weii ni yo entiendo esa mamada:u
Answer:
b = -7 please mark brainliest :))))
Step-by-step explanation:
Answer:
Step-by-step explanation:
From the given information:
The uniform distribution can be represented by:
![f_x(x) = \dfrac{1}{1500} ; o \le x \le \ 1500](https://tex.z-dn.net/?f=f_x%28x%29%20%3D%20%5Cdfrac%7B1%7D%7B1500%7D%20%3B%20o%20%5Cle%20x%20%5Cle%20%20%20%5C%20%201500)
The function of the insurance is:
![I(x) = \left \{ {{0, \ \ \ x \le 250} \atop {x -20 , \ \ \ \ \ 250 \le x \le 1500}} \right.](https://tex.z-dn.net/?f=I%28x%29%20%3D%20%5Cleft%20%5C%7B%20%7B%7B0%2C%20%5C%20%5C%20%5C%20x%20%5Cle%20250%7D%20%5Catop%20%7Bx%20-20%20%2C%20%5C%20%5C%20%20%5C%20%5C%20%5C%20250%20%5Cle%20x%20%5Cle%201500%7D%7D%20%5Cright.)
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)
![= \dfrac{1}{1500 } \dfrac{(x - 250)^2}{2} \Big |^{1500}_{250}](https://tex.z-dn.net/?f=%3D%20%5Cdfrac%7B1%7D%7B1500%20%7D%20%5Cdfrac%7B%28x%20-%20250%29%5E2%7D%7B2%7D%20%5CBig%20%7C%5E%7B1500%7D_%7B250%7D)
![\dfrac{5}{12} \times 1250](https://tex.z-dn.net/?f=%5Cdfrac%7B5%7D%7B12%7D%20%5Ctimes%201250)
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)
![= \dfrac{1}{1500 } \dfrac{(x - 250)^3}{3} \Big |^{1500}_{250}](https://tex.z-dn.net/?f=%3D%20%5Cdfrac%7B1%7D%7B1500%20%7D%20%5Cdfrac%7B%28x%20-%20250%29%5E3%7D%7B3%7D%20%5CBig%20%7C%5E%7B1500%7D_%7B250%7D)
![\dfrac{5}{18} \times 1250^2](https://tex.z-dn.net/?f=%5Cdfrac%7B5%7D%7B18%7D%20%5Ctimes%201250%5E2)
∴
![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:
![= \sqrt{Var(I(x))}](https://tex.z-dn.net/?f=%3D%20%5Csqrt%7BVar%28I%28x%29%29%7D)
![= 1250 \sqrt{\dfrac{5}{48}}](https://tex.z-dn.net/?f=%3D%201250%20%5Csqrt%7B%5Cdfrac%7B5%7D%7B48%7D%7D)
≅ 404
Answer:
see below
Step-by-step explanation:
Find the magnitude (-1)^2 + (sqrt3)^2 = m^2 m = 2
find the angle arctan (sqrt3/-1) = 120 degrees
the cube root would then be
(cos 40 + isin 40)
A, x-intercept is when y=0, so in this case when y=0, the equation becomes 5x=0. Therefore, x must equal 0 for 5x=0. so y=0 and x=0 so the answer it (0,0), which is A.