Answer:
yes
Step-by-step explanation:
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:
Slope of PV = -4
Step-by-step explanation:
Coordinates of the vertices of the given triangle PSV,
P → (2, 4)
S → (3, 4)
V → (3, 0)
Slope of the segment between two points
and
is given by,
Slope = 
Therefore, slope of segment PV will be,
Slope = 
= -4
Slope of PV = -4
In this case, the vector is passing through a right angle as shown with the little square on the left side. (all right angles equal 90 degrees) so you would simply subtract 64 from 90. (90 - 64 = 26) so x equals 26 degrees.
Answer:
a:
10
b:
-10 and 10
c:
8
Step-by-step explanation:
i could be wrong on c tho