Answer:
6546 students would need to be sampled.
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
.
The margin of error is:

The dean randomly selects 200 students and finds that 118 of them are receiving financial aid.
This means that 
90% confidence level
So
, z is the value of Z that has a p-value of
, so
.
If the dean wanted to estimate the proportion of all students receiving financial aid to within 1% with 90% reliability, how many students would need to be sampled?
n students would need to be sampled, and n is found when M = 0.01. So






Rounding up:
6546 students would need to be sampled.
Answer:
1.27/50
2.3/4
3.31/100
4.25/46
5.25/48
I know this because I got them right when I took the quiz
Step-by-step explanation:
Step-by-step explanation:
3c + 3s = 12.50
4c + 2s = 10.00
3S = 12.50-3c
s = 4.17 - c
4c + 2 (4.17-c) = 10.00
2c + 8.34 = 10.00
2c = 1.66
chips = 0.83
soda = 3.34