Answer:
a) The margin of error for a 90% confidence interval when n = 14 is 18.93.
b) The margin of error for a 90% confidence interval when n=28 is 12.88.
c) The margin of error for a 90% confidence interval when n = 45 is 10.02.
Step-by-step explanation:
The t-distribution is used to solve this question:
a) n = 14
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 14 - 1 = 13
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 13 degrees of freedom(y-axis) and a confidence level of
. So we have T = 1.7709
The margin of error is:

In which s is the standard deviation of the sample and n is the size of the sample.
The margin of error for a 90% confidence interval when n = 14 is 18.93.
b) n = 28
27 df, T = 1.7033

The margin of error for a 90% confidence interval when n=28 is 12.88.
c) The margin of error for a 90% confidence interval when n = 45 is
44 df, T = 1.6802

The margin of error for a 90% confidence interval when n = 45 is 10.02.
Hi! You need 3 boxes because if you divide 21 1/4 / 10 you get 2 1/8 or 2.125 so you at least have two boxes. You will still have some left over after the two boxes so you will need another box for those. I hope this helped, Goodluck :)
<span> x − 2y = −1 => x = 2y -1
2x + y = −12
2(2y - 1) + y = -12
4y - 2 + y = -12
5y = -10
y = -2
x = 2</span>·(-2) - 1
x = -4 - 1
x = -5
Answer <span>D (-5,-2)</span>
Attached the solution with work shown.