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3 years ago

If two lines don’t intersect, then is it parrell (( true or false ))

1 answer:

7 0

The definition of parallel lines states that they don't intersect, so that would always be true. (In Euclidean geometry, anyway.)

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Answer:

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Answer:

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The blue figure is a translation of the black image. Write a rule to describe the translation.

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3 years ago

What is the length of the rectangle?

gregori [183]

Step-by-step explanation:

$area \: of \: rectangle \: = 60 \: {yards}^{2} \\ \therefore \: (w + 4) \times \: w = 60\\ \\ \therefore \: {w}^{2} + 4w -60 = 0 \\ \\ \therefore \: {w}^{2} + 10w - 6w -60 = 0 \\ \\ \therefore \: w({w} + 10) - 6(w + 10)= 0 \\ \\ \therefore \: ({w} + 10) (w- 6)= 0 \\ \therefore \: {w} + 10 = 0 \: or \: w- 6 = 0 \\ \therefore \: {w} = - 10 \: or \: w = 6 \\ \because \: sides \: of \: rectangle \: cant \: be \: negative \\ \therefore \: w \neq \: - 10 \\ \\ \therefore \: w = 6 \\ \\ \therefore \: w + 4 = 6 + 4 = 10 \\ \\ length \: of \: rectangle \: = 10 \: yards.$

4 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