I think it’s 18/5 maybe idk try it!
Answer:
A sample of 1068 is needed.
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 is:

95% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population proportion?
We need a sample of n.
n is found when M = 0.03.
We have no prior estimate of
, so we use the worst case scenario, which is 
Then






Rounding up
A sample of 1068 is needed.
$4.20 / 36 = $0.11667 per potato<span>
</span>
Finance charge
(5,834.53−150)×(0.204÷12)=96.64
New balance
5,834.53−150+96.64+325
=6,106.17
2/3 times x-4 = 6
thats the answer :)