Answer is P=3.5 or p=7/2
They are both correct
29% would be the answer if you are rounding up
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.
Answer:
549
Step-by-step explanation:
A calculator can give you an equivalent expression.
I might work from the middle outward:
= 486 -3 +33×2
= 483 +66
= (480 +60) + (3 +6)
= 540 + 9
= 549
Well 32 divided by 8 would be 4 so... 32 x 4 would be 128 so 128 times. If it’s not correct I’m really sorry.