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:
The largest number =5 3/4
The smallest number =1 2/5
Required difference = 5 3/4 − 1 2/5 = 4 7/2
I could be very off. hope it helps somehow
The answer is 24 you just multiply the initial by the posibilities
Answer:
0.1875
Step-by-step explanation:
NO EXPLANATION
°^°
Answer:
D and C
Step-by-step explanation:
1) You can cross check using desmos.
2) cos(120) is - 0.5