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GarryVolchara [31]

2 years ago

13

5 Vildia arranges cards for a game into 3 rows of 12 cards. What are two other ways she can arrange the cards so that there are

the same number of cards in each row? Show your work.

1 answer:

atroni [7]2 years ago

3 0

Hopefully this should help

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Someone please answer !!!!

Goryan [66]

The answer is C

By the way, that expression means "n choose k". Hope I helped!

7 0

3 years ago

Read 2 more answers

HELP! 8TH GRADE MATH PLS LOL

LekaFEV [45]

<h2><u>Given</u><u>:</u><u>-</u></h2>

${ \large{➢ \: \: \bf{x = - 8}}}$

<h2><u>Solution</u><u>:</u><u>-</u></h2>

${ \large{➢ \: \: \bf{ \dfrac{2y}{3} = 3 \times - 8 + 34}}}$

${ \large{➢ \: \: \bf{ \dfrac{2y}{3} = - 24 + 34}}}$

${ \large{ \bf{ ➢ \: \: \cancel{2}y = 3 \times { \cancel{(10)} {} \: \: ^{5} }}}}$

${ \large{ \therefore{ \bf{y = 15 \: \: Ans.}}}}$

3 0

3 years ago

A rectangular field is 240m wide. The width of the field is 3/8 the length. What is the area of the field in square kilometers r

andrey2020 [161]

This means you should do 240*3/8 which is 90. so 90 times 240=21600
the answer is 216000km^2

3 0

4 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

The X- and y-coordinates of point P are each to be chosen at random from the set of integers 1 through 10.

bagirrra123 [75]

Answer:

Ok, as i understand it:

for a point P = (x, y)

The values of x and y can be randomly chosen from the set {1, 2, ..., 10}

We want to find the probability that the point P lies on the second quadrant:

First, what type of points are located in the second quadrant?

We should have a value negative for x, and positive for y.

But in our set; {1, 2, ..., 10}, we have only positive values.

So x can not be negative, this means that the point can never be on the second quadrant.

So the probability is 0.

3 0

3 years ago