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3 years ago

The satellite photos seemed to indicate

Mathematics

1 answer:

Dafna1 [17]3 years ago

8 0

Answer:the warehouse's width is 100 feet.

Step-by-step explanation:

Let L represent the length of the rectangular warehouse.

Let W represent the width of the rectangular warehouse.

The rectangular warehouse was 20 feet longer than 3 times its width. This means that

L = 3W + 20

The perimeter of a rectangle is

the sum of all its sides. Therefore, the formula for determining the perimeter of a rectangle expressed as

Perimeter = (L + W)

If the perimeter of the warehouse was 840 feet, it means that

2(L + W) = 840

Dividing through by 2, it becomes

L + W = 420- - - - - - - - 1

Substituting L = 3W + 20 into equation 1, it becomes

3W + 20 + W = 420

4W = 420 - 20

4W = 400

W = 400/4 = 100

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3 years ago

the first three steps of completing the square to solve the quadratic equation X2+4X-6=0 are shown below. what re the next 3 ste

Vilka [71]

Answer:

see the explanation

Step-by-step explanation:

<u><em>The complete question is</em></u>

The first three steps of completing the square to solve the quadratic equation x2 +4x -6= 0, are shown below.

Step 1: x2 +4x= 6

Step 2: x2 +4x+4=6+4

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we have

step 4

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$\sqrt{(x+2)^2} =\sqrt{10}$

step 5

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step 6

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2 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:

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

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