Answer:
This might not help but I know 10 to the power of 3 is 1000 and the answer might be 10×10×10. 10×10×10 is equal to 1000 and is the same as 10 to the power of 3.
If you put 5 which means its 2 1/2 ( I think) Because two 1/2 makes one.
Answer:
I can't see three but I'll answer the first two.
Step-by-step explanation:
Number one: The first, third, and fifth options are correct,
Number two: A is correct because if you divide 3/4 you get 0.75. Check by multiplying 0.75 for one by 4 and you get three dollars.
So for three, one pound of pears costs two dollars. That is consistent with the graphs. Pick the answer that states that the best.
Answer:
P(A∣D) = 0.667
Step-by-step explanation:
We are given;
P(A) = 3P(B)
P(D|A) = 0.03
P(D|B) = 0.045
Now, we want to find P(A∣D) which is the posterior probability that a computer comes from factory A when given that it is defective.
Using Bayes' Rule and Law of Total Probability, we will get;
P(A∣D) = [P(A) * P(D|A)]/[(P(A) * P(D|A)) + (P(B) * P(D|B))]
Plugging in the relevant values, we have;
P(A∣D) = [3P(B) * 0.03]/[(3P(B) * 0.03) + (P(B) * 0.045)]
P(A∣D) = [P(B)/P(B)] [0.09]/[0.09 + 0.045]
P(B) will cancel out to give;
P(A∣D) = 0.09/0.135
P(A∣D) = 0.667
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:
