Mathematical Ideas (13th Edition) - Standalone book
Mathematical Ideas (13th Edition) - Standalone book
13th Edition
ISBN: 9780321977076
Author: Charles D. Miller, Vern E. Heeren, John Hornsby, Christopher Heeren
Publisher: PEARSON
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Chapter 15.3, Problem 5E

Solve each problem.

New Trees for Wisconsin Parks The schoolchildren of Wisconsin began a statewide effort one year to raise money to purchase new trees for five state parks. Altogether, the students raised enough money to purchase 239 trees. A committee decided that the trees should be apportioned based on the amount of land in each of the five parks as shown.

State park a b c d e
Acres 1429 8639 7608 6660 5157

(a) Find the total number of acres in the five state parks, and compute the standard divisor for the apportionment of the 239 trees.

(b) Use the Hamilton method to apportion the trees.

(c) Use the Jefferson method to apportion the trees. As always, the modified divisor needed for the Jefferson apportionment is found by slowly decreasing the value of the standard divisor computed in part (a). Use m d = 122 .

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Chapter 15 Solutions

Mathematical Ideas (13th Edition) - Standalone book

Ch. 15.1 - Applying Four Voting Methods to a Voter Profile...Ch. 15.1 - Applying Four Voting Methods to a Voter Profile...Ch. 15.1 - Applying Four Voting Methods to a Voter Profile...Ch. 15.1 - Applying Four Voting Methods to a Voter Profile...Ch. 15.1 - Applying Four Voting Methods to a Voter Profile...Ch. 15.1 - Applying Four Voting Methods to a Voter Profile...Ch. 15.1 - Applying Four Voting Methods to a Voter Profile...Ch. 15.1 - Applying Four Voting Methods to a Voter Profile...Ch. 15.1 - Holding a Runoff Election One common solution to...Ch. 15.1 - Prob. 20ECh. 15.1 - Prob. 21ECh. 15.1 - Prob. 22ECh. 15.1 - Prob. 23ECh. 15.1 - Prob. 24ECh. 15.1 - Prob. 25ECh. 15.1 - Prob. 26ECh. 15.1 - Prob. 27ECh. 15.1 - The Pairwise Comparison Method Each table...Ch. 15.1 - Prob. 29ECh. 15.1 - Prob. 30ECh. 15.1 - The Borda Method Each table represents a Borda...Ch. 15.1 - Prob. 32ECh. 15.1 - Prob. 33ECh. 15.1 - Prob. 34ECh. 15.1 - Prob. 35ECh. 15.1 - Prob. 36ECh. 15.1 - The Coombs Method The Coombs method of voting is a...Ch. 15.1 - Prob. 38ECh. 15.1 - Prob. 39ECh. 15.1 - Prob. 40ECh. 15.2 - Identifying Violations of the Majority Criterion...Ch. 15.2 - Identifying Violations of the Majority Criterion...Ch. 15.2 - Identifying Violations of the Majority...Ch. 15.2 - Identifying Violations of the Majority Criterion...Ch. 15.2 - Identifying Violations of the Condorcet...Ch. 15.2 - Identifying Violations of the Condorcet Criterion...Ch. 15.2 - Identifying Violations of the Condorcet Criterion...Ch. 15.2 - Identifying Violations of the Condorcet Criterion...Ch. 15.2 - Prob. 9ECh. 15.2 - Prob. 10ECh. 15.2 - Prob. 11ECh. 15.2 - Prob. 12ECh. 15.2 - Prob. 13ECh. 15.2 - Prob. 14ECh. 15.2 - Prob. 15ECh. 15.2 - Prob. 16ECh. 15.2 - Prob. 17ECh. 15.2 - Prob. 18ECh. 15.2 - Prob. 19ECh. 15.2 - Irrelevant Alternatives in a Hare Method Election...Ch. 15.2 - 21. Explain why a violation of the majority...Ch. 15.2 - Prob. 22ECh. 15.2 - Prob. 23ECh. 15.2 - Prob. 24ECh. 15.2 - Prob. 25ECh. 15.2 - Prob. 26ECh. 15.2 - Prob. 27ECh. 15.2 - Prob. 28ECh. 15.2 - Prob. 29ECh. 15.2 - Prob. 30ECh. 15.2 - Prob. 31ECh. 15.2 - Prob. 32ECh. 15.2 - Prob. 33ECh. 15.2 - Prob. 34ECh. 15.3 - Find each quantity (to the nearest whole number)...Ch. 15.3 - Find each quantity (to the nearest whole number)...Ch. 15.3 - Find each quantity (to the nearest whole number)...Ch. 15.3 - Find each quantity (to the nearest whole number)...Ch. 15.3 - Solve each problem. 5. New Trees for Wisconsin...Ch. 15.3 - Apportioning Computers to Schools Enrollments for...Ch. 15.3 - Assigning Faculty to Courses The English...Ch. 15.3 - 8. Apportioning Sailboats to Resorts The number of...Ch. 15.3 - Prob. 9ECh. 15.3 - 10. Show that the Webster method apportionment of...Ch. 15.3 - Prob. 11ECh. 15.3 - Prob. 12ECh. 15.3 - Prob. 13ECh. 15.3 - Prob. 14ECh. 15.3 - Prob. 15ECh. 15.3 - Find the Huntington-Hill cutoff point for rounding...Ch. 15.3 - Creating a Profile of School Bus Riders Create a...Ch. 15.3 - Prob. 18ECh. 15.3 - Prob. 19ECh. 15.3 - Prob. 20ECh. 15.3 - The standard quotas rounded up to the nearest...Ch. 15.3 - Prob. 22ECh. 15.3 - Prob. 23ECh. 15.3 - Prob. 24ECh. 15.4 - Quota Rule Violations with the Jefferson Method In...Ch. 15.4 - Quota Rule Violations with the Jefferson Method In...Ch. 15.4 - Quota Rule Violations with the Jefferson Method In...Ch. 15.4 - Quota Rule Violations with the Jefferson Method In...Ch. 15.4 - Alabama Paradox with the Hamilton Method In each...Ch. 15.4 - Alabama Paradox with the Hamilton Method In each...Ch. 15.4 - Alabama Paradox with the Hamilton Method In each...Ch. 15.4 - Alabama Paradox with the Hamilton Method In each...Ch. 15.4 - Population Paradox with the Hamilton Method In...Ch. 15.4 - Population Paradox with the Hamilton Method In...Ch. 15.4 - Population Paradox with the Hamilton Method In...Ch. 15.4 - Population Paradox with the Hamilton Method In...Ch. 15.4 - New States Paradox with the Hamilton Method In...Ch. 15.4 - New States Paradox with the Hamilton Method In...Ch. 15.4 - New States Paradox with the Hamilton Method In...Ch. 15.4 - New States Paradox with the Hamilton Method In...Ch. 15.4 - Violations of the Quota Rule? For each...Ch. 15.4 - Violations of the Quota Rule? For each...Ch. 15.4 - Violations of the Quota Rule? For each...Ch. 15.4 - Prob. 20ECh. 15.4 - Prob. 21ECh. 15.4 - Prob. 22ECh. 15.4 - Prob. 23ECh. 15.4 - Prob. 24ECh. 15.4 - Prob. 25ECh. 15.4 - 26. The Jefferson and Adams methods are both...Ch. 15 - How many different complete rankings are possible...Ch. 15 - Prob. 2TCh. 15 - Prob. 3TCh. 15 - Prob. 4TCh. 15 - Prob. 5TCh. 15 - Why is the irrelevant alternatives criterion an...Ch. 15 - Prob. 7TCh. 15 - Prob. 8TCh. 15 - Prob. 9TCh. 15 - Prob. 10TCh. 15 - Prob. 11TCh. 15 - Prob. 12TCh. 15 - Prob. 13TCh. 15 - Prob. 14TCh. 15 - Prob. 15TCh. 15 - Prob. 16TCh. 15 - Prob. 17TCh. 15 - Prob. 18TCh. 15 - Prob. 19TCh. 15 - Prob. 20TCh. 15 - Prob. 21TCh. 15 - Prob. 22TCh. 15 - Prob. 23TCh. 15 - Prob. 24TCh. 15 - Prob. 25TCh. 15 - One hundred seats are to be apportioned to 4...Ch. 15 - Prob. 27TCh. 15 - Prob. 28TCh. 15 - Prob. 29TCh. 15 - Explain the Alabama paradox.Ch. 15 - Prob. 31TCh. 15 - Prob. 32T
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