As we do not know the other cars avg speed we can not tell the answer. Please repeat the question.
Answer:
D. is the correct one.........
Answer:
A sample size of 345 is needed so that the confidence interval will have a margin of error of 0.07
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
.
The margin of error of the interval is given by:

In this problem, we have that:

99.5% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.07?
This is n when M = 0.07. So







A sample size of 345 is needed so that the confidence interval will have a margin of error of 0.07
Blank #1 is 0.67 and Blank #2 is 13.3. I am not sure about Blank #3. Here is a tip: Mean absolute deviation is the average of the absolute deviations. Tell me if I am right ok?
First off, you really don't need to write "what percent???" here.
25% of 80 is 0.25(80) = 20.