Answer:
The sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.
Step-by-step explanation:
The (1 - <em>α</em>)% confidence interval for population mean (<em>μ</em>) is:

The margin of error of a (1 - <em>α</em>)% confidence interval for population mean (<em>μ</em>) is:

The information provided is:
<em>σ</em> = $60
<em>MOE</em> = $2
The critical value of <em>z</em> for 95% confidence level is:

Compute the sample size as follows:

![n=[\frac{z_{\alpha/2}\times \sigma }{MOE}]^{2}](https://tex.z-dn.net/?f=n%3D%5B%5Cfrac%7Bz_%7B%5Calpha%2F2%7D%5Ctimes%20%5Csigma%20%7D%7BMOE%7D%5D%5E%7B2%7D)
![=[\frac{1.96\times 60}{2}]^{2}](https://tex.z-dn.net/?f=%3D%5B%5Cfrac%7B1.96%5Ctimes%2060%7D%7B2%7D%5D%5E%7B2%7D)

Thus, the sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.
Let T = total cost
Let 8c = total amount of children tickets
Let 12a = total amount of adult tickets
T = 8c + 12a
Answer is <span>C. 80
hope that helps
75% are full with 20 seats empty so </span>20 seats empty = 25%
75% * 20 / 25% = 60
Answer:-7x(to the power of 2)-3y+4x-2y(to the 3rd power)-7x( to the third power)
Step-by-step explanation:
Collect like terms