<u>EXPLANATION</u><u>:</u>
Given that
sin θ = 1/2
We know that
sin 3θ = 3 sin θ - 4 sin³ θ
⇛sin 3θ = 3(1/2)-4(1/2)³
⇛sin 3θ = (3/2)-4(1/8)
⇛sin 3θ = (3/2)-(4/8)
⇛sin 3θ = (3/2)-(1/2)
⇛sin 3θ = (3-1)/2
⇛sin 3θ = 2/2
⇛sin 3θ = 1
and
cos 2θ = cos² θ - sin² θ
⇛cos 2θ = 1 - sin² θ - sin² θ
⇛cos 2θ = 1 - 2 sin² θ
Now,
cos 2θ = 1-2(1/2)²
⇛cos 2θ = 1-2(1/4)
⇛cos 2θ = 1-(2/4)
⇛cos 2θ = 1-(1/2)
⇛cos 2θ = (2-1)/2
⇛cos 2θ = 1/2
Now,
The value of sin 3θ /(1+cos 2θ
⇛1/{1+(1/2)}
⇛1/{(2+1)/2}
⇛1/(3/2)
⇛1×(2/3)
⇛(1×2)/3
⇛2/3
<u>Answer</u> : Hence, the req value of sin 3θ /(1+cos 2θ) is 2/3.
<u>also</u><u> read</u><u> similar</u><u> questions</u><u>:</u> If sin Θ = 2/3 and tan Θ < 0, what is the value of cos Θ?
brainly.com/question/12618768?referrer
if cos θ -sin θ =1, find θ cos θ +sin θ?
brainly.com/question/88125?referrer
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:
Answer:
The formula for area is length times width.
it's 8m long, and 6m wide. When we multiply it, we get 48m that is for one side of the cube, so we count how many sides there is. There's 6, so we multiply 6 times 48 and the complete surface area for the whole cube is 288m
The least common multiple (LCM) of a set of numbers is the lowest value that all of the numbers can be divided by and have a result of a whole number.
In this case, the least common multiple is 21 because 21/7=3 and 21/21=1.
There are 6 sides on a dice. This means each number has a 1/6 chance of appearing. (1 number out of 6 sides). This includes 6, so the answer to a is 1/6. If you roll or do not roll a six, this means every side of the dice counts. The answer to be, by extension, is 6/6. Finally, c is b (the probability of anything) minus a (the probability of a six), representing every situation that is not a 6, so 5/6.
