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:
Step-by-step explanation:

Solve for r.
You want to get r by itself on one side on the equal sign.
bh + hr = 25
Subtract bh from both sides.
hr = 25 - bh
Divide h on both sides.
r = 25 - bh / h
The two h's cancel each other out.
r = 25 - b
Hope this helps!
178.
If number CLXXVIII is within to text or sentence it should be read in its equivalent in arabic numbers, in this case 178.
:)
Answer:
( 6, pi/6)
Step-by-step explanation:
( 3 sqrt(3), 3)
To get r we use x^2 + y ^2 = r^2
( 3 sqrt(3) )^2 + 3^2 = r^2
9 *3 +9 = r^2
27+9 = r^2
36 = r^2
Taking the square root of each side
sqrt(36) = sqrt(r^2)
6 =r
Now we need to find theta
tan theta = y/x
tan theta = 3 / 3 sqrt(3)
tan theta = 1/ sqrt(3)
Taking the inverse tan of each side
tan ^-1 ( tan theta) = tan ^ -1 ( 1/ sqrt(3))
theta = pi /6