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.
Both equations are the same
<span>y=−4x+4 ----> y+4x=4,
so </span><span>consistent dependent</span>
Convert 5% to .05 multiply that by total and youll get the tax amount and add the to the total then for the tip you multiply .15 by the total and shabam theres your answer
The answer to this would be $309.4