sol11 - AMS 301 (Spring 2011) Homework Set # 11 Solution...

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Unformatted text preview: AMS 301 (Spring 2011) Homework Set # 11 Solution Notes Problem A: Let M be the math club and C the chess club, then N ( M ) = 15, N ( C ) = 12. Since 13 members belong to exactly one of the clubs, we know that N ( M C ) N ( M C ) = 13. Also, N ( M C ) = N ( M ) + N ( C ) N ( M C ). Subtracting N ( M C ) from both sides of the equation we get N ( M C ) N ( M C ) = N ( M ) + N ( C ) 2 N ( M C ). Plugging in the values we know, we get: 13 = 15 + 12 2 N ( M C ), so N ( M C ) = (15 + 12 13) / 2 = 7. # 16, 8.1: (a). Yes. If 20 like all 3, then 25 = 45 20 like any pair, (but not all three) therefore, 60 20 25 25 = 10 like only tennis, and this is impossible! (b). Let A be the percent that like all. We must have that A 45, and then the number of people that like only chess is 50 45 45 + A = A 40 0, so A 40....
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This note was uploaded on 05/28/2011 for the course AMS 301 taught by Professor Estie during the Spring '11 term at University of Florida.

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