Answer : -4a+10
Substitute (a-3) for x
-4(a-3)-2
-4a+12-2
-4a+10 but can also be written as -2(2a-5)
Answer:
Linear: The first and third one. (0,5) & (1,1)
Exponential: The second one (above). (-2, 1/16)
Quadratic function: The last one (below). (-3,35)
Step-by-step explanation:
The graph is below the dark pink is where both equations overlap
Answer:
-764.28
Step-by-step explanation:
Given the joint cumulative distribution of X and Y as
![F(x,y) = \frac{xy(x+y)}{2000000}\ \ \ \, \ 0\leq x100, 0\leq y\leq 100](https://tex.z-dn.net/?f=F%28x%2Cy%29%20%3D%20%5Cfrac%7Bxy%28x%2By%29%7D%7B2000000%7D%5C%20%5C%20%5C%20%5C%2C%20%5C%200%5Cleq%20x100%2C%200%5Cleq%20y%5Cleq%20100)
#First find
and probability distribution function ,
:
![F_x(x)=F(x,100)\\\\\\=\frac{100x(x+100)}{2000000}\\\\\\\\=\frac{100x^2+10000x}{2000000}\\\\\\=>f_x(x)=\frac{x}{10000}+\frac{1}{200}](https://tex.z-dn.net/?f=F_x%28x%29%3DF%28x%2C100%29%5C%5C%5C%5C%5C%5C%3D%5Cfrac%7B100x%28x%2B100%29%7D%7B2000000%7D%5C%5C%5C%5C%5C%5C%5C%5C%3D%5Cfrac%7B100x%5E2%2B10000x%7D%7B2000000%7D%5C%5C%5C%5C%5C%5C%3D%3Ef_x%28x%29%3D%5Cfrac%7Bx%7D%7B10000%7D%2B%5Cfrac%7B1%7D%7B200%7D)
#Have determined the probability distribution unction ,
, we calculate the Expectation of the random variable X:
![E(X)=\int\limits^{100}_0 \frac{x^2}{10000}+\frac{x}{200} dx \\\\\\\\=|\frac{x^3}{30000}+\frac{x^2}{400}|\limits^{100}_0\\\\=58.33\\\\](https://tex.z-dn.net/?f=E%28X%29%3D%5Cint%5Climits%5E%7B100%7D_0%20%5Cfrac%7Bx%5E2%7D%7B10000%7D%2B%5Cfrac%7Bx%7D%7B200%7D%20%20dx%20%5C%5C%5C%5C%5C%5C%5C%5C%3D%7C%5Cfrac%7Bx%5E3%7D%7B30000%7D%2B%5Cfrac%7Bx%5E2%7D%7B400%7D%7C%5Climits%5E%7B100%7D_0%5C%5C%5C%5C%3D58.33%5C%5C%5C%5C)
#We then calculate
:
![E(X^2)=\int\limits^{100}_0 \frac{x^3}{10000}+\frac{x^2}{200}\ dx\\\\=\frac{x^4}{40000}+\frac{x^3}{600}|\limits^{100}_0=4166.67\\\\Var(X)=E(X^2)-(E(X))^2=4166.67-58.33^2\\\\Var(X)=764.28](https://tex.z-dn.net/?f=E%28X%5E2%29%3D%5Cint%5Climits%5E%7B100%7D_0%20%5Cfrac%7Bx%5E3%7D%7B10000%7D%2B%5Cfrac%7Bx%5E2%7D%7B200%7D%5C%20dx%5C%5C%5C%5C%3D%5Cfrac%7Bx%5E4%7D%7B40000%7D%2B%5Cfrac%7Bx%5E3%7D%7B600%7D%7C%5Climits%5E%7B100%7D_0%3D4166.67%5C%5C%5C%5CVar%28X%29%3DE%28X%5E2%29-%28E%28X%29%29%5E2%3D4166.67-58.33%5E2%5C%5C%5C%5CVar%28X%29%3D764.28)
Hence, the Var(X) is 764.28
It is 4 I believe sorry if I’m wrong