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:
Below.
Step-by-step explanation:
4x^2 - 4x - 1 = 0
This will not factor.
But 4x^2 - 4x + 1 = 0 will:
= (2x - 1)(2x - 1) = 0
x = 0.5 (*2)
-30-6r=36
We move all terms to the left:
-30-6r-(36)=0
We add all the numbers together, and all the variables
-6r-66=0
We move all terms containing r to the left, all other terms to the right
-6r=66
r=66/-6
r=-11
$1500 divided by 12 months = $125 per month. 0.75% x 120 (10 years x 12 months in each) = 90%. $125 x 120 = $15,000 + ($125 x 90%= $11,250) = $26,250. $26,250 + $2500 = $28,750 overall in the account after 10 years
Can u just post the whole question or elaborate further
