example; (9,2,2), (2,3,6), (18,2,1), (4,3,3), (6,6,1), (12,3,1) etc ...

== Question 24 ==
Considering the second candle took 3 hours to burn out, half the time it took the first candle to burn out (6 hours). I made each candle into a standard and equal fraction of 6/6. Then I began by reducing the first candle to 5/6. Since the first candle lasted twice as long as the second candle, it is safe to assume that the second candle would go down 2 fractional points compared to the first candle, making it 4/6. You do this for one more interval and you find that when the first candle is equal to 4/6 the second candle is down to 2/6, half the value of the first. So doing some simple adjustments, the first candle lasted for 6 hours, and it is double the height of the second candle at fractions 4/6, therefore it took 2 hours before the first candle was twice as big as the second candle.

== Question 25 ==
For this question, I started out by figuring out how long each candle had lasted. The L + 1 candle lasted for 6 hours, whereas the L candle lasted for 4 hours. It could then be concluded that the L + 1 candle lasted 3/2 longer than the L candle. And since (L + 1)/L = 6/4, by cross-multiplying this equation, you see that 6L = 4(L + 1), which in turn becomes 6L = 4L + 4. By subtracting 4L from both sides to get rid of it on the right side, the equation then becomes 2L = 4. With some simple division, you discover that L = 2

Course:MATH110/Archive/2010-2011/003/Groups/Group 11

2010-12-01T00:29:54Z

ShannonLee: /* Basic Skills Project: Trigonometric Functions */

{{Infobox MATH110 Groups
| group number = 11
| member 1 = [[User:AbdulSherifAbdulrahaman|Abdul-Sherif Abdul-Rahaman]]
| member 2 = [[User:ShannonLee|Shannon Lee]]
| member 3 = [[User:AmirKushairyMohamadNazri|Amir Mohamad Nazri]]
| member 4 = [[User:MattVetter|Matt Vetter]]
| member 5 =
| member 6 =
}}
==Basic Skills Project: Trigonometric Functions==
This page will focus on trigonometric function. Group 11 hopes that through providing examples, and step-by-step solution guides we can eliminate the struggles many of us face in regards to solving problems involving trigonometry. Hopefully after reviewing this page you will be able to relate trigonometric functions to the study of everyday occurrences, some of which include: waves, daily temperature averages through the seasons, alternating current, electronic signals, as well as other cyclic phenomena. In the "Just-in-Time" text book, chapter 7, titled "Cyclic Phenomena: The Six Basic Trigonometric Functions," can provide further assistance when reviewing trig functions.

A trigonometric function, also known as a circular function, is a function that relates the lengths of the sides of a right triangle with the angles of the triangle. The basic trig functions are sine, cosine, tangent, cotangent, secant and cosecant.

f(x)=sin x and f(x)=cos x are functions defined for any real number x.
The other basic trigonometric functions we also have to remember are

1)tangent: tan x = sin x/ cos x
2)cotangent: cot x = cos x/sin x
3)secant: 1/cos x
4)cosecant: 1/sin x

[[File:Right_triangle_.png|frame|Caption: Right Triangle.]]

{| class=wikitable style="margin-left:1em"
! style="text-align:left" | '''Function'''
! style="text-align:left" | '''Abbreviation'''
! style="text-align:left" | '''Description'''
! style="text-align:left" | '''Trigonometric Identities using Radians'''
|- style="background-color:#FFFFFF"
| '''Sine'''
| sin
| <math>\frac {\text{opposite}} {\text{hypotenuse}} </math>
| <math>\sin \theta \equiv \cos \left(\frac{\pi}{2} - \theta \right) \equiv \frac{1}{\csc \theta}</math>
|- style="background-color:#FFFFFF"
| '''Cosine'''
| cos
| <math>\frac {\text{adjacent}} {\text{hypotenuse}} </math>
| <math>\cos \theta \equiv \sin \left(\frac{\pi}{2} - \theta \right) \equiv \frac{1}{\sec \theta}\,</math>
|- style="background-color:#FFFFFF"
| '''Tangent'''
| tan
|align=center| <math>\frac {\text{opposite}} {\text{adjacent}} </math>
| <math>\tan \theta \equiv \frac{\sin \theta}{\cos \theta} \equiv \cot \left(\frac{\pi}{2} - \theta \right) \equiv \frac{1}{\cot \theta} </math>
|- style="background-color:#FFFFFF"
| '''Cotangent'''
| cot
|align=center| <math>\frac {\text{adjacent}} {\text{opposite}} </math>
| <math>\cot \theta \equiv \frac{\cos \theta}{\sin \theta} \equiv \tan \left(\frac{\pi}{2} - \theta \right) \equiv \frac{1}{\tan \theta} </math>
|- style="background-color:#FFFFFF"
| '''Secant'''
| sec
| <math>\frac {\text{hypotenuse}} {\text{adjacent}} </math>
| <math>\sec \theta \equiv \csc \left(\frac{\pi}{2} - \theta \right) \equiv\frac{1}{\cos \theta} </math>
|- style="background-color:#FFFFFF"
| '''Cosecant'''
| csc
| <math>\frac {\text{hypotenuse}} {\text{opposite}} </math>
| <math>\csc \theta \equiv \sec \left(\frac{\pi}{2} - \theta \right) \equiv\frac{1}{\sin \theta} </math>
|}{{clearright}}

To Begin learning about trigonometric functions, lets review the unit circle:

[[File:UnitCircle2.gif|center|frame|Caption: The Unit Circle.]]

The x coordinate on the unit circle is cos(x), and the y coordinate is sin(x)
(x,y) = (cos(x), sin(x))

Because cos(x) and sin(x) is a point on the unit circle, its coordinate must satisfy the equation of the circle: x^2+y^2 = 1

Therefore, Sin^2(x)+cos^2(x) = 1

The unit circle has a radius = 1
The circumference of the unit circle = 2π radians

each complete rotation of the unit circle will add 2π or 360°, for instance 2 complete revolutions of the unit circle = 4π or 720°,

On the Unit Circle, x and x + 2π correspond to the same point, this fact means that the graphs of y=sin(x) and y=cos(x) repeat over and over again, and gives us sin(x) = sin(x + 2kπ), where k is any whole number and represents the number of revolutions. From this we can say that sin(x) has a period of 2π. This can also be said about cos(x).

Converting Degrees to Radians:
Angle *(2π/360°)
for example: Convert 27° into radians:
27°*(2π/360°)= (3/40)*2π = 3π/20

Converting Radians to Degrees:
Radian * n(π), where π is equal to 180°
for example: Convert 3π into degrees
3π = 3(180°) = 540°

'''Sine graphs'''

If sin(x) is plotted on a set of "x" and "y" axis, the graph of sin(x) would be drawn as shown below:

http://i56.tinypic.com/2mmxnq0.jpg

The demonstration above is to show how the values from the unit circle can be plotted onto a set of "x" and "y" axis. The graph is not limited, it depends on a given scope. In other words, as long as the range of the sin function is not given, sin(x) graph will repeat as a wave until infinite. The sin(x) graph below shows for x values from 0 to 3π(540°).In order to calculate the points where slope = 0, we must make equate the derivate of sin(x) = 0.

http://i54.tinypic.com/okbkpd.jpg

'''Cosine graphs'''

Cosine have the exact same characteristics to that of a sine graph. The only difference between the two is that cosine function shifts to the right by a factor of 90° or π/2 in radians:

y = cos(x)

http://i55.tinypic.com/34hh2t2.jpg

As mentioned previously, cosine functions have the exact same behavior as the sine graph, and the only difference between them is that cosine functions shifts to the right by a factor of 90° or π/2 in radians. Mathematically speaking, if we were to expressed cos(x) in terms of sin(x), it would be written as; cos(x) = (sin(x+π/2)). The comparison between a sine and a cosine function is demonstrated as below;

http://i52.tinypic.com/s5920j.jpg

'''Tangent Graphs'''

http://i53.tinypic.com/2vd058x.jpg

==Basic Skills Project-Group Contribution==
http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_11/Basic_Skills

== Homeworks ==

HOMEWORK 4[[http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_11/Homework_4]]

[[Course:MATH110/003/Groups/Group 11/Information on Basic Skills|Information on Basic Skills]]

== Homework 3==

==Question 1==
Well in both instance the travel time were exactly the same. One was given in hour and minute and the second was given in minutes. We know that an hour is compose of 60 minutes so at 1hour and 20 minute we have 80 total minutes. The Second trip back was 80 minutes so both trips took equal time.

==Question 2==
The the reason the police did not stop the woman was because she was walking on the street, she was not driving. If the officer was to not stop her as she drove on to oncoming traffic he would not be doing his job. If the officer did not try stopping her it is obvious she was not breaking any laws.

==Question 3==
Well in the first two boxes you could pick a fruit up and figured it is either apples or oranges, it is a homogenous mixture in the first two boxes, but the third is a heterogenous mixture. I would personally grab into each box three times so i figure it is all the same or a mixture of both

==Question 4==
He is a brother to the fiddler because he too is blind.

==Question 5==
The moving quarter has made one full revolution around the fixed coin.

== Question 6==
If we let the 3 kinds of apples be X, Y and Z respectively, we know from the question that all these types of apples will sum up to a value; let that value be a.
Our goal is to have at least two of the same type of apple, which leads to our question-how many apples do we need to draw in order to get a matching pair of apples?

In our attempt to find a solution we found the probability of drawing any type of apple out of the basket, concluding that it is necessary to draw 3 times minimally to have any chance of a matching pair being taken out.

On the 1st draw:
1/a chances of getting any of the types of apples.

1 apple drawn from a basket filled with a apples.

On the 2nd draw:
1/(a-1) chances of getting any of the types of apples, however since you have drawn one out already, one of the 3 types is short an apple, reducing the total by 1.

Another apple drawn from the basket filled with (a-1) apples.

On the 3rd draw:
1/(a-2) chances of getting any type of the apples, with one group short of an apple, possible the same group that you drew an apple from in the 1st and/or 2nd group, resulting in a matching pair of apples.

A third apple drawn from the basket filled with (a-2) apples.

The necessary information that are lacking in this question are the total number of apples in the basket and the total number of each type of apple. Polya's method was only useful in sorting out our thinking. In order to apply probability and combinatoric concepts and further Step 3 of Polya's method, more information is required.--[[User:BernadetteHii|BernadetteHii]] 17:55, 16 October 2010 (UTC)
== Question 7==
We know that there are 80 socks in a drawer, 40 blue socks and 40 brown socks.

i.) Our goal in the first half of the question is to draw out a matching pair, how many socks need to be taken from the drawer to ensure a matching pair is found?

Similarly to the preceding question, we use probability of drawing any of the socks out of the drawer, concluding that it is necessary to draw 3 socks, minimally, to have a pair of matching socks.

On the 1st draw:
The probability drawing a blue or brown sock is 1/80.

1/40 chance of drawing a blue sock
1/40 chance of drawing a brown sock

On the 2nd draw:
1/79 chance of drawing a blue or brown sock, however the amount of one colour of socks is not reduced by 1.

If we take the two scenarios:
a.) 1st draw=blue sock, you now have a 1/39 chance of drawing a blue sock and a 1/40 chance of drawing a brown one on the 2nd draw.
b.) 1st draw=brown sock, you now have a 1/39 chance of drawing a brown sock and a 1/40 chance of drawing a blue one on the 2nd draw.

On the 3rd draw:
1/78 chance of drawing a blue or brown sock, however the amount of one of the colours has been reduced again by 1 sock.

a.) 2nd draw=blue sock, you now have a 1/38 chance of drawing a blue sock on the 3rd draw, assuming the 1st draw also was a blue sock
b.)2nd draw=brown sock, you now have a 1/38 chance of drawing a bluce sock on the 3rd draw assuming the 1st draw was also a brown sock.

As more socks are taken from the drawer, one by one, the amount of one colour of sock reduces. Since the picking is at random one colour will not necessarily decrease faster than the other, assuming that they are mixed in the drawer not nicely laid out. Therefore, to ensure a matching pair is picked, 3 socks minimally need to be taken from the drawer.

ii.) To draw an unmatched pair of socks from this drawer, simply draw two socks. Picking is at random and assuming the socks are mixed up and not laid out within the drawer, drawing two socks is sufficient enough to two different socks are taken out.

1st draw:
1/40 chance of getting a blue sock
1/40 chance of getting a brown sock

On the first draw there is an equal chance of drawing either a blue or brown sock.

2nd draw:
1/40 chance of getting a brown sock
1/40 chance of getting a blue sock

On the second draw there is also an equal chance of drawing either a blue or brown sock, assuming that the sock draw the first time, is the opposite colour of the sock drawn the second time.--[[User:BernadetteHii|BernadetteHii]] 17:55, 16 October 2010 (UTC)
== Question 8==
If we assume Reuben's birthday is December 31, then on December 31 he turned 21 years old.
If it is January 1, 2 days ago the would have been December 30. On that day he was still 20 years old. By the end of this year he will turn 22 years old. One year from the day January 1 he will be 23.

Polya's method seems to help organize the wording in this question making it easier to interpret the information the question gives us.--[[User:BernadetteHii|BernadetteHii]] 17:55, 16 October 2010 (UTC)
== Question 9==
From the question we know that the distance between each rung is 1 foot. If 10 rungs are showing we can assume that these 10 rungs are hanging off the side of the boat. Also, we assume that these 10 rungs are not submerged in water yet. If the tide rises 5 feet, this is equivalent to the distance between the bottom rung and the 5th rung of the ladder, therefore 5 rungs are now submerged in water, leaving the 5 upper rungs out of the water.

Similar to question 8, Polya's method is helpful to organize the wording in this question and to deduce from the question an appropriate scenario, in this case, 10 rungs of the ladder on the side of a boat, not covered by water.--[[User:BernadetteHii|BernadetteHii]] 17:55, 16 October 2010 (UTC)
== Question 10==
Polya's method helps start the question off by helping us interpret what the question is telling us.
i.) one-half the population of women can be written as
(1/2) (1/2)=(1/4)

If we know half the population are women, then it follows the other half of the population are men.
One-half the population also equals
(1/2)(1/2)=(1/4)

Knowing that half the population are chocolate eaters, we an take (1/4) of male chocolate eaters and (1/4) of female chocolate eaters, adding these together we get

(1/4)+(1/4)=(1/2), which is consistent with the information given in the question.

ii.) One-half of all men is
(1/2)(1/2)=(1/4)

therefore it follows that 1/4 the population of men are chocolate eaters.--[[User:BernadetteHii|BernadetteHii]] 17:55, 16 October 2010 (UTC)

== Question 11 ==
A woman, her older brother, her son, and her daughter are chess players. The worst player's twin, who is one of the four players, and the best player are of opposite sex. The worst player and best have the same age. If this is possible, who is the worst player?

The first step in solving this problem is understanding what the question is asking us and determining what information we are given. The question is asking us to solve out of four possible players, which on is the worst.

The information we are given is:
-The twin of the worst player is the opposite sex of the best player
-the worst and best player have the same age

Through this information we can determine that it is not possible to solve for the worst player, if one of the twins is the worst player and the best player is the same age and opposite gender of the other twin, there is no way of determining which twin is the worst player; also, no other player fits the description of the best player.

== Question 12 ==
A Manhattan fellow had a girlfriend in the Bronx and a girlfriend in Brooklyn. He decided which girlfriend to visit by arriving randomly at he train station and taking the first of the Bronx or Brooklyn trains that arrived. The trains to Brooklyn and the Bronx each arrived regularly every 10 minutes. Not long after he began his scheme the man's Bronx girlfriend left him because he rarely visited. Give a (logical) explanation.

This question is asking us to conclude why it was that the man rarely visited his Bronx girlfriend in order to determine why he was broken up with.

The man's daily routine probably meant he arrived at the train station at a similar point every time, the train's also run on a consistent schedule. Therefore, it is likely that the Brooklyn train schedule matched the Manhattan fellow's typical schedule more often; consequently, he rarely visited his Bronx girlfriend.

== Question 13 ==
If a clock takes 5 seconds to strike 5:00 (with 5 equally spaced chimes), how long does it take to strike 10:00 (with 10 equally spaced chimes)?

x= #chimes/second

x= 5 seconds/ 5chimes = 1 chime/ second

from this we can conclude that the clock chimes at a rate of 1 chime per second; therefore, if we want to solve for the time it would take to strike 10:00 (with 10 equally spaced chimes), we divide number of chimes by the rate of chimes:

10chimes/ 1chime/s = 10 seconds

== Question 14 ==
One day in the maternity ward, the name tags for four girl babies become mixed up.
i)How many different ways could two of the babies be tagged correctly and two incorrectly?

The question is asking us to find the number of ways two correct and two incorrect names can be arrangged. There are four names for the four babies. Let the correct order be ABCD. If there are two correct name orders and two incorrect these are our possible arrangements: ABDC, ADCB, ACBD, BACD, DBCA, CBAD. There are 6 possible arrangements given that 2 of the names are correct and the other two are not.

ii) In how many ways could three of the babies be tagged correctly and one baby be tagged incorrectly?

This question wants to know how many, if at all, ways can three of the baby names be correct and one incorrect. If the correct order of the names is ABCD, and ABC are correct the only possible name assignment for the fourth baby is D; therefore there are no possible arrangements for three correctly ordered names and one incorrect.

== Question 15 ==
Alex says to you, "I'll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck." Should you accept his bet?

This question is asking if Alex's assumption that the number of red cards in the first half of a split deck as there are black cards in the second half of a split deck is correct.

To solve, let
R= red cards
B= black cards

R(in first half deck) = B(in second half deck)

52 cards = R+B, half a deck = 26 cards

if,
1st half of deck has 16R + 10B, that leaves 16B +10R for the 2nd half of the deck

or,
1st half of deck has 2R+24B, that leaves 2B +24R for the 2nd half of the deck

No, you should not accept his bet because you would owe him a lot of money. Alex's statement that the number of red cards in the first half of the deck is equal to the number of black cards in the second half of the deck is true. Half a deck = 26 cards, so if you have 14 red cards in your first half of the deck that means the remaining 12 cards are black; therefore, in the second half of the deck, there are 14 black cards and 12 red cards remaining. So the number of red in the first half of the deck is equal to the number of black in the second half of the deck.

== Qustion 16 ==
Suppose that each daughter in your family has the same number of brothers as she has sisters, and each son in your family has twice as many sisters as he has brothers. How many sons and daughters are in the family?

Assume that total kids: 7
Boys: 3
Girls: 4
From this each son would have twice as many sisters as he has brothers.

== Question 17 ==
The zero point on a bathroom scale is set incorrectly, but otherwise the scale is accurate. It shows 60 kg when Dan stands on the scale, 50 kg when Sarah stands on the scale, but 105 kg when Dan and Sarah both stand on the scale. Does the scale read too high or too low? Explain.

Theoretically, if both Dan and Sarah stand on the scale, it should have read “110” kg. Since it reads 105kg instead of 110 kg, it shows that the scale reads too high since it should reduce about 2.5 kg for each individual for it to be 105kg. Therefore, the actual reading for Dan should be 57.5 kg and as for Sarah’s it should be 47.5 kg. Both of these weights make up to 105kg.

== Question 18 ==
Alice takes one-third of the pennies from a large jar. Then Bret takes one-third of the remaining pennies from the jar. Finally, Carla takes one-third of the remaining pennies from the jar, leaving 40 pennies in the jar. How many pennies were in the jar at the start?

In order to calculate the total pennies in the jar, one has to work backwards. Starting from the time Carla takes 1/3 of the pennies from the jar. In order to find the previous amount of pennies in the jar before Carla took 1/3 of it, the amount left must be divided by a factor of 1/3. This can be written below as “x” represents the amount of pennies before Carla took 1/3 of it;
1/3 of pennies = 40
Pennies = 120
This number would then be divided by a factor of 1/3 to calculate the amount before Bret took 1/3 of it;
1/3 of pennies = 120
Pennies = 360
Then 360 would be divided again by the same factor of 1/3 to find the original amount of pennies in the jar;
1/3 pennies = 360
Pennies = 1080
In conclusion, there were 1080 pennies originally in the jar.

== Question 19 ==
One morning each member of Angela's family drank an eight-ounce cup of coffee and milk, with the (nonzero) amounts of coffee and milk varying from cup to cup. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. What is the least number of people in the family?

If Angela consumes a quarter of the total amount of milk, that means there could only be a minimum of 4 number of people in the family. Since (1/4) x 4 makes 1 carton of milk, it is reasonable to conclude that the least number of people in Angela’s family would be 4 members.

== Question 20 ==
Of two clocks next to each other, one runs 5 min per hour fast and the other runs 5 min per hour slow. At midnight the clocks show the same time. At what time are they are one hour apart?

At every hour, the first clock runs about 5 minutes faster, and the second clock runs 5 minutes per hour slow. Therefore, for each consecutive hour, the first clock would gain 5 minutes while the second clock would lose 5 minutes. As a result, the total of minutes gained by the first clock for every hour is 10 minutes compared to the second clock. Because there is 60 minutes in an hour, and the first clock gains a lead of 10 minutes every hour, the two clocks would be an hour apart once the first clock strikes 18:00.

== Question 21 ==
Sven was in ninth place among seventeen runners in the marathon. It states that Sven was in the exact middle of placement in the marathon, which automatically narrowed the number of racers down to an odd number. This is because any single odd number is surrounded by the same amount of even numbers, so in this case of 9th, there are 8 racers before him, and 8 racers after him, leading the number to seventeen racers. The insert about Dan finishing in 10th place ruled out 19 as another potential number of racers going with the same logic as before. And Lars finishing in 16th place made it obvious there was more than 16 racers. Now there could not have been more racers as with Sven finishing ahead of Dan in 10th place, and being in the exact middle, there could not possibly have been more than 17 racers.

== Question 22 ==
Given that the vacation can be no shorter than 13 days, as it states that the rain cannot endure for the entire day. In the morning if it rains, it is followed by a sunny afternoon, and if it rains in the afternoon it is sunny in the morning. There are two distinct time frames, morning and afternoon, per day. Since there are only 2 slots per day, it makes sense to add the numbers of each type of day; 13 periods of rain, 12 sunny afternoons, and 11 bright mornings. After the addition you come out to the number of 36, and since there are 2 time slots per day, 36/2 is equal to 18 days of vacation.

== Question 23 ==
Yes you would be able to find the ages of the children. It would be necessary however to know the date at least. The number set for dates alone is much to factor in, from the 1 - 31 at most. However definitely not as high as 31 since the product of the three children's ages is 36. The date narrows it down by quite a bit. It would be simple and very possible to come up with a list of viable ages through simple calculations, but to narrow that list down the date would help. For example; (9,2,2), (2,3,6), (18,2,1), (4,3,3), (6,6,1), (12,3,1) etc ...

== Question 24 ==
Considering the second candle took 3 hours to burn out, half the time it took the first candle to burn out (6 hours). I made each candle into a standard and equal fraction of 6/6. Then I began by reducing the first candle to 5/6. Since the first candle lasted twice as long as the second candle, it is safe to assume that the second candle would go down 2 fractional points compared to the first candle, making it 4/6. You do this for one more interval and you find that when the first candle is equal to 4/6 the second candle is down to 2/6, half the value of the first. So doing some simple adjustments, the first candle lasted for 6 hours, and it is double the height of the second candle at fractions 4/6, therefore it took 2 hours before the first candle was twice as big as the second candle.

== Question 25 ==
For this question, I started out by figuring out how long each candle had lasted. The L + 1 candle lasted for 6 hours, whereas the L candle lasted for 4 hours. It could then be concluded that the L + 1 candle lasted 3/2 longer than the L candle. And since (L + 1)/L = 6/4, by cross-multiplying this equation, you see that 6L = 4(L + 1), which in turn becomes 6L = 4L + 4. By subtracting 4L from both sides to get rid of it on the right side, the equation then becomes 2L = 4. With some simple division, you discover that L = 2