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3 years ago

How many??? Please help

Mathematics

2 answers:

laiz [17]3 years ago

7 0

They bought 2 cat toys and 3 dog toys.

Anika [276]3 years ago

4 0

3 dog toys and 2 cat toys

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Which of the following complete the proof above?

svp [43]

Answer:

(A)

Step-by-step explanation:

To be honest, since all of them are different, as long as you get the first one right, you will get the others right. A is your answer, for the following reasons:

Reflexive Property: The Reflexive property states that the mirrored version of itself will always be congruent with itself. Generally, to make it easier, they will write it opposite (as they did in the given proof):

<em>line</em> AC ≅ <em>line</em> CA

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3 years ago

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Which graph represents the function?

mario62 [17]

For this function we can find y-intercept.
x=0, y=-2
This graph is on the top, right.

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3 years ago

Find the value of 7w-2 given that - 2w-4=6 .

Vesnalui [34]

Answer:

w= -5

Step-by-step explanation:

-2w - 4 = 6

-2w - 4 + 4 = 6 + 4
-2w = 10

Isolate "w":

$\frac{-2w}{-2} = \frac{10}{-2}$

w = -5

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