Math+103+Section+11+Exam+_2

Math+103+Section+11+Exam+_2 - Math 103 Exam #2 April 19,...

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Math 103 Exam #2 April 19, 2011 Name________________________ There are 11 problems and 2 extra credit problem. Good luck and have fun! Show me all you’ve learned ! 1. 10 points Alex, Judy, and Lois are dividing the 18 inch long sub which has a 12 inch vegetarian piece and a 6 inch meat piece. The sub is shown in the above picture. The three friends are dividing the sub using the lone-chooser method . Alex likes the vegetarian and meatball parts equally well. Judy likes the meatball part four times as much as she likes the vegetarian part. Lois likes the meatball part twice as much as she likes the vegetarian part. Alex and Judy are the dividers and Lois is the chooser. In the first division, Judy divides the sub into two shares (a left share s1 and a right share s2). a. After Judy cuts, Alex gets to choose. Specify which of the two shares Alex should choose, and give the values of the shares to Judy and Alex. Put your answers in the following table:
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Shares s1=[ , ] s2=[ , ] value to Judy (as a fraction) value to Alex (as a fraction) a. What are s1 and s2 in interval notation? That is, s1=[ , ]? s2=[ , ]? Alex chooses s1 or s2? b. Describe how Judy would then subdivide her share into three pieces. Give the three pieces using interval notation . c. Describe how Alex would then subdivide his share into three pieces. Give the three pieces using interval notation . d. Describe the final fair division of the sub and give the value of each player’s share (as a fraction of the total value of the sub in the eyes of the player receiving it) by filling in the following table: Player Shares received at end Value as a fraction Judy Alex Lois 6 points
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2. a. In the divider-chooser method, why do we start with a coin toss? It is better to be a chooser than a divider. The coin toss is used to determine who is the chooser. b. True or false. ( You do not need to give a reason.) In the divider-chooser method, the chooser can end up with a value as small as 50% and as large as 100%. True The chooser is always guaranteed a 50% value. Depending on how the divider divides, the chooser may end up with a 100% value. For example, suppose that the divider is dividing a half chocolate, half vanilla cake and the divider likes each part equally, and the divider cuts the cake so that one piece is all chocolate and the other piece is all vanilla. Now suppose that the chooser is allergic to chocolate.
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This note was uploaded on 01/20/2012 for the course MATH 103 taught by Professor Berkowitz during the Spring '07 term at Rutgers.

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Math+103+Section+11+Exam+_2 - Math 103 Exam #2 April 19,...

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