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3 years ago

Answers...........................?

Mathematics

1 answer:

Lemur [1.5K]3 years ago

7 0

1.25/5 = 0.25

Your calculator can help with this if you don't happen to know already that 1/5 of 5/4 = 1/4.

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sue and 4 friends buy a box of 362 base ball cards at a yard sale. If they share the cards equally,how many cards will each pers

Radda [10]

They should receive 72 each 362 divided by 5 (number people)

4 0

3 years ago

What is the value of 5.4?

Galina-37 [17]

The value is .54 hope this helped

5 0

3 years ago

Angelina wants to buy the same number of pens and pencils. Pens come in 8 packs and pencils come in packs of 9. What is the leas

andrew11 [14]

Answer:

Angelina should buy 9 packs of pens and 8 packs of pencils.

Step-by-step explanation:

Each pack of pens = 8 pens

Each pack of pencils = 9 pencils.

Also we know that, Commutative Property of Multiplication states that

a x b = b x a for any two numbers a and b.

So, 8 x 9 = 72 = 9 x 8

Hence, she should buy 9 packs of pens, so she will have 9 x 8 = 72 pens.

And she should buy 8 packs of pencils, she will have 8 x 9 = 72 pencils.

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

3 years ago

I need help!!! Please help me cuties<br> REPORTING LINKS!!

vitfil [10]

Answer:

1. 15.4

2. 9.66

3. 13.16

4. 12.32

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers