Answer:
5 x 4 + 2 - 3 / 1
Step-by-step explanation:
Took me a lot of tries but got it! :) Hope this helps
Answer:
![s^2 = \frac{(1.48-1.502)^2 +(1.45-1.502)^2 +(1.54-1.502)^2 +(1.52-1.502)^2 +(1.52-1.502)^2}{5-1} =0.00132](https://tex.z-dn.net/?f=s%5E2%20%3D%20%5Cfrac%7B%281.48-1.502%29%5E2%20%2B%281.45-1.502%29%5E2%20%2B%281.54-1.502%29%5E2%20%2B%281.52-1.502%29%5E2%20%2B%281.52-1.502%29%5E2%7D%7B5-1%7D%20%3D0.00132)
And for the deviation we have:
![s= \sqrt{0.00132}=0.0363](https://tex.z-dn.net/?f=%20s%3D%20%5Csqrt%7B0.00132%7D%3D0.0363)
And that value represent the best estimator for the population deviation since:
Step-by-step explanation:
For this case we have the following data:
1.48,1.45,1.54,1.52,1.52
The first step for this cae is find the sample mean with the following formula:
![\bar X =\frac{\sum_{i=1}^n X_}{n}](https://tex.z-dn.net/?f=%5Cbar%20X%20%3D%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5En%20X_%7D%7Bn%7D)
And replacing we got:
![\bar X= \frac{1.48+1.45+1.54+1.52+1.52}{5} = 1.502](https://tex.z-dn.net/?f=%20%5Cbar%20X%3D%20%5Cfrac%7B1.48%2B1.45%2B1.54%2B1.52%2B1.52%7D%7B5%7D%20%3D%201.502)
And now we can calculate the sample variance with the following formula:
![s^2 =\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}](https://tex.z-dn.net/?f=%20s%5E2%20%3D%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5En%20%28X_i%20-%5Cbar%20X%29%5E2%7D%7Bn-1%7D)
And replacing we got:
![s^2 = \frac{(1.48-1.502)^2 +(1.45-1.502)^2 +(1.54-1.502)^2 +(1.52-1.502)^2 +(1.52-1.502)^2}{5-1} =0.00132](https://tex.z-dn.net/?f=s%5E2%20%3D%20%5Cfrac%7B%281.48-1.502%29%5E2%20%2B%281.45-1.502%29%5E2%20%2B%281.54-1.502%29%5E2%20%2B%281.52-1.502%29%5E2%20%2B%281.52-1.502%29%5E2%7D%7B5-1%7D%20%3D0.00132)
And for the deviation we have:
![s= \sqrt{0.00132}=0.0363](https://tex.z-dn.net/?f=%20s%3D%20%5Csqrt%7B0.00132%7D%3D0.0363)
And that value represent the best estimator for the population deviation since:
The product of 19 +85 is 104
I believe it should be 36.00$
Since the court is 9 meters wide we would have to multiply that by 4 because each meter is 4$