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Course:MATH110/Archive/2010-2011/003/Groups/Group 15/Homework 4

2010-10-20T08:30:52Z

PeterDabrowski: /* Question 3 */

== Question 1 ==

'''Five persons named their pets after each other. From the following clues, can you decide which pet belongs to Suzan's mother? Tosh owns a cat, Bianca owns a frog. Jaela owns a parrot. Jun owns a snake. Suzan is the name of the frog. The cat is named Jun. The name by which they call the turtle is the name of the woman whose pet is Tosh. Finally, Suzan's mother's pet is Bianca.'''

Polya’s law is about ways to tackle a problem successfully so it should apply here as well. Step one: Read the question carefully. The question, when broken down, implicitly asks us two questions in place of one, because they both need to be answered before we can answer the original question. They are: 

1. Who is Suzan’s mother? 
2. What pet does Suzan’s mother own? (which is also the “ultimate” question) 

From reading the question we are already given a useful clue. That is, Suzan’s mother, must not be Suzan. We can forget about Suzan and narrow the list of five candidates down to four (though this doesn't prove particularly useful for the problem at hand as you will be able to see). We then work down the list of other clues, basically taking them, a better description wanting, at face value (and easy, too!). By simple logic we know that Suzan’s pet is a turtle (because it is the only pet left without an owner). So we get the following data in the form of a sorted table (Polya’s Step 2: Plan and map out your strategy in a chart, table, graph, etc.). 
[[File:Picture1.png]]

We then try to find out the name of Suzan’s turtle. It is the name of the person whose pet is called Tosh. Tosh can be Jaela’s parrot or Jun’s snake, but not Tosh’s cat nor Bianca’s frog. (It cannot be Suzan’s turtle either unless the question is playing tricks with our head... A rather dumb trick that would be too, wouldn’t it?) So, if we express this in our table: 
[[File:Picture2.png]]

So, is it Jaela or Jun? Do we have enough data to determine that yet? ( Step 1: Identify if enough data is given to solve the problem) Turns out we do. Jun cannot be Suzan’s turtle’s name, because we already know that Tosh’s cat is called Jun. Therefore, the turtle’s name must be Jaela. 

That leaves two more unknowns on our table to be determined: the names of Jaela’s parrot and Jun’s snake. They can only be either Tosh / Bianca (because that’s the only two names left – simple!) To find the next clue, we re-visit the original clues – it is important to revisit the question since as you are reading anything, as more information emerges the original material seems to present new meanings as you read it again because now you can read into deeper layers of meaning with additional perspective. So, since the turtle’s name is also the name of the woman who’s pet is Tosh, and Jaela being the turtle’s name, we conclude that the name of Jaela’s parrot is Tosh. That leaves Jun’s pet, whose name must be Bianca. 
[[File:Picture3.png]]

I believe you can now confidently answer the original question, which is “what pet belongs to Suzan’s mother”. We know from the beginning that Suzan’s mother’s pet is Bianca. When we look for Bianca in the table we’ve drawn up to tackle the problem, we can answer with a lot of confidence that Suzan’s mother is Jun, and perhaps more importantly (to answer the question directly), Suzan’s mother’s pet is a snake ("don't mess with him")...apparently a boy snake with a girly name!

== Question 2 ==

Bohao, Tim, Dylan, Chan and Stewart

5 players

3 of them are right handed
2 of them are left handed

3 of them are under 2m
2 of them are over 2m

We are looking for the centre player who is left handed and also over 2m.

-Tim or Chan must be right handed, because Dylan and Bohao are both right handed and Stewart is left handed.
-Bohao is over 2m tall so this means that either Dylan or Tim must be the same height as Chan and Stewart who are both under 2m.
-You can already see a common trend developing, the fact that we are trying to find both the variables for Tim's height as well as his handedness.
-This then narrows down to Dylan and TIm to be over 2m
-Dylan is right handed though

This leads us to the conclusion that Tim is the centre player because he is the only valid option, Dylan who is over 2m is right handed and therefor does not fit the description.

== Question 3 ==

Adam, Bobo, Charles, Ed, Hassan, Jason, Mathieu, Pascal, and Sung have formed a basketball team. The facts are in BOLD and their interpretations after them:

'''Adam does not like the cather,'''
-The catcher isnt adam
'''Ed's sister is engaged to the second baseman,'''
-ED isnt the second baseman
'''The center fielder is taller than the right fielder,'''
-Look at last point
'''Hassan and the third baseman live in the same building,'''
-Hassan aint the third baseman
'''Pascal and Charles each won $20 from the pitcher at a poker game,'''
-Pascal and Charles are not the pitcher
'''Ed and the outfielders play cards during their free time,'''
-ED is not an outfielder
'''The pitcher's wife is the third baseman's sister,'''
-The pitcher is married
'''All the battery and infield except Charles, Hassan and Adam are shorter than Sung,'''
-Charles, Hassan and Adam are not the outfielder but could be the battery or infield
'''Pascal, Adam and the shortstop lost $100 each at the race track,'''
-Pascal and Adam are not the shortstop
'''The second baseman beat Pascal, Hassan, Bobo and the catcher at billiards,'''
-So Pascal, Hassan and Bobo are not the second basemen and the catcher
'''Sung is in the process of getting a divorce,'''
-Sung is married
'''The catcher and the third baseman each have two legitimate children,'''
-They are both married
'''Ed, Pascal, Jason, the right fielder and the center fielder are bachelors, the others are all married,'''
-Ed, Pascal, Jason right fielder, center fielder are all not married.
-Ed, Pascal, Jason are neither the right fielder or center fielder.
-The pitcher the catcher are married
-Ed is not the third basemen, the pitcher, or the catcher
-Jason is not the third basemen, the pitcher, or the catcher
-Sung is not the right outfield, the out field center, or second basemen
'''The shortstop, the third baseman and Bobo all attended the fight,'''
-Bobo is neither the shortstop or the third baseman
'''Mathieu is the shortest player of the team,'''
-So he is obviously not the center fielder

If you plot all of the above on a table you find out that:

Catcher = Charles
First Baseman = Ed
Center Fielder = Bobo
Pitcher = Hassan
Second Baseman = Jason
Right Fielder = Mathieu
Left Fielder = Pascal
Shortstop = Sung
3rd Baseman = Adam

== Question 4 ==
'''Six players - Petra, Carla, Janet, Sandra, Li and Fernanda - are competing in a chess tournament over a period of five days. Each player plays each of the others once. Three matches are played simultaneously during each of the five days. The first day, Carla beats Petra after 36 moves. The second day, Carla was again victorious when Janet failed to complete 40 moves within the required time limit. The third day had the most exciting match of all when Janet declared that she would checkmate Li in 8 moves and succeeded in doing so. On the fourth day, Petra defeated Sandra. Who played against Fernanda on the fifth day?'''

1). First of all we sort out the details given by the question:
6 players-> P, C, J, S, L and F.

Tournament= 5 days

->each player plays each of others ONCE.

->3 matches are played in 5 days

Therefore, we know that 2 people play against each other in each day.

1st day: '''C''' vs P ( C wins) J,S,L,F

2nd day: '''C''' vs J ( C wins) P,S,L,F

3rd day: '''J''' vs L (J wins) P,C,S,F

4th day: '''P''' vs S (P wins) C,J,L,F

5th day: F vs ???

Then, we use the info given, try to pair the unknown ones up:

1st day: CP, ( JS, ''L''F )

2nd day: CJ, ( PL, ''S''F)

3rd day: JL, ( ''P''F, CS)

4th day: PS, ( CL, ''J''F)

5th day: F havn't played with C according to the info above.

Therefore, Fernanda is playing against '''Carla''' on the 5th day.

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

By following the rule : “No one of the three noisemakers was quiet for three consecutive days” and “no pair of them made noise on more than one day” Tuesday, Thursday and friday are all unknown.
As we can see we could put it like below.

Saturday --> Salesman
Sunday --> Dog
Monday --> Salesman + Construction
Tuesday --> Sleep in.
Wednesday - Salesman + Dog
Thursday --> Construction
Friday --> Salesman
Saturday --> Construction
Sunday --> Dog

Course:MATH110/Archive/2010-2011/003/Groups/Group 15/Homework 4

2010-10-20T08:29:00Z

PeterDabrowski: /* Question 3 */

== Question 1 ==

'''Five persons named their pets after each other. From the following clues, can you decide which pet belongs to Suzan's mother? Tosh owns a cat, Bianca owns a frog. Jaela owns a parrot. Jun owns a snake. Suzan is the name of the frog. The cat is named Jun. The name by which they call the turtle is the name of the woman whose pet is Tosh. Finally, Suzan's mother's pet is Bianca.'''

Polya’s law is about ways to tackle a problem successfully so it should apply here as well. Step one: Read the question carefully. The question, when broken down, implicitly asks us two questions in place of one, because they both need to be answered before we can answer the original question. They are: 

1. Who is Suzan’s mother? 
2. What pet does Suzan’s mother own? (which is also the “ultimate” question) 

From reading the question we are already given a useful clue. That is, Suzan’s mother, must not be Suzan. We can forget about Suzan and narrow the list of five candidates down to four (though this doesn't prove particularly useful for the problem at hand as you will be able to see). We then work down the list of other clues, basically taking them, a better description wanting, at face value (and easy, too!). By simple logic we know that Suzan’s pet is a turtle (because it is the only pet left without an owner). So we get the following data in the form of a sorted table (Polya’s Step 2: Plan and map out your strategy in a chart, table, graph, etc.). 
[[File:Picture1.png]]

We then try to find out the name of Suzan’s turtle. It is the name of the person whose pet is called Tosh. Tosh can be Jaela’s parrot or Jun’s snake, but not Tosh’s cat nor Bianca’s frog. (It cannot be Suzan’s turtle either unless the question is playing tricks with our head... A rather dumb trick that would be too, wouldn’t it?) So, if we express this in our table: 
[[File:Picture2.png]]

So, is it Jaela or Jun? Do we have enough data to determine that yet? ( Step 1: Identify if enough data is given to solve the problem) Turns out we do. Jun cannot be Suzan’s turtle’s name, because we already know that Tosh’s cat is called Jun. Therefore, the turtle’s name must be Jaela. 

That leaves two more unknowns on our table to be determined: the names of Jaela’s parrot and Jun’s snake. They can only be either Tosh / Bianca (because that’s the only two names left – simple!) To find the next clue, we re-visit the original clues – it is important to revisit the question since as you are reading anything, as more information emerges the original material seems to present new meanings as you read it again because now you can read into deeper layers of meaning with additional perspective. So, since the turtle’s name is also the name of the woman who’s pet is Tosh, and Jaela being the turtle’s name, we conclude that the name of Jaela’s parrot is Tosh. That leaves Jun’s pet, whose name must be Bianca. 
[[File:Picture3.png]]

I believe you can now confidently answer the original question, which is “what pet belongs to Suzan’s mother”. We know from the beginning that Suzan’s mother’s pet is Bianca. When we look for Bianca in the table we’ve drawn up to tackle the problem, we can answer with a lot of confidence that Suzan’s mother is Jun, and perhaps more importantly (to answer the question directly), Suzan’s mother’s pet is a snake ("don't mess with him")...apparently a boy snake with a girly name!

== Question 2 ==

Bohao, Tim, Dylan, Chan and Stewart

5 players

3 of them are right handed
2 of them are left handed

3 of them are under 2m
2 of them are over 2m

We are looking for the centre player who is left handed and also over 2m.

-Tim or Chan must be right handed, because Dylan and Bohao are both right handed and Stewart is left handed.
-Bohao is over 2m tall so this means that either Dylan or Tim must be the same height as Chan and Stewart who are both under 2m.
-You can already see a common trend developing, the fact that we are trying to find both the variables for Tim's height as well as his handedness.
-This then narrows down to Dylan and TIm to be over 2m
-Dylan is right handed though

This leads us to the conclusion that Tim is the centre player because he is the only valid option, Dylan who is over 2m is right handed and therefor does not fit the description.

== Question 3 ==

Adam, Bobo, Charles, Ed, Hassan, Jason, Mathieu, Pascal, and Sung have formed a basketball team. The facts are in BOLD and their interpretations below them:

'''Adam does not like the cather,'''
-The catcher isnt adam
'''Ed's sister is engaged to the second baseman,'''
-ED isnt the second baseman
'''The center fielder is taller than the right fielder,'''
-Look at last point
'''Hassan and the third baseman live in the same building,'''
-Hassan aint the third baseman
'''Pascal and Charles each won $20 from the pitcher at a poker game,'''
-Pascal and Charles are not the pitcher
'''Ed and the outfielders play cards during their free time,'''
-ED is not an outfielder
'''The pitcher's wife is the third baseman's sister,'''
-The pitcher is married
'''All the battery and infield except Charles, Hassan and Adam are shorter than Sung,'''
-Charles, Hassan and Adam are not the outfielder but could be the battery or infield
'''Pascal, Adam and the shortstop lost $100 each at the race track,'''
-Pascal and Adam are not the shortstop
'''The second baseman beat Pascal, Hassan, Bobo and the catcher at billiards,'''
-So Pascal, Hassan and Bobo are not the second basemen and the catcher
'''Sung is in the process of getting a divorce,'''
-Sung is married
'''The catcher and the third baseman each have two legitimate children,'''
-They are both married
'''Ed, Pascal, Jason, the right fielder and the center fielder are bachelors, the others are all married,'''
-Ed, Pascal, Jason right fielder, center fielder are all not married.
-Ed, Pascal, Jason are neither the right fielder or center fielder.
-The pitcher the catcher are married
- Ed is not the third basemen, the pitcher, or the catcher
- Jason is not the third basemen, the pitcher, or the catcher
- Sung is not the right outfield, the out field center, or second basemen
'''The shortstop, the third baseman and Bobo all attended the fight,'''
-Bobo is neither the shortstop or the third baseman
'''Mathieu is the shortest player of the team,'''
-So he is obviously not the center fielder

If you plot all of the above on a table you find out that:

Catcher = Charles
First Baseman = Ed
Center Fielder = Bobo
Pitcher = Hassan
Second Baseman = Jason
Right Fielder = Mathieu
Left Fielder = Pascal
Shortstop = Sung
3rd Baseman = Adam

== Question 4 ==
'''Six players - Petra, Carla, Janet, Sandra, Li and Fernanda - are competing in a chess tournament over a period of five days. Each player plays each of the others once. Three matches are played simultaneously during each of the five days. The first day, Carla beats Petra after 36 moves. The second day, Carla was again victorious when Janet failed to complete 40 moves within the required time limit. The third day had the most exciting match of all when Janet declared that she would checkmate Li in 8 moves and succeeded in doing so. On the fourth day, Petra defeated Sandra. Who played against Fernanda on the fifth day?'''

1). First of all we sort out the details given by the question:
6 players-> P, C, J, S, L and F.

Tournament= 5 days

->each player plays each of others ONCE.

->3 matches are played in 5 days

Therefore, we know that 2 people play against each other in each day.

1st day: '''C''' vs P ( C wins) J,S,L,F

2nd day: '''C''' vs J ( C wins) P,S,L,F

3rd day: '''J''' vs L (J wins) P,C,S,F

4th day: '''P''' vs S (P wins) C,J,L,F

5th day: F vs ???

Then, we use the info given, try to pair the unknown ones up:

1st day: CP, ( JS, ''L''F )

2nd day: CJ, ( PL, ''S''F)

3rd day: JL, ( ''P''F, CS)

4th day: PS, ( CL, ''J''F)

5th day: F havn't played with C according to the info above.

Therefore, Fernanda is playing against '''Carla''' on the 5th day.

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

By following the rule : “No one of the three noisemakers was quiet for three consecutive days” and “no pair of them made noise on more than one day” Tuesday, Thursday and friday are all unknown.
As we can see we could put it like below.

Saturday --> Salesman
Sunday --> Dog
Monday --> Salesman + Construction
Tuesday --> Sleep in.
Wednesday - Salesman + Dog
Thursday --> Construction
Friday --> Salesman
Saturday --> Construction
Sunday --> Dog

Course:MATH110/Archive/2010-2011/003/Groups/Group 15

2010-10-13T08:32:11Z

PeterDabrowski: /* Question 21 */

Group members:
* Danny Choi
* Peter David Dabrowski
* Allie Miller
* Anabelle Tory
* Vivian Zhang

----

Homework 3

== '''Question 1''' ==

'''A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain. '''

1hour and 20min is identical to 80min, there is nothing else to explain.

== '''Question 2''' ==

== '''Question 3''' ==

== '''Question 4''' ==

== '''Question 5''' ==

== '''Question 6''' ==

== '''Question 7''' ==

== '''Question 8''' ==

== '''Question 9''' ==

== '''Question 10''' ==

== '''Question 11''' ==

== '''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 the 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.'''

The Manhattan fellow likely has a daily schedule along with a set time allotted to visiting his girlfriends. The trains also have a schedule so statistically speaking there will be a point of where both schedules meet up in favor of the Brooklyn train.

== '''Question 13''' ==

== '''Question 14''' ==

== '''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?'''

That depends if we are using the Jokers in the deck (54 cards), if we are then I would accept his bet, if we are not (52 cards) then I would not accept his bet. This is because if we were using the jokers, they are both black cards so the total ratio of black to red is 28Black to 26Red as a result it is only possible to have a different amount of black cards to red cards in each half of the deck. However if there were no jokers in the deck, then there would be a perfect ratio of 26Black to 26Red cards. If you were to shuffle them and split them in half there will be different amounts of red and black in each stack, however the amount of one color in one stack will always be the same as the opposite color in the opposite stack. The scenario without jokers is explained below.

There are 52 cards in total and <math> \tfrac{52}{2}=26 </math> cards in each half. Half of all cards are red and half of all cards are black.

In the first half there are ''x'' red cards and ''y'' black cards.

''x''+''y''=26

where

''x''=26-''y''

and

''y''=26-''x''

In the second half there are the red cards which are not in the first half 26-''x'' which is equal to ''y''. And there are the black cards which are not in the first half 26-''y'' which is equal to ''x''. Therefore there will always be as many red cards in the first half of the deck as there are black cards in the second deck.

== '''Question 16''' ==

== '''Question 17''' ==

== '''Question 18''' ==

== '''Question 19''' ==

== '''Question 20''' ==

== '''Question 21''' ==

'''Sven placed exactly in the middle among all runners in a race. Dan was slower than Sven, in 10th place, and Lars was in 16th place. How many runners were in the race?'''

There were 17 runners in the race. This is because Sven must be in less then 10th place (this is given). So assuming that he is in 9th place there are 8 people ahead of him, and because he is exactly in the middle there would have to be 8 people behind him all the way to 17th place (9+8) which would satisfy the given statement that there was somebody in 16th place. It would be impossible for Sven to place any lower then 9th place because then there would not be a 16th place for Lars to take.

== '''Question 22''' ==

== '''Question 23''' ==

== '''Question 24''' ==

== '''Question 25''' ==

Course:MATH110/Archive/2010-2011/003/Groups/Group 15

2010-10-13T07:48:29Z

PeterDabrowski: /* Question 15 */

Group members:
* Danny Choi
* Peter David Dabrowski
* Allie Miller
* Anabelle Tory
* Vivian Zhang

----

Homework 3

== '''Question 1''' ==

'''A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain. '''

1hour and 20min is identical to 80min, there is nothing else to explain.

== '''Question 2''' ==

== '''Question 3''' ==

== '''Question 4''' ==

== '''Question 5''' ==

== '''Question 6''' ==

== '''Question 7''' ==

== '''Question 8''' ==

== '''Question 9''' ==

== '''Question 10''' ==

== '''Question 11''' ==

== '''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 the 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.'''

The Manhattan fellow likely has a daily schedule along with a set time allotted to visiting his girlfriends. The trains also have a schedule so statistically speaking there will be a point of where both schedules meet up in favor of the Brooklyn train.

== '''Question 13''' ==

== '''Question 14''' ==

== '''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?'''

That depends if we are using the Jokers in the deck (54 cards), if we are then I would accept his bet, if we are not (52 cards) then I would not accept his bet. This is because if we were using the jokers, they are both black cards so the total ratio of black to red is 28Black to 26Red as a result it is only possible to have a different amount of black cards to red cards in each half of the deck. However if there were no jokers in the deck, then there would be a perfect ratio of 26Black to 26Red cards. If you were to shuffle them and split them in half there will be different amounts of red and black in each stack, however the amount of one color in one stack will always be the same as the opposite color in the opposite stack. The scenario without jokers is explained below.

There are 52 cards in total and <math> \tfrac{52}{2}=26 </math> cards in each half. Half of all cards are red and half of all cards are black.

In the first half there are ''x'' red cards and ''y'' black cards.

''x''+''y''=26

where

''x''=26-''y''

and

''y''=26-''x''

In the second half there are the red cards which are not in the first half 26-''x'' which is equal to ''y''. And there are the black cards which are not in the first half 26-''y'' which is equal to ''x''. Therefore there will always be as many red cards in the first half of the deck as there are black cards in the second deck.

== '''Question 16''' ==

== '''Question 17''' ==

== '''Question 18''' ==

== '''Question 19''' ==

== '''Question 20''' ==

== '''Question 21''' ==

== '''Question 22''' ==

== '''Question 23''' ==

== '''Question 24''' ==

== '''Question 25''' ==

Course:MATH110/Archive/2010-2011/003/Groups/Group 15

2010-10-13T07:37:54Z

PeterDabrowski: /* Question 12 */

Course:MATH110/Archive/2010-2011/003/Groups/Group 15

2010-10-13T07:34:33Z

PeterDabrowski: /* Question 1 */

User:PeterDabrowski

2010-09-20T08:04:43Z

PeterDabrowski:

Hi, I'm Peter and i'm majoring in economics!

Pythagorean Theorem:

The Pythagorean theorem is named after the Greek mathematician Pythagoras. Which is probably due to the convenience that we know his name as there is much evidence that this theory was known by the human race in the time of the Babylonian and Mesopotamian empires. This theory is very simple yet an extremely important accomplishment of the human race and a tribute to the power of the human mind to comprehend such ideas. It applies strictly to right triangles as opposed to the wrong ones and states that the longest side times itself is equal to the sum the the other two sides multiplied by themselves. c^2=a^2+b^2. this formula can also be rearranged algebraically to find the lengths of the other two sides. This formula has applications in a wide range of fields that include architecture, engineering, physics, mathematics etc. And I should be at 150 words right about now!

User:PeterDabrowski

2010-09-20T07:20:32Z

PeterDabrowski: Created page with 'Hi, I'm Peter and i'm majoring in economics!'

Hi, I'm Peter and i'm majoring in economics!