Answer:
Step-by-step explanation:
From the given information:
The uniform distribution can be represented by:

The function of the insurance is:

Hence, the variance of the insurance can also be an account forum.
![Var [I_{(x}) = E [I^2(x)] - [E(I(x)]^2](https://tex.z-dn.net/?f=Var%20%5BI_%7B%28x%7D%29%20%3D%20E%20%5BI%5E2%28x%29%5D%20-%20%5BE%28I%28x%29%5D%5E2)
here;
![E[I(x)] = \int f_x(x) I (x) \ sx](https://tex.z-dn.net/?f=E%5BI%28x%29%5D%20%3D%20%5Cint%20f_x%28x%29%20I%20%28x%29%20%5C%20sx)
![E[I(x)] = \dfrac{1}{1500} \int ^{1500}_{250{ (x- 250) \ dx](https://tex.z-dn.net/?f=E%5BI%28x%29%5D%20%3D%20%5Cdfrac%7B1%7D%7B1500%7D%20%5Cint%20%5E%7B1500%7D_%7B250%7B%20%28x-%20250%29%20%5C%20dx)


Similarly;
![E[I^2(x)] = \int f_x(x) I^2 (x) \ sx](https://tex.z-dn.net/?f=E%5BI%5E2%28x%29%5D%20%3D%20%5Cint%20f_x%28x%29%20I%5E2%20%28x%29%20%5C%20sx)
![E[I(x)] = \dfrac{1}{1500} \int ^{1500}_{250{ (x- 250)^2 \ dx](https://tex.z-dn.net/?f=E%5BI%28x%29%5D%20%3D%20%5Cdfrac%7B1%7D%7B1500%7D%20%5Cint%20%5E%7B1500%7D_%7B250%7B%20%28x-%20250%29%5E2%20%5C%20dx)


∴
![Var {I(x)} = 1250^2 \Big [ \dfrac{5}{18} - \dfrac{25}{144}]](https://tex.z-dn.net/?f=Var%20%7BI%28x%29%7D%20%3D%201250%5E2%20%5CBig%20%5B%20%5Cdfrac%7B5%7D%7B18%7D%20-%20%5Cdfrac%7B25%7D%7B144%7D%5D)
Finally, the standard deviation of the insurance payment is:


≅ 404
Answer:
15/164
Step-by-step explanation:
5/8 ÷ 6 5/6 = 15 / 164
Grouping is when you have to terms that are the same inside the parenthesis and a different or same number outside the parenthesis and then the one inside the parenthesis u are going to group them not combined them and write them once and the numbers inside the parenthesis group them together by adding them and put parenthesis between them
Ex
5(4x+1)10(4x+1)
(5+10)(4x+1)
Answer:
um michelles broke that couldnt happen
Step-by-step explanation:
Answer:
Step-by-step explanation:
3/5 + 1/3
taking lcm we get 15
=<u> 3 * 3 + 1 * 5</u>
15
= <u>9 + 5</u>
15
=14/ 15