Since 36x² + 60x + 25 is the perfect square trinomial, we use this formula a² + 2ab + b² = (a + b)². Hence, the solution is (6x + 5)²
Answer:
1 : 1
Step-by-step explanation:
When a circle is inscribed in a cylinder, the height of the cylinder is equal to the diameter of the sphere and the radius of the cylinder is same as that of the sphere.
Let the radius of sphere is r.
height of cylinder, h = 2r
radius of cylinder = r
Surface area of sphere, A = 4πr²
lateral surface area of cylinder, A' = 2 πrh
A' = 2πr x 2r = 4πr²
The ratio of surface area of sphere to the lateral surface area of cylinder is 1 : 1.
Answer:
Its x=-10
Step-by-step explanation:
С = πD
С = 3,14×40
С = 126 in ← <span>to the nearest inch </span>
Answer:
(a)![E[X+Y]=E[X]+E[Y]](https://tex.z-dn.net/?f=E%5BX%2BY%5D%3DE%5BX%5D%2BE%5BY%5D)
(b)
Step-by-step explanation:
Let X and Y be discrete random variables and E(X) and Var(X) are the Expected Values and Variance of X respectively.
(a)We want to show that E[X + Y ] = E[X] + E[Y ].
When we have two random variables instead of one, we consider their joint distribution function.
For a function f(X,Y) of discrete variables X and Y, we can define
![E[f(X,Y)]=\sum_{x,y}f(x,y)\cdot P(X=x, Y=y).](https://tex.z-dn.net/?f=E%5Bf%28X%2CY%29%5D%3D%5Csum_%7Bx%2Cy%7Df%28x%2Cy%29%5Ccdot%20P%28X%3Dx%2C%20Y%3Dy%29.)
Since f(X,Y)=X+Y
![E[X+Y]=\sum_{x,y}(x+y)P(X=x,Y=y)\\=\sum_{x,y}xP(X=x,Y=y)+\sum_{x,y}yP(X=x,Y=y).](https://tex.z-dn.net/?f=E%5BX%2BY%5D%3D%5Csum_%7Bx%2Cy%7D%28x%2By%29P%28X%3Dx%2CY%3Dy%29%5C%5C%3D%5Csum_%7Bx%2Cy%7DxP%28X%3Dx%2CY%3Dy%29%2B%5Csum_%7Bx%2Cy%7DyP%28X%3Dx%2CY%3Dy%29.)
Let us look at the first of these sums.
![\sum_{x,y}xP(X=x,Y=y)\\=\sum_{x}x\sum_{y}P(X=x,Y=y)\\\text{Taking Marginal distribution of x}\\=\sum_{x}xP(X=x)=E[X].](https://tex.z-dn.net/?f=%5Csum_%7Bx%2Cy%7DxP%28X%3Dx%2CY%3Dy%29%5C%5C%3D%5Csum_%7Bx%7Dx%5Csum_%7By%7DP%28X%3Dx%2CY%3Dy%29%5C%5C%5Ctext%7BTaking%20Marginal%20distribution%20of%20x%7D%5C%5C%3D%5Csum_%7Bx%7DxP%28X%3Dx%29%3DE%5BX%5D.)
Similarly,
![\sum_{x,y}yP(X=x,Y=y)\\=\sum_{y}y\sum_{x}P(X=x,Y=y)\\\text{Taking Marginal distribution of y}\\=\sum_{y}yP(Y=y)=E[Y].](https://tex.z-dn.net/?f=%5Csum_%7Bx%2Cy%7DyP%28X%3Dx%2CY%3Dy%29%5C%5C%3D%5Csum_%7By%7Dy%5Csum_%7Bx%7DP%28X%3Dx%2CY%3Dy%29%5C%5C%5Ctext%7BTaking%20Marginal%20distribution%20of%20y%7D%5C%5C%3D%5Csum_%7By%7DyP%28Y%3Dy%29%3DE%5BY%5D.)
Combining these two gives the formula:

Therefore:
![E[X+Y]=E[X]+E[Y] \text{ as required.}](https://tex.z-dn.net/?f=E%5BX%2BY%5D%3DE%5BX%5D%2BE%5BY%5D%20%5Ctext%7B%20%20as%20required.%7D)
(b)We want to show that if X and Y are independent random variables, then:

By definition of Variance, we have that:
![Var(X+Y)=E(X+Y-E[X+Y]^2)](https://tex.z-dn.net/?f=Var%28X%2BY%29%3DE%28X%2BY-E%5BX%2BY%5D%5E2%29)
![=E[(X-\mu_X +Y- \mu_Y)^2]\\=E[(X-\mu_X)^2 +(Y- \mu_Y)^2+2(X-\mu_X)(Y- \mu_Y)]\\$Since we have shown that expectation is linear$\\=E(X-\mu_X)^2 +E(Y- \mu_Y)^2+2E(X-\mu_X)(Y- \mu_Y)]\\=E[(X-E(X)]^2 +E[Y- E(Y)]^2+2Cov (X,Y)](https://tex.z-dn.net/?f=%3DE%5B%28X-%5Cmu_X%20%20%2BY-%20%5Cmu_Y%29%5E2%5D%5C%5C%3DE%5B%28X-%5Cmu_X%29%5E2%20%20%2B%28Y-%20%5Cmu_Y%29%5E2%2B2%28X-%5Cmu_X%29%28Y-%20%5Cmu_Y%29%5D%5C%5C%24Since%20we%20have%20shown%20that%20expectation%20is%20linear%24%5C%5C%3DE%28X-%5Cmu_X%29%5E2%20%20%2BE%28Y-%20%5Cmu_Y%29%5E2%2B2E%28X-%5Cmu_X%29%28Y-%20%5Cmu_Y%29%5D%5C%5C%3DE%5B%28X-E%28X%29%5D%5E2%20%20%2BE%5BY-%20E%28Y%29%5D%5E2%2B2Cov%20%28X%2CY%29)
Since X and Y are independent, Cov(X,Y)=0

Therefore as required:
