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Write your answer on your own sheet of paper and either scan it or take a picture of it and upload it or type your answers into a word document As we've learned, the weights of the players can be deceiving when it comes to determining the amount of power each individual player has. By manipulating the quota, one can make the balance of power be whatever one wants. We’re going to work with this weighted voting system: [q: 5, 4, 3] Determine what to use for the quota to get the indicated BPI values. In each case, explain how you determined your answer by either writing a sentence or two to explain your thought process, or show the work to find the BPIs to prove that your quota actually works. 1) BPI(P1) = 33.33%   BPI(P2) = 33.33%    BPI(P3) = 33.33% [All players have equal power] 2) BPI(P1) = 60% BPI(P2) = 20% BPI(P3) = 20% 3) BPI(P1) = 50% BPI(P2) = 50% BPI(P3) = 0% [P3 is a dummy]

Write your answer on your own sheet of paper and either scan it or take a picture of it and upload it or type your answers into a word document As we've learned, the weights of the players can be deceiving when it comes to determining the amount of power each individual player has. By manipulating the quota, one can make the balance of power be whatever one wants. We’re going to work with this weighted voting system: [q: 5, 4, 3] Determine what to use for the quota to get the indicated BPI values. In each case, explain how you determined your answer by either writing a sentence or two to explain your thought process, or show the work to find the BPIs to prove that your quota actually works. 1) BPI(P1) = 33.33%   BPI(P2) = 33.33%    BPI(P3) = 33.33% [All players have equal power] 2) BPI(P1) = 60% BPI(P2) = 20% BPI(P3) = 20% 3) BPI(P1) = 50% BPI(P2) = 50% BPI(P3) = 0% [P3 is a dummy]

Big Ideas Math A Bridge To Success Algebra 1: Student Edition 2015
Big Ideas Math A Bridge To Success Algebra 1: Student Edition 2015
1st Edition
ISBN:9781680331141
Author:HOUGHTON MIFFLIN HARCOURT
Publisher:HOUGHTON MIFFLIN HARCOURT
Chapter11: Data Analysis And Displays
Section11.3: Shapes Of Distributions
Problem 1Q
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Write your answer on your own sheet of paper and either scan it or take a picture of it and upload it or type your answers into a word document

As we've learned, the weights of the players can be deceiving when it comes to determining the amount of power each individual player has.

By manipulating the quota, one can make the balance of power be whatever one wants.

We’re going to work with this weighted voting system: [q: 5, 4, 3]

Determine what to use for the quota to get the indicated BPI values.

In each case, explain how you determined your answer by either writing a sentence or two to explain your thought process, or show the work to find the BPIs to prove that your quota actually works.

1) BPI(P1) = 33.33%   BPI(P2) = 33.33%    BPI(P3) = 33.33% [All players have equal power]

2) BPI(P1) = 60% BPI(P2) = 20% BPI(P3) = 20%

3) BPI(P1) = 50% BPI(P2) = 50% BPI(P3) = 0% [P3 is a dummy]

 

 

Safari File Edit View History Bookmarks Window Help 53% Sun Oct 10 7:47 PM A uwmil.instructure.com Chapter Two Project b Ask a Question | bartleby UWM = MATH 103-201 > Assignments > Chapter Two Project MIL Fall 2021 Chapter Two Project Start Assignment Account Home Announcements Due Oct 3 by 11:59pm Points 10 Submitting a file upload Dashboard Office 365 File Types doc, docx, pdf, jpg, and png Available until Dec 16 at 11:59pm Discussions Courses Here's a video with me explaining the instructions for this project e Grades Write your answer on your own sheet of paper and either scan it or take a picture of it and upload it or type your Calendar Zoom answers into a word document 画 As we've learned, the weights of the players can be deceiving when comes to determining the amount of power Inbox each individual player has. By manipulating the quota, one can make the balance of power be whatever one wants. History We're going to work with this weighted voting system: [q: 5, 4, 3] Help Determine what to use for the quota to get the indicated BPI values. In each case, explain how you determined your answer by either writing a sentence or two to explain your thought process, or show the work to find the BPIS to prove that your quota actually works. 1) BPI(P1) = 33.33% BPI(P2) = 33.33% BPI(P3) = 33.33% [All players have equal power] 2) BPI(P1) = 60% BPI(P2) = 20% BPI(P3) = 20% %3D 3) BPI(P1) = 50% BPI(P2) = 50% BPI(P3) = 0% [P3 is a dummy] • Previous Next
Transcribed Image Text:Safari File Edit View History Bookmarks Window Help 53% Sun Oct 10 7:47 PM A uwmil.instructure.com Chapter Two Project b Ask a Question | bartleby UWM = MATH 103-201 > Assignments > Chapter Two Project MIL Fall 2021 Chapter Two Project Start Assignment Account Home Announcements Due Oct 3 by 11:59pm Points 10 Submitting a file upload Dashboard Office 365 File Types doc, docx, pdf, jpg, and png Available until Dec 16 at 11:59pm Discussions Courses Here's a video with me explaining the instructions for this project e Grades Write your answer on your own sheet of paper and either scan it or take a picture of it and upload it or type your Calendar Zoom answers into a word document 画 As we've learned, the weights of the players can be deceiving when comes to determining the amount of power Inbox each individual player has. By manipulating the quota, one can make the balance of power be whatever one wants. History We're going to work with this weighted voting system: [q: 5, 4, 3] Help Determine what to use for the quota to get the indicated BPI values. In each case, explain how you determined your answer by either writing a sentence or two to explain your thought process, or show the work to find the BPIS to prove that your quota actually works. 1) BPI(P1) = 33.33% BPI(P2) = 33.33% BPI(P3) = 33.33% [All players have equal power] 2) BPI(P1) = 60% BPI(P2) = 20% BPI(P3) = 20% %3D 3) BPI(P1) = 50% BPI(P2) = 50% BPI(P3) = 0% [P3 is a dummy] • Previous Next
A uwmil.instructure.com Chapter Two Readings: Math 103 Fall 2021 Online b Ask a Question | bartleby Announcements UWM 2.1: Weighted Voting Introduction Office 365 Minimize File Preview Discussions Page 2 > of 5 ZOOM + Grades Meaningful Voting Systems Асcount Zoom There are some limitations on what the quota could be. Dashboard Here are some examples that show some problems with choosing just any number for the quota. [4: 3, 2, 1, 1, 1] Courses This system has 5 players. It has total weight of 3+2+1+1+1 = 8. 4 votes are required to pass a motion. Calendar Suppose that Player One (3 votes) and Player Three (1 vote) decide to vote in favor, but Player Two (2 votes), Player Four (1 vote) and Player Five (1 vote) all oppose the motion. Inbox Then it is 4 votes in favor (3 + 1) and 4 votes against (2 + 1 + 1). Nothing is decided. In order to avoid this problem of ties, we announce that the quota must be at least a majority of the total weight. History Here's another example where things can go wrong. Help [7: 2, 2, 1] This system has 3 players. It has total weight 2 + 2+ 1 = 5. 7 votes are required to pass a motion. In this case, even if all three players vote "Yes", they will only have 5 votes, and they will never pass any motions. To avoid this problem, we announce that the quota must be less than or equal to the total weight. Algebraically, this looks like this: In a weighted voting system with N players: w,+w,+w3+...+w, 1 |<q<w,+w2+w 3+...+w, 2
Transcribed Image Text:A uwmil.instructure.com Chapter Two Readings: Math 103 Fall 2021 Online b Ask a Question | bartleby Announcements UWM 2.1: Weighted Voting Introduction Office 365 Minimize File Preview Discussions Page 2 > of 5 ZOOM + Grades Meaningful Voting Systems Асcount Zoom There are some limitations on what the quota could be. Dashboard Here are some examples that show some problems with choosing just any number for the quota. [4: 3, 2, 1, 1, 1] Courses This system has 5 players. It has total weight of 3+2+1+1+1 = 8. 4 votes are required to pass a motion. Calendar Suppose that Player One (3 votes) and Player Three (1 vote) decide to vote in favor, but Player Two (2 votes), Player Four (1 vote) and Player Five (1 vote) all oppose the motion. Inbox Then it is 4 votes in favor (3 + 1) and 4 votes against (2 + 1 + 1). Nothing is decided. In order to avoid this problem of ties, we announce that the quota must be at least a majority of the total weight. History Here's another example where things can go wrong. Help [7: 2, 2, 1] This system has 3 players. It has total weight 2 + 2+ 1 = 5. 7 votes are required to pass a motion. In this case, even if all three players vote "Yes", they will only have 5 votes, and they will never pass any motions. To avoid this problem, we announce that the quota must be less than or equal to the total weight. Algebraically, this looks like this: In a weighted voting system with N players: w,+w,+w3+...+w, 1 |<q<w,+w2+w 3+...+w, 2
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