Mathematics: A Practical Odyssey
8th Edition
ISBN: 9781305104174
Author: David B. Johnson, Thomas A. Mowry
Publisher: Cengage Learning
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Question
Chapter 6.2, Problem 19E
To determine
(a)
To find:
The apportionment of legislative of seats by using Jefferson’s Method for the given condition.
To determine
(b)
To find:
The apportionment of legislative of seats by using Adams’s Method for the given condition.
To determine
(c)
To find:
The apportionment of legislative of seats by using Webster’s Method for the given condition.
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Consider the apportionment problem for College Town.
North : 5,500 , South : 4,500 , East : 7,600, West : 7,400
Suppose each council member is to represent approximately 2,100 citizens. Use Jefferson's plan assuming there must be 10 representatives.
Consider the apportionment problem.
NORTH: 18,200 SOUTH: 12,500 EAST: 17,800 WEST: 13,400
Use Webster's Plan assuming there must be 16 representatives.
Consider the following apportionment problem, table[]North, 18, 400, East: 17,400 (South: 12,200, West: 14,000, Use the apportionment plan requested below assuming that there must be zs representatives. Jefferson's plan North South East West
Chapter 6 Solutions
Mathematics: A Practical Odyssey
Ch. 6.1 - Four candidates, Alliotti, Baker, Cruz, and Daud,...Ch. 6.1 - Prob. 2ECh. 6.1 - Prob. 3ECh. 6.1 - Prob. 4ECh. 6.1 - Prob. 5ECh. 6.1 - Prob. 6ECh. 6.1 - Prob. 7ECh. 6.1 - Number of Ballots Cast 6 8 7 10 15 1st choice C C...Ch. 6.1 - Prob. 9ECh. 6.1 - Prob. 10E
Ch. 6.1 - Prob. 11ECh. 6.1 - Four candidates, Harrison H, Lennon L, McCartney...Ch. 6.1 - Prob. 13ECh. 6.1 - Five candidates, Fino F, Gempler G, Holloway H,...Ch. 6.1 - Prob. 15ECh. 6.1 - Prob. 16ECh. 6.1 - Prob. 17ECh. 6.1 - Prob. 18ECh. 6.1 - Prob. 19ECh. 6.1 - Prob. 20ECh. 6.1 - In an election, there are seven candidates. a....Ch. 6.1 - Prob. 22ECh. 6.1 - Prob. 23ECh. 6.1 - Prob. 24ECh. 6.1 - Prob. 25ECh. 6.1 - Prob. 26ECh. 6.1 - Prob. 27ECh. 6.1 - Prob. 28ECh. 6.1 - Prob. 29ECh. 6.1 - Prob. 30ECh. 6.1 - What is Arrows Impossibility Theorem?Ch. 6.2 - A small country consists of three states, A, B and...Ch. 6.2 - A small country consists of three states, A, B and...Ch. 6.2 - Suppose that the governors of three Middle...Ch. 6.2 - Suppose that the governors of three midwestern...Ch. 6.2 - Prob. 5ECh. 6.2 - Prob. 6ECh. 6.2 - Prob. 7ECh. 6.2 - Prob. 8ECh. 6.2 - A local school district contains four middle...Ch. 6.2 - A local school district contains four elementary...Ch. 6.2 - Prob. 11ECh. 6.2 - In J.R.R. Tolkiens Middle Earth, the regions in...Ch. 6.2 - Prob. 13ECh. 6.2 - Prob. 14ECh. 6.2 - Prob. 15ECh. 6.2 - Prob. 16ECh. 6.2 - Prob. 17ECh. 6.2 - Prob. 18ECh. 6.2 - Prob. 19ECh. 6.2 - Prob. 20ECh. 6.2 - Prob. 21ECh. 6.2 - Prob. 22ECh. 6.2 - Prob. 23ECh. 6.2 - Prob. 24ECh. 6.2 - Prob. 25ECh. 6.2 - Prob. 26ECh. 6.2 - Prob. 27ECh. 6.2 - Prob. 28ECh. 6.2 - Prob. 29ECh. 6.2 - Prob. 30ECh. 6.2 - Prob. 31ECh. 6.2 - Prob. 32ECh. 6.2 - Use the Hill-Huntington Method to determine how...Ch. 6.2 - Prob. 34ECh. 6.2 - Prob. 35ECh. 6.2 - Prob. 36ECh. 6.2 - Prob. 37ECh. 6.2 - Prob. 38ECh. 6.2 - Prob. 39ECh. 6.2 - Prob. 40ECh. 6.2 - Prob. 41ECh. 6.2 - Prob. 42ECh. 6.2 - Prob. 43ECh. 6.2 - Prob. 44ECh. 6.2 - Prob. 45ECh. 6.2 - Prob. 46ECh. 6.2 - Prob. 47ECh. 6.2 - Prob. 48ECh. 6.2 - What is apportionment?Ch. 6.2 - Prob. 50ECh. 6.2 - Prob. 51ECh. 6.2 - Prob. 52ECh. 6.2 - Prob. 53ECh. 6.2 - Prob. 54ECh. 6.2 - Prob. 55ECh. 6.2 - Prob. 56ECh. 6.2 - Prob. 57ECh. 6.2 - Since the founding of the United States, what...Ch. 6.2 - Prob. 59ECh. 6.2 - Prob. 60ECh. 6.3 - A small country consists of three states: A, B,...Ch. 6.3 - Prob. 2ECh. 6.3 - Prob. 3ECh. 6.3 - Prob. 4ECh. 6.3 - A small country consists of four states: A, B, C,...Ch. 6.3 - Prob. 6ECh. 6.3 - Prob. 7ECh. 6.3 - Prob. 8ECh. 6.3 - Prob. 9ECh. 6.3 - Prob. 10ECh. 6.3 - A small country consists of three states:A, B, and...Ch. 6.3 - Prob. 12ECh. 6.3 - Prob. 13ECh. 6.3 - Prob. 14ECh. 6.3 - Prob. 15ECh. 6.3 - What is the Alabama Paradox? Explain its meaning.Ch. 6.3 - Prob. 17ECh. 6.3 - Prob. 18ECh. 6.3 - Prob. 19ECh. 6.CR - Prob. 1CRCh. 6.CR - Prob. 2CRCh. 6.CR - Prob. 3CRCh. 6.CR - Prob. 4CRCh. 6.CR - Prob. 5CRCh. 6.CR - Prob. 6CRCh. 6.CR - Prob. 7CRCh. 6.CR - Prob. 8CRCh. 6.CR - Prob. 9CRCh. 6.CR - Prob. 10CRCh. 6.CR - Prob. 11CRCh. 6.CR - Prob. 12CRCh. 6.CR - Prob. 13CRCh. 6.CR - Prob. 14CRCh. 6.CR - Prob. 15CRCh. 6.CR - Prob. 16CRCh. 6.CR - Prob. 17CRCh. 6.CR - A small country consists of three states: A, B,...Ch. 6.CR - Prob. 19CRCh. 6.CR - Prob. 20CRCh. 6.CR - Prob. 21CRCh. 6.CR - Prob. 22CRCh. 6.CR - Prob. 23CRCh. 6.CR - Prob. 24CRCh. 6.CR - Prob. 25CRCh. 6.CR - Prob. 26CRCh. 6.CR - Prob. 27CRCh. 6.CR - Prob. 28CRCh. 6.CR - What is the New States Paradox? Explain its...Ch. 6.CR - Prob. 30CRCh. 6.CR - Prob. 31CRCh. 6.CR - Prob. 32CRCh. 6.CR - Prob. 33CRCh. 6.CR - Prob. 34CRCh. 6.CR - What method of appointment for the House of...
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- 1) find the mode of regions for the states that voted for the Democratic and for the Republic respectively and explain the findings.arrow_forwardConsider the apportionment problem. North: 14,300 South: 12,500 East: 18,400 West: 11,200 Use Webster's plan assuming there must be 16 representatives. N ____________S_________ E__________ W_______ __________ __________ __________ _________arrow_forwardConsider the apportionment problem for College Town. North: 5,100 South: 9,300 East: 6,800 West: 8,800 Suppose each council member is to represent approximately 3,000 citizens. Use Jefferson's assuming there must be 10 representatives. N __________S__________ E__________ W __________ __________ __________ _________arrow_forward
- Which of the following is a flaw in apportionment methods? Jefferson Plan Hare System Hamilton Plan Alabama Paradoxarrow_forwardConsider the apportionment of 60 doctors for a physicians organization. The apportionment using Hamilton's method is shown in the table below. Does the Alabama paradox occur using Hamilton's method if the number of doctors is increased from 60 to 61? Clinic A B C D E Total Patients 656 536 515 549 602 2858 Standard quota 13.77 11.25 10.81 11.53 12.64 60.00 Lower quota 13 11 10 11 12 57 Hamilton's apportionment 14 11 11 11 13 60 Complete the table below with the new apportionment for clinics A, B, C, D, and E using a standard divisor rounded to two decimal places. Clinic A B C D E Total Patients 656 536 515 549 602 2858 Hamilton's apportionment ? ? ? ? ? 61arrow_forward
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