Use each of the digits 2, 3, 4, 6, 7, 8 exactly once to

Steve, 18 years old, is three times as old as his brother Daniel. How old will Steve be when he will be twice as old as Daniel?

gtnhenbr [62]

18 ÷ 3 is 6, which means that Daniel is 6, and there is a 12 year gap between Steve and Daniel.

12 x 2 is 24, so therefore when Steve is 24, Daniel will be 12, so Steve is twice as old as Daniel:)

4 0

2 years ago

What is -3x+13=16 help me with each step.

hoa [83]

-3x+13=16

Move +13 to the other side. Sign changes from +13 to -13.

-3x+13-13=16-13

-3x=16-13

-3x=3

Divide by -3 for both sides.

-3/-3x=3/-3

Cross out -3 and -3, divide by -3, then becomes 1*1*x=x

x=-1

Answer: x=-1

6 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

Can someone help me with this

Alborosie

✌HEYA!!!
HERE IS YOUR ANSWER=
THERE ARE ONLY TWO CASES WHEN COMPUTER PICKED A 1 AND A 2 OUT OF 25..
CASES= (1,2);(2,1)

P (E)=2/25
O.O8
WHICH IS OPTION D.

HOPE IT HELPS YOU '_'

7 0

3 years ago

15 points! What is the scale factor of the dilation?

maw [93]

Answer: $\frac{1}{2}$