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3 years ago

What is the value of X

Mathematics

1 answer:

user100 [1]3 years ago

7 0

Answer:

See below for answers and explanations

Step-by-step explanation:

4) 35° and x° are vertical angles, so they are congruent to each other. Therefore, x°=35°. Since y°+y°+35°=180° since they are all supplementary to each other, then y°=72.5° when solved.

5) A is true. Think of two adjacent angles measured 60° and 30° that are next to each other. They add up to 90° as well.

B isn't true. Vertical angles are congruent to each other, not supplementary.

C isn't true. The statement contradicts itself. Complementary angles add up to 90° while supplementary angles add up to 180°

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May someone please help me with this :)

AlladinOne [14]

Answer: 16 pi or ≈50.27

Explanation: Formula for a cylinder:
V = pi r ^2 h
V = pi (2)^2 4
V = 4 pi (4)
V = 16pi

8 0

2 years ago

Ms. Jan brought in cookies for her class. She gave out half of them in the morning. At lunch, she gave out 12 more. She then had

Olegator [25]

Answer:

Step-by-step explanation:

if she had 10 left at the end and you add the other half from lunch which is 12, then that gives you 22. Since before lunch she had half (22), all you gotta do is add the other half which is 22. 22+22=44 and that's your answer.

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3 years ago

I am trying to solve for x

padilas [110]

Here how i found my answer.

5 0

3 years ago

Please help ASAP! i will mark brainliest

Kitty [74]

Use two pints on the graph to find the slope:

(0,0) and (5,1)

Slope = Change in Y over the change in x.

Slope = (1-0) / (5-0) = 1/5 = 0.20

The slope of the given equation is 0.25 which is greater than 0.20.

The unit rate is greater in the equation.

8 0

3 years ago

Let X and Y be discrete random variables. Let E[X] and var[X] be the expected value and variance, respectively, of a random vari

Ulleksa [173]

Answer:

(a) $E[X+Y]=E[X]+E[Y]$

(b) $Var(X+Y)=Var(X)+Var(Y)$

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

$E[f(X,Y)]=\sum_{x,y}f(x,y)\cdot P(X=x, Y=y).$

Since f(X,Y)=X+Y

$E[X+Y]=\sum_{x,y}(x+y)P(X=x,Y=y)\\=\sum_{x,y}xP(X=x,Y=y)+\sum_{x,y}yP(X=x,Y=y).$

Let us look at the first of these sums.

$\sum_{x,y}xP(X=x,Y=y)\\=\sum_{x}x\sum_{y}P(X=x,Y=y)\\\text{Taking Marginal distribution of x}\\=\sum_{x}xP(X=x)=E[X].$

Similarly,

$\sum_{x,y}yP(X=x,Y=y)\\=\sum_{y}y\sum_{x}P(X=x,Y=y)\\\text{Taking Marginal distribution of y}\\=\sum_{y}yP(Y=y)=E[Y].$

Combining these two gives the formula:

$\sum_{x,y}xP(X=x,Y=y)+\sum_{x,y}yP(X=x,Y=y) =E(X)+E(Y)$

Therefore:

$E[X+Y]=E[X]+E[Y] \text{ as required.}$

(b)We want to show that if X and Y are independent random variables, then:

$Var(X+Y)=Var(X)+Var(Y)$

By definition of Variance, we have that:

$Var(X+Y)=E(X+Y-E[X+Y]^2)$

$=E[(X-\mu_X +Y- \mu_Y)^2]\\=E[(X-\mu_X)^2 +(Y- \mu_Y)^2+2(X-\mu_X)(Y- \mu_Y)]\\$Since we have shown that expectation is linear$\\=E(X-\mu_X)^2 +E(Y- \mu_Y)^2+2E(X-\mu_X)(Y- \mu_Y)]\\=E[(X-E(X)]^2 +E[Y- E(Y)]^2+2Cov (X,Y)$

Since X and Y are independent, Cov(X,Y)=0

$=Var(X)+Var(Y)$

Therefore as required:

$Var(X+Y)=Var(X)+Var(Y)$

7 0

3 years ago