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2 years ago

Solve the inequality −8 ≤ z + 6.4

Mathematics

1 answer:

drek231 [11]2 years ago

5 0

Answer:

-14.4<or=z

Step-by-step explanation:

-8-6.4or = z+6.4-6.4

-14.4_<z

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One side of a rectangle is 3ft shorter than twice the other side. Find the sides if the perimeter is 24ft

Svetllana [295]

Answer:

X=7, Y=5

Step-by-step explanation:

In the attached file

3 0

3 years ago

Steven paid $3.50 for bowling shoes and then $1.50 per game after that. Write an equation that models this.

Korvikt [17]

Answer:

T = 3.50 + 1.50G

Step-by-step explanation:

G = number of games

T = total cost

total cost = cost of shoes + cost per game

T = 3.50 + 1.50G

I'm not 100% positive, but it's my best guess.

3 0

2 years ago

F(x)=11(-4)^2-5(-4)+13<br> =-143<br> (-4, -143)???

lora16 [44]

Answer:

(-4,169)

Step-by-step explanation:

11 (-4)^2 - 5(-4)+13

11(16)- 20 +13

176 -20+13

169

7 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

2 years ago

The probability that a randomly selected point on the grid below is in the blue area is 9/16

DochEvi [55]

Answer:

The correct option is 1.

Step-by-step explanation:

it is given that the probability of a randomly selected point on the grid below is in the blue area is 9/16.

In the given grid we have only two colors that are blue and white.

Let A be the event of a randomly selected point on the given grid is in the blue area.

$P(A)=\frac{9}{16}$

If the randomly selected point on the given grid does not lie in the blue area, it means it lies on white area.

We will calculate P(A'), to find the probability that a randomly selected point is in the white on the grid.

We know that the sum of the probability is 1.

$P(A)+P(A')=1$

$P(A')=1-P(A)$

It is given that $P(A)=\frac{9}{16}$

$P(A')=1-\frac{9}{16}$

Therefore option 1 is correct.

8 0

2 years ago

Read 2 more answers