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YOU ARE NOT REQUIRED TO DO THE MATH PROBLEM JUST RESPOND TO THE STUDENT POST WITH relevant contributions to solving or understanding the problem. This submission is about responding to five other student post from my discussion board. The responses don’t have to be long just relevant contributions to solving or understanding the problem. Please see attachment.Please respond to the 5 student Post listed below.Communicate clearly. Do not use abbreviations common in texting.All your responses must make important contributions to solving or understanding the problem. These posts must be relevant, unique (not repeating ideas from someone else’s post), and justified (you explain your reasoning in a way that others can understand). You do not need to write extensively but contribute thoughtfully in ways that help us all learn about important mathematical ideas. Thoughtful questions are valid contributions.Please see attachment for the student post and remember they don’t have to be long just relevant contributions to solving or understanding the problem.
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Christa Williams
Module 5 – M5D1 A Hamilton Apportionment – MAT101 Group 3
Question: In this discussion, you will work in small groups in online discussions to solve a mathematics
problem involving apportioning representatives. This group discussion will help prepare you for activity
M5A2, in which you will write and submit an individual complete solution to a related problem.
Solve the following problem and post your solutions to the discussion forum.
The Clarkstown Central School District covers 4 towns. There are 22 members of the school board, and the
4 towns have the populations shown in the following table:
Population
Town A
9,000
Town B
9,100
Town C
25,475
Town D
56,425
1. The school district uses the Hamilton method to apportion its 22 board members to the 4 towns. How
many board members are assigned to each town, using this method?
2. The following year, 1,000 people move out of town A and into town D. Now, how many board
members does each town have?
3. Compare the results from the 2 years. Do you think they make sense? Are they fair? Why or Why not?
1
Christa Williams
Module 5 – M5D1 A Hamilton Apportionment – MAT101 Group 3
The Clarkstown Central School District covers 4 towns. There are 22 members of the school board, and the 4 towns have the populations
shown in the following table:
Population
9,000
9,100
25,475
56,425
Town A
Town B
Town C
Town D
1. The school district uses the Hamilton method to apportion its 22 board members to the 4 towns. How many board
members are assigned to each town, using this method?
My calculation of the board members assigned to each town using the Hamilton method are as follow.
𝑡𝑜𝑡𝑎𝑙 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛
Standard divisor =𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑏𝑜𝑎𝑟𝑑 𝑚𝑒𝑚𝑏𝑒𝑟𝑠
Standard divisor =
9,000+9,100+25,475+56,425=100,000
22
=
100000
=
22
4545.45
School District
Town A
Town B
Town C
Town D
Population
Standard
Quota
Minimum
Quota
Fractional
Remainder
Final
Apportionment
9,000
9,100
25,475
25475
= 5.60
56,425
56425
= 12.41
100,0000
21.99
9,000
4545.45
= 1.98
9100
4545.45
= 2.00
4545.45
Total
4545.45
1
2
5
12
20
0.98
0
.60
.41
1.99
2
2
6
12
22
The standard divisor is the total population of 4 towns represented. The standard quota is the number of towns that
covers Clarkstown Central School District. Town C and D have a larger population and according to the chart above,
Town D has the majority of board members assigned.
2. The following year, 1,000 people move out of town A and into town D. Now, how many board members does each
town have?
Town A and B has 2 board members, C has 6 and D has 13 which includes 1 extra board member which should go to
Town A since it has the largest fractional remainder.
School District
Town A
Town B
Town C
Town D
Population
Standard
Quota
9,000-1,000=8,000
8000
=
4545.45
1.76000176
1
9,100
25,475
25475
=
4545.45
5.604505605
5
56,425+1,000=57,425
57425
= 12.63351263
100000
22.00002041
12
20
Minimum
Quota
Fractional
Remainder
Final
Apportionment
.76000176
9100
=
2.002002002
2
4545.45
Total
4545.45
.002002002
.604505605
.63351263
2.000021997
2
6
13
23
(largest fractional remainder)
2
2
Christa Williams
Module 5 – M5D1 A Hamilton Apportionment – MAT101 Group 3
3. Compare the results from the 2 years. Do you think they make sense? Are they fair? Why or Why not? Be sure to
answer all parts of the question above. It is possible that you will not be successful at answering all parts of the
question at this point, and that is okay. Describe what you were able to solve and what questions you have
remaining.
When comparing the results from the 2 years, Town D has the largest population. However, the Hamilton’s method
doesn’t look at the population at large. Instead, its determining factor is the largest fractional remainder. The largest
fractional remainder will receive the extra board member even though it has a smaller population. I believe this method
is unjust because it overlooks the majority of the population and standard quota to grant an area with a lesser volume or
constituents the apportionment as shown for Town A vs. Town D.
3
YEAR 1:
Population
A
B
C
D
Totals
9.000
9.100
25.475
56.425
100.000
St. Quota
1,98
2
5,6
12,41
21,99
Min. Qota
1
2
5
12
20
Total:
2
2
6
12
22
This first thing that needs to be done is to find the standard divisor
To do that add all of the town populations, and then divide by the number of seats that you need to fill
The total population of all the towns is 100,000. The number of eats to fill is 22
The divison is 1000,000/22= 4545.45
Now divide each towns population by 4545.45 See the table above for the final divison.
After dividing no round down. Finally you see that you don’t have all of the seats filled. We will then compare, and
YEAR 2:
Population
St. Quota
Min. Qota
Total:
A
8.000
1,76
1
2
B
9.100
2
2
2
C
25.475
5,6
5
5
D
56.425
12,63
12
13
Totals
100.000
21,99
20
22
In year two we learn that the population shifted from Town A by 1,000 to Town D
I have reflected this in the chart above. I used all of the same methods.
will then compare, and alocate the remaining seats based on each towns standard quota.
Question: In this discussion, you will work in small groups in online discussions to solve a
mathematics problem involving apportioning representatives. This group discussion will help
prepare you for activity M5A2, in which you will write and submit an individual complete
solution to a related problem.
Solve the following problem and post your solutions to the discussion forum.
The Clarkstown Central School District covers 4 towns. There are 22 members of the school
board, and the 4 towns have the populations shown in the following table:
Population
Town A
9,000
Town B
9,100
Town C
25,475
Town D
56,425
1. The school district uses the Hamilton method to apportion its 22 board members to the 4
towns. How many board members are assigned to each town, using this method?
2. The following year, 1,000 people move out of town A and into town D. Now, how many
board members does each town have?
3. Compare the results from the 2 years. Do you think they make sense? Are they fair? Why
or Why not?
Total population: 100,000
22 members on the school board coving four towns.
𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝐷𝑖𝑣𝑖𝑠𝑜𝑟 =
100,000/22= 4,545 standard divisor
Population
# of items to be allocated
4,545 =
100,000
22
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑞𝑢𝑜𝑡𝑎 =
9,000
Town A: 1.98 = 4,545
Town B: 2.00 =
9,1000
4,545
population
𝑠𝑡𝑑 𝑑𝑖𝑣
Town C: 5.61 =
25,475
4,545
Town D: 12.41 =
56,425
4,545
20 Board members to be assigned, 2 remain.
Town A is priority #1 .98 and Town C is priority #2 .61
How many school board members:
(Town A: 2)
(Town B: 2)
(Town C: 6)
(Town D: 12)
= 22 members
I seat still remains, Town B is the priority town with .4 decimal left over and gets the extra council
member.
Year 2, 1,000 people move form Town A to Town D
8,000
Town A: 1.76 = 4,545
Town B: 2.00 =
9,1000
4,545
Town C: 5.61 =
25,475
4,545
Town D: 12.63 =
57,425
4,545
Two seats are remaining. Town A with .76 is priority #1 and Town D with .63 is priority #2.
How many school board members:
(Town A: 2)
(Town B: 2)
(Town C: 5)
(Town D: 13)
= 22 members
I think these result are fair, however, if I lived in Town C I would be unhappy. Town C has 44% the
population of Town D but has only 38% of the representation of Town D.
For comparison if we used the Jefferson method (4,300 as the divisor) for second year. Towns C and D
would get the extra council member. The Hamilton system seems to help the smaller town more often,
but the larger towns can still benefit. In my opinion, this is the only fair way to settle these type of
societal problems. Removing the human factor and let the numbers and mathematics speak for
themselves.
John
Question: In this discussion, you will work in small groups in online discussions to solve a
mathematics problem involving apportioning representatives. This group discussion will help
prepare you for activity M5A2, in which you will write and submit an individual complete
solution to a related problem.
Solve the following problem and post your solutions to the discussion forum.
The Clarkstown Central School District covers 4 towns. There are 22 members of the school
board, and the 4 towns have the populations shown in the following table:
Population
Town A
9,000
Town B
9,100
Town C
25,475
Town D
56,425
1. The school district uses the Hamilton method to apportion its 22 board members to the 4
towns. How many board members are assigned to each town, using this method?
2. The following year, 1,000 people move out of town A and into town D. Now, how many
board members does each town have?
3. Compare the results from the 2 years. Do you think they make sense? Are they fair? Why
or Why not?
The Clarkstown Central School District covers 4 towns. There are 22 members of the school
board, and the 4 towns have the populations shown in the following table:
Town A: 9000
Town B: 9100
Town C: 25475
Town D: 56425
Standard Divisor:
Total Pop/#of seats
9000+9100+25475+56425 = 100000/22 = 4545.4545
Standard quota for each town:
State pop/standard divisor
Town A: 1.9800
Town B: 2.0020
Town C: 5.6045
Town D: 12.4135
Rounded-down Q:
Town A: 1
Town B: 2
Town C: 5
Town D: 12
There are 2 board members to be apportioned.
A: 2
B: 2
C: 6
D: 12
Town A: 8000 (standard quota) = 1.7600
Town B: 9100 = 2.0020
Town C: 25475 = 5.6045
Town D: 57425 = 12.6335
Rounded-down Q:
A: 1 + 1= 2
B: 2
C: 5
D: 12 +1 = 13
Compare the results from the 2 years. Do you think they make sense? Are they fair? Why or
Why not?
With adding the additional $1000 to town D, it gained another board member and town A
still has the 2 members. I do think the more populated towns need the most members and I
think the amount dispersed for each town is appropriate.
Question: In this discussion, you will work in small groups in online discussions to solve a
mathematics problem involving apportioning representatives. This group discussion will help
prepare you for activity M5A2, in which you will write and submit an individual complete
solution to a related problem.
Solve the following problem and post your solutions to the discussion forum.
The Clarkstown Central School District covers 4 towns. There are 22 members of the school
board, and the 4 towns have the populations shown in the following table:
Population
Town A
9,000
Town B
9,100
Town C
25,475
Town D
56,425
1. The school district uses the Hamilton method to apportion its 22 board members to the 4
towns. How many board members are assigned to each town, using this method?
2. The following year, 1,000 people move out of town A and into town D. Now, how many
board members does each town have?
3. Compare the results from the 2 years. Do you think they make sense? Are they fair? Why
or Why not?
What I know:
-The Clarks Town Central School District covers 4 towns.
-There are 22 members of the school board, and the 4 towns have the populations shown
below:
–Town A – 9000
–Town B – 9100
–Town C – 25,475
–Town D – 56,425
What I have to answer:
-The school district uses the Hamilton method to apportion its 22 board members to the 4
towns. How many board members are assigned to each town, using this method?
Board Members: 22
Total Population: 9000 + 9100 + 25475 + 56425 = 100,000
Board Members / Total Population = Quota
22 / 100,000 = .00022
.00022 x 9000 = 1.98
.00022 x 9100 = 2.002
.00022 x 25,475 = 5.60
.00022 x 56,425 = 12.41
This is as far as I have gotten. Hopefully I am on the right track. I look forward to
comparing our work. Good luck group 3!
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