40.5
Trust me i did it and got it right!! if its wrong lmk and i can explain it for u
I believe the answer to be only sequence A.
-Hope this helps, have a nice day! :)
Answer:
The 99% confidence interval for the proportion of readers who would like more coverage of local news is (0.3685, 0.4315).
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 zscore that has a pvalue of
.
For this problem, we have that:

99% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:

The upper limit of this interval is:

The 99% confidence interval for the proportion of readers who would like more coverage of local news is (0.3685, 0.4315).
Answer:
$3.85? 12/3.12 = $3.85
The answer would have came out of 3.84615384615
Then round up the answer to 3.85. So the unit price is $3.85.