#### Answer

(a) Using Hamilton's method, each school is apportioned the following number of computers:
School A is apportioned 20 computers.
School B is apportioned 18 computers.
School C is apportioned 16 computers.
School D is apportioned 3 computers.
(b) The Alabama paradox occurs. Initially, with 57 computers, School D was allocated 3 computers. After the number of computers increased to 58, School D was allocated only 2 computers.

#### Work Step by Step

(a) We can find the standard divisor.
$standard~divisor = \frac{total ~enrollment}{number~of~ computers}$
$standard~divisor = \frac{14,250}{57}$
$standard~divisor = 250$
We can find each school's standard quota. The standard quota of each school is the school's enrollment divided by the standard divisor.
School A:
$standard ~quota = \frac{enrollment}{standard~divisor}$
$standard~quota = \frac{5040}{250}$
$standard~quota = 20.16$
School B:
$standard ~quota = \frac{enrollment}{standard~divisor}$
$standard~quota = \frac{4560}{250}$
$standard~quota = 18.24$
School C:
$standard ~quota = \frac{enrollment}{standard~divisor}$
$standard~quota = \frac{4040}{250}$
$standard~quota = 16.16$
School D:
$standard ~quota = \frac{enrollment}{standard~divisor}$
$standard~quota = \frac{610}{250}$
$standard~quota = 2.44$
Hamilton's method is an apportionment method that involves rounding each standard quota down to the nearest whole number. Surplus computers are given, one at a time, to the schools with the largest decimal parts in their standard quotas until there are no more surplus computers.
Initially, each school is apportioned its lower quota.
School A is apportioned 20 computers.
School B is apportioned 18 computers.
School C is apportioned 16 computers.
School D is apportioned 2 computers.
We can find the total number of computers which have been apportioned.
total = 20 + 18 + 16 + 2 = 56 computers
Since there is a total of 57 computers, there is one surplus computer. One more computer is given to School D because it has the largest decimal part (0.44) in its standard quota.
Using Hamilton's method, each school is apportioned the following number of computers:
School A is apportioned 20 computers.
School B is apportioned 18 computers.
School C is apportioned 16 computers.
School D is apportioned 2 + 1 = 3 computers.
(b) We can find the standard divisor.
$standard~divisor = \frac{total ~enrollment}{number~of~ computers}$
$standard~divisor = \frac{14,250}{58}$
$standard~divisor = 245.69$
We can find each school's standard quota. The standard quota of each school is the school's enrollment divided by the standard divisor.
School A:
$standard ~quota = \frac{enrollment}{standard~divisor}$
$standard~quota = \frac{5040}{245.69}$
$standard~quota = 20.51$
School B:
$standard ~quota = \frac{enrollment}{standard~divisor}$
$standard~quota = \frac{4560}{245.69}$
$standard~quota = 18.56$
School C:
$standard ~quota = \frac{enrollment}{standard~divisor}$
$standard~quota = \frac{4040}{245.69}$
$standard~quota = 16.44$
School D:
$standard ~quota = \frac{enrollment}{standard~divisor}$
$standard~quota = \frac{610}{245.69}$
$standard~quota = 2.48$
Hamilton's method is an apportionment method that involves rounding each standard quota down to the nearest whole number. Surplus computers are given, one at a time, to the schools with the largest decimal parts in their standard quotas until there are no more surplus computers.
Initially, each school is apportioned its lower quota.
School A is apportioned 20 computers.
School B is apportioned 18 computers.
School C is apportioned 16 computers.
School D is apportioned 2 computers.
We can find the total number of computers which have been apportioned.
total = 20 + 18 + 16 + 2 = 56 computers
Since there is a total of 58 computers, there are two surplus computers. The first computer is given to School B because it has the largest decimal part (0.56) in its standard quota. The second computer is given to School A because it has the second largest decimal part (0.51) in its standard quota.
Using Hamilton's method, each school is apportioned the following number of computers:
School A is apportioned 20 + 1 = 21 computers.
School B is apportioned 18 + 1 = 19 computers.
School C is apportioned 16 computers.
School D is apportioned 2 computers.
We can see that the Alabama paradox occurs. Initially, with 57 computers, School D was allocated 3 computers. After the number of computers increased to 58, School D was allocated only 2 computers. Therefore, the Alabama paradox occurs.