The point P has coordinates (x,y) = (-2,6) so x = -2 and y = 6
Replace x and y with those values into the rule given
So,
(x,y) ---> (x-2, y-16)
turns into
(-2,6) ---> (-2-2, 6-16) = (-4,-10)
P = (-2,6)
P ' = (-4,-10)
The answer is -10 because your teacher just wants the y coordinate of point P'
(B)
if <em>f</em>(a) = K, then that means <em>f </em>sends (a) to (K).
the inverse function, <em>f</em>-1, goes back where you started, so <em>f</em>-1 <em />sends (K) to (a).
<em>f</em>-1(K) = a.
It is a whole number and integer.
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
The answer is -34 cause if x is -7 and y is 3 then you know you switch em out and do the thing and get the answer.
