The sample of students required to estimate the mean weekly earnings of students at one college is of size 96.04.
For the population mean (μ) , we have the (1 - α)% confidence interval as:
X ± Zₐ / 2 + I / √n
margin of error = MOE = Zₐ / 2 ×I / √n
We are given:
σ = $10
MOE = $2
The critical value of z for 95% confidence level is
Zₐ / 2 = Zₓ = 1.96 ( for x as 0.025)
n = (1.96 (10))²
n = 96.04
Thus, the sample of students required to estimate the mean weekly earnings of students at one college is of size, 96.04.
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24. C
25. D
I honestly don't know.
I think the answer is m=1/5
Answer:

Step-by-step explanation:
From the question we are told that:
Loudness function 
Sound of a person 
Number of persons talking 
Generally the Relative intensity for 20 people is mathematically given by
Relative intensity for

Relative intensity for

Generally the loudness for 20 people is mathematically given by




Therefore increase in loudness X is given as


Answer:
x = -2
Step-by-step explanation:
- 3 x - 6x = -18x
- 4 - 18x + 9 = 41
9 - 4 = 5
- 18x + 5 = 41
- 18x = 41 - 5
- 18x = 36
36 divided by -18 is -2