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3 years ago

HELP ME PLEASE

Mathematics

2 answers:

aivan3 [116]3 years ago

7 0

Answer:

Answer is A

Step-by-step explanation:

Len [333]3 years ago

4 0

Answer:

Step-by-step explanation:

Answer is C because if swimmer A does 14 laps to start an 3 laps every day after that and on the chart swimmer B does 11 laps to start and 2 laps every day after that you would compare.

on day 3 swimmer B has 17 (shown on chart)

on day 3 swimmer A has 23 (14, 17, 20, 23)

if you would want it to complicate it 14:3 and 11:2

14 ÷ 14 is 1

3 ÷ 14 is .214

11 ÷ 11 is 1

2 ÷ 11 is .181

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Which is the property that justifies the action?

lidiya [134]

A) Action: multiply both sides by 4.

Have a look:

(x/4)x4=16x4
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As you can see, the value of x can be found by multiplying 4 on both sides.

Hope I helped :)

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4 years ago

Read 2 more answers

If you wanted to make the graph of y = 8x - 10 less steep, as well as shift the

azamat

Answer:

4x-15=y

Step-by-step explanation:

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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$\\=E(X-\mu_X)^2 +E(Y- \mu_Y)^2+2E(X-\mu_X)(Y- \mu_Y)]\\=E[(X-E(X)]^2 +E[Y- E(Y)]^2+2Cov (X,Y)$