Answer:
The 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.

In which
z is the z-score that has a p-value of
.
A store randomly samples 603 shoppers over the course of a year and finds that 142 of them made their visit because of a coupon they'd received in the mail.
This means that 
95% confidence level
So
, z is the value of Z that has a p-value of
, so
.
The lower limit of this interval is:

The upper limit of this interval is:

The 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694).
The answer is 12 because 576 divided 48 is 12 .
Well first you would subtract 9 from both sides to isolate your variable
so it would look like
1/4y =-81/2
now you divide 1/4 from both sides
so you're left with
y = -34