Factor as the product of two binomials <br> X^2 +8x +12 = ?

The perimeter of a rectangle is 50 cm and the length is 4 cm longer than the width. Write and solve the system of equations usin

laila [671]

I hope this helps you

3 0

3 years ago

( − 35 − 42 − 63 ) ∶ ( + 7 )

olga nikolaevna [1]

(-35-42-63)/7= (-140)/7=-20

7 0

2 years ago

Determine wether the function is increasing, decreasing, or constant

evablogger [386]

Answer:

decreasing

Step-by-step explanation:

"Increasing" means the graph is going up from left to right.

"Decreasing" means the graph is going down from left to right.

"Constant" means the graph is "flat" (this is not a technical term) it is keeping the same y value, neither going up nor going down.

What can be super confusing is the

(2.2, 5) mentioned in the question. THIS IS NOT A POINT. It is an interval and points and intervals unfortunately have the same notation sometimes.

An "interval" is a section of the graph, here: FROM 2.2 not including 2.2, TO 5 not including 5. These are like the address on the x-axis. If you look at your graph at 2.2 on the x-axis, it is a peak(relative maximum) and it goes down to the right to where x is 5 where it bottoms out (relative minimum) So on that interval, from 2.2 to 5, the graph is DECREASING.

4 0

2 years ago

16+ greater or less than+19

saul85 [17]

By 16+ do you mean positive 16 because if you do then it is less than 19+

7 0

3 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

Factor as the product of two binomials X^2 +8x +12 = ?