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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Answer:
5/8
Step-by-step explanation:
5/8 has the greatest value
F+b= x
(f+3)+(b+3)=x
(f-2)-(b-2)=x
Answer:
Equation: 65+1.50m=78.50
Solution to the equation= 9 minutes
Step-by-step explanation:
The parents have to give 65 dollars for their children to be under care till 6 o'clock. AFter 6 o'clock, they had to pay 1.50 dollars for one minute. This parent gives a total of 78.50 dollar for that day. So, the equation is :
65+1.50m=78.50
(btw, m stands for minutes)
So to solve the equation:
1.50m=13.50
m= 9
