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 23E
To determine
The best method out of the given choice between the five appointment methods, Hamilton, Jefferson, Webster, Adams and the Huntington-Hill method.
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Now the city expands to cover a new district called district C. As a result, the city hires
10 new police officers to cover the new district. This brings the total number of officers
to 50. The total population of the city is shown in the table. Complete the table using
Hamilton's method to apportion the police offices for the populations after 2000. Round
all standard quotas and the divisor to 7 decimal places if needed. You can use a
spreadsheet to help you figure out an answer or calculate using a calculator.
Find the divisor. Enter your answer in the space provided below.
Divisor:
Population
2000
Initial Quotas
Final Quotas
State
Standard
Quotas
А
6,550
11,520
4,698
22,768
B
C
Total
21.)
7.) Did a new state paradox occur? Explain your answer.
Write your response below:
Suppose a nation has 5 states, with populations shown in the chart below. The
representative body had 150 seats.
Find:
the standard divisor,
the modified divisor, and
the distribution of representatives to each state.
Then complete the table. Round all standard and modified quotas to 7 decimal places if
needed.
You can use a spreadsheet to help you figure out an answer or calculate using a
calculator
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Chapter 15 Solutions
Mathematical Ideas (13th Edition) - Standalone book
Ch. 15.1 - Choosing a Poster Dog by the Plurality Method A...Ch. 15.1 - Choosing a Poster Dog by the Plurality Method A...Ch. 15.1 - Choosing a Poster Dog by Alternative Methods For...Ch. 15.1 - Choosing a Poster Dog by Alternative MethodsFor...Ch. 15.1 - Observing the Effect of the Number of Candidates...Ch. 15.1 - Observing the Effect of the Number of Candidates...Ch. 15.1 - Observing the Effect of the Number of Candidates...Ch. 15.1 - Observing the Effect of the Number of Candidates...Ch. 15.1 - Observing the Effect of the Number of Candidates...Ch. 15.1 - Observing the Effect of the Number of Candidates...
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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- The population for the following district is subdivided by region: Region 15 has 317 students Region 16 has 148 students Region 17 has 221 students If an organization selects 3 students to represent the district, calculate the standard divisor needed to be eligible for representation of the region. Round your answer to 2 decimal places.arrow_forwardA country consists of 20 states. There are 522 seats in the Congress which are to be apportioned among the 20 states to their respective population. The following table shows the population of the 20 states. Use a spreadsheet program to answer the following questions. Round all standard quotas, modified quotas, and the divisor to 5 decimal places if needed. You can use a spreadsheet to help you figure out an answer or calculate using a calculator. Stat Populati Standa Adams' Webste Modifi Modifi Adams' Webste Advantage on rd S r's ed ed S r's or e Quotas Rounde Rounde Quotas Quotas Modifi Modifi Disadvant d Up Quotas Quotas Adam Webste Rounde Rounde d for for ed ed age A 413,875 B 500,032 397,004 D 360,874 E 193,993 F 131,279 G 76,121 H 297,022 I 94,239 J 423,566 K 565,883 L 473,525 M 399,170 N 403,576 100,904 P 73,143 Q 360,997 R 410,181 S 456,755 T 139,361 Tota 6,271,50 Seats: 522arrow_forwardA country consists of 20 states. There are 386 seats in the Congress which are to be apportioned among the 20 states to their respective population. The following table shows the population of the 20 states. Use a spreadsheet program to answer the following questions. Round all standard quotas, modified quotas, and the divisor to 5 decimal places if needed. You can use a spreadsheet to help you figure out an answer or calculate using a calculator. Seats: 386 Hamilton Jefferson's Modified Modified Lower Quotas Disadvantage Jefferson State Population Standard Initial Benefit and Quotas Quota without Quotas decimal A 538,723 61,837 C 346,146 448,119 D E 500,521 F 89,678 G 221,803 H 314,404 I 41,754 J 373,154 K 94,310 399,190 M 297,083 391,098 163,551 218,689 Q 329,896 R 77,150 S 113,753 T 101,301 Total 5,122,220arrow_forward
- Twenty sections of bilingual math courses, taught in both English and Spanish, are to be offered in introductory algebra, intermediate algebra, and liberal arts math. The preregistration figures for the number of students planning to enroll in these bilingual sections are given in the following table. Use Webster's method with d = 29.6 to determine how many bilingual sections of each course should be offered. Course Introductory Algebra| Intermediate Algebra Liberal Arts Math Enrollment 139 271 182 ..... Course Introductory Algebra Number of Apportioned Sections 5 (Type an integer.) Course Intermediate Algebra Number of Apportioned Sections (Type an integer.)arrow_forwardplease answerarrow_forwardThe following table shows the number of fifth and sixth grade teachers in a school district and the number of students in each of those grades. The number of teachers for each of the grade levels was determined by using the Huntington-Hill apportionment method. The district has decided to hire a new teacher for either the fifth or sixth grade. Fifth grade Sixth grade Number of teachers 21 22 Number of students 649 730 (a) Use the apportionment principle to determine to which grade the new teacher should be assigned. O fifth grade O sixth grade (b) Use the Huntington-Hill apportionment principle to determine to which grade the new teacher should be assigned. O fifth grade O sixth grade How does this result compare with the result in part (a)? O same result O different resultarrow_forward
- A company has four divisions with 560, 1230, 1490, and 1760 people, respectively. A total of 18 IT workers must be allocated to each division according to their size. Complete parts (a) through (d) to find the apportionment using the Hill-Huntington method. Find the standard divisor. Find the standard quota and modified standard quota for each division. Using a modified standard divisor of 275, find the modified quota for each division. Find the geometric mean for each division. Show the rounded quota (according to the Hill-Huntington method) for each division and the final apportionment of IT workers. PLEASE WATCH GRAMMAR, PUNCTUATION, AND HAVE CLEAR FORMATTING. Thank you.arrow_forwardJefferson's method, Webster's method, and Adams's method require using a _______ quota. arrow_forwardUse Adamss's method of apportionment to solve the following problem. 1. Suppose a nation has 6 states, with populations shown in the chart below. The representative body had 200 seats. Find: the standard divisor, the modified divisor, and the distribution of representatives to each state. Then complete the table. Round all standard and modified quotas to 7 decimal places if needed. You can use a spreadsheet to help you figure out an answer or calculate using a calculator. Representative Seats State E F Total Standard Divisor Modified Divisor 200 Population Standard Rounded Modified Quotas Up Quotas Quotas 1,598,400 1,236,300 5,460,200 826,900 965,500 1,112,700 11,200,000 Modified Upper Quotasarrow_forward
- Use Adamss's method of apportionment to solve the following problem. 1. Suppose a nation has 6 states, with populations shown in the chart below. The representative body had 200 seats. Find: the standard divisor, the modified divisor, and the distribution of representatives to each state. Then complete the table. Round all standard and modified quotas to 7 decimal places if needed. You can use a spreadsheet to help you figure out an answer or calculate using a calculator. Representative Seats 200 Rounded Modified Modified Upper Quotas State Population Standard Quotas Up Quotas Quotas 1,598,400 1,236,300 5,460,200 826,900 965,500 1,112,700 11,200,000 А В C F Total Standard Divisor Modified Divisorarrow_forwardA small country is comprised of five states, A, B, C, D, and E. The population of each state is given in the following table. Congress will have 55 seats divided among the five states according to their respective populations. Use Jefferson's method with d = 32,920 to apportion 55 seats. State Number of Apportioned Seats (Type an integer.) State Number of Apportioned Seats (Type an integer.) State Number of Apportioned Seats (Type an integer.) State Number of Apportioned Seats (Type an integer.) State Number of Apportioned Seats (Type an integer.) A C D E State A B Population 113,752 186,914 (...) с 348.754 D 504,691 E 745,459arrow_forwardConsider the Republic of Ashbury and apportion 250 seats among the five states using Huntington-Hill method. Use the population from the table below. You can use a spreadsheet to help you figure out an answer or calculate using a calculator. Find: ● the standard divisor, ● the modified divisor, and the distribution of representatives to each state. ● Representative Seats State A B C D E Total Standard Divisor Modified Divisor a Ob Oc Od 250 Population 1,230,520 1,230,600 3,500,230 725,000 875,500 7,561,850 Standard Quota Modified Quotas Standard divisor: 30,247.40 Modified divisor: 30,320 States: A:41, B: 41, C: 115, D:24, E:29 Standard divisor: 30,247.40 Modified divisor: 30,305 States: A:41, B: 41, C: 116, D:24, E:29 Standard divisor: 30,247.40 Modified divisor: 30,386 States: A:40, B: 41, C: 115, D:24, E:29 Standard divisor: 30,247.40 Modified divisor: 30,320 States: A:41, B: 40, C: 115, D:24, E:29 Round Round down up Geometric Mean Hill- Huntington Methodarrow_forward
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