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103s11chap4Asolutions

# 103s11chap4Asolutions - Math 103 Spring 2011 Solutions to...

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Math 103, Spring 2011, Solutions to Chapter 4A homework 8 points 4. The “states” in any apportionment problem are the entities that will have “seats” assigned to them according to a share rule. In this case, the states are the 4 shifts and the seats are the 225 nurses. (a) The standard divisor equals the total population divided by the number of seats, i.e. the total number of patients divided by the number of nurses, which 871 + 1029 + 610 + 190 12 225 = = (b) The standard divisor represents the average number of patients per nurse . Another way to phrase this is that if every nurse cared for exactly the same number of patients, that number of patients would be 12. (c) A has standard quota 871 72.583 12 = ; B has standard quota 1029 85.75 12 = ; C has standard quota 610 50.833 12 = ; D : 190 15.833 12 = 6 points 14. Applying Hamilton's method leads to the following results: Shift Standard quota Lower standard quota Fractional part Hamilton apportionment (number of nurses) A 72.583 72 .583 72 B 85.75 85 .75 86 C 50.833 50 .833 51 D 15.833 15 .833 16 Total: 225 222 3 225 The last column indicates how many nurses are apportioned to each shift under Hamilton's method.

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103s11chap4Asolutions - Math 103 Spring 2011 Solutions to...

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