<img src="https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B28%7D%20" id="TexFormula1" title=" \sqrt[3]{28} " alt=" \sqrt[3]{28} " ali

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Elena-2011 [213]

3 years ago

gn="absmiddle" class="latex-formula">
28

Mathematics

1 answer:

zloy xaker [14]3 years ago

6 0

I’m not understanding your question maybe explain it differently?

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Please help!<br><br> 22- 25<br><br> Thank you!

Diano4ka-milaya [45]

Answer:

-3

Step-by-step explanation:

7 0

2 years ago

For f(x) = 2x – 3 find f(-1/4)

Firdavs [7]

9514 1404 393

Answer:

-3 1/2

Step-by-step explanation:

Put the argument where the variable is and do the arithmetic.

f(-1/4) = 2(-1/4) -3 = -1/2 -3

f(-1/4) = -(3 1/2)

5 0

3 years ago

Can sum1 help me w this plz

mel-nik [20]

Answer: x = 1.28 or -1.28

5 0

2 years ago

Read 2 more answers

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

Would be great if someone can help with this.

Neporo4naja [7]

Answer:

144

Step-by-step explanation:

In a regular decagon like the one shown in the picture each angle is 144 degrees and it all adds up to 1,440. If you want proof search up "angle of a decagon"

8 0

2 years ago