Answer:
We need a sample size of 564.
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:

The margin of error is:

95% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
Based upon a 95% confidence interval with a desired margin of error of .04, determine a sample size for restaurants that earn less than $50,000 last year.
We need a sample size of n
n is found when 
So






Rounding up
We need a sample size of 564.
Answer:
I think it's C and F. Don't depend on me tho I'm only in the 7th grade but I just told what I think is more logical
Divide 4 on both sides, x=6.4/4 which is 1.6
Answer:
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Your answer is n = 9/8
Brainliest? Thx.
45 / 3 = 15
45 - 15 = 30
30 / 5 = 6
6 x 2 = 12
30 - 12 = 18
The answer is D:18