7.5 min per walk around the block
It would be 9.54x10^6 because the nine is 6 spaces away from the decimal point if you get what i mean.
(2,-5),(-1,4)
slope = (4 - (-5) / (-1 - 2) = (4 + 5) / -3 = -9/3 = -3 <==
Just divide the 83,280 by 12, when you do that you get 6940. So 6940 is how much she makes per month then you take that 6940 and divide it by 4 to get what she makes per week, when you do that you get 1735.
So, Susan Johnson would earn 6940$ per month if she would be paid per month and if she was paid per week she would earn 1735$ per week.
<h2><u>
Answer with explanation:</u></h2>
The confidence interval for population mean (when population standard deviation is unknown) is given by :-
![\overline{x}-t^*\dfrac{s}{\sqrt{n}}< \mu](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D-t%5E%2A%5Cdfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%3C%20%5Cmu%3C%5Coverline%7Bx%7D%2Bz%5E%2A%5Cdfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D)
, where n= sample size
= Sample mean
s= sample size
t* = Critical value.
Given : n= 25
Degree of freedom : ![df=n-1=24](https://tex.z-dn.net/?f=df%3Dn-1%3D24)
![s=\ $19.95](https://tex.z-dn.net/?f=s%3D%5C%20%2419.95)
Significance level for 98% confidence interval : ![\alpha=1-0.98=0.02](https://tex.z-dn.net/?f=%5Calpha%3D1-0.98%3D0.02)
Using t-distribution table ,
Two-tailed critical value for 98% confidence interval :
![t^*=t_{\alpha/2,\ df}=t_{0.01,\ 24}=2.4922](https://tex.z-dn.net/?f=t%5E%2A%3Dt_%7B%5Calpha%2F2%2C%5C%20df%7D%3Dt_%7B0.01%2C%5C%2024%7D%3D2.4922)
⇒ The critical value that should be used in constructing the confidence interval = 2.4922
Then, the 95% confidence interval would be :-
![93.36-(2.4922)\dfrac{19.95}{\sqrt{25}}< \mu](https://tex.z-dn.net/?f=93.36-%282.4922%29%5Cdfrac%7B19.95%7D%7B%5Csqrt%7B25%7D%7D%3C%20%5Cmu%3C93.36%2B%282.4922%29%5Cdfrac%7B19.95%7D%7B%5Csqrt%7B25%7D%7D)
![=93.36-9.943878< \mu](https://tex.z-dn.net/?f=%3D93.36-9.943878%3C%20%5Cmu%3C93.36%2B9.943878)
![=93.36-9.943878< \mu](https://tex.z-dn.net/?f=%3D93.36-9.943878%3C%20%5Cmu%3C93.36%2B9.943878)
![=83.416122< \mu](https://tex.z-dn.net/?f=%3D83.416122%3C%20%5Cmu%3C103.303878%5Capprox83.4161%3C%5Cmu%3C103.3039)
Hence, the 98% confidence interval for the mean repair cost for the dryers. = ![83.4161](https://tex.z-dn.net/?f=83.4161%3C%5Cmu%3C103.3039)