How do you subtract negative numbers using counters

Solve the system of two linear inequalities graphically.

Sladkaya [172]

Answer:

See attachment

Step-by-step explanation:

Isolate y in the first inequality:

$2x + 4y < -16\\4y < - 16 -2 x\\y < -4 -\frac{1}{2} x\\y < -\frac{x+8}{2} \\$

Now, with both x and y inequalities found, graph it.

7 0

1 year ago

(-3+x) divided by 2= negative one half

Arlecino [84]

Answer:

are u trying to find x?

Step-by-step explanation:

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

Solve the system equation

den301095 [7]

<h2>Answer:</h2>

We will use substitution to solve this system for x.

$x - 2y = 27\\y = 5x\\\\x - 2(5x) = 27\\\\x - 10x = 27\\\\-9x = 27\\\\x = -3$

Now that we know what x is, we can determine y by pluging in -3 for x in one of the original equations. I will use the first one.

$-3 - 2y = 27\\\\-2y = 30\\\\y = -15\\$

We now have our solution.

D. $(-3, -15)$

We can check our answer by pluging the solution into one of the original equations to see if both sides equal each other. I will use the first equation.

$-3 - 2(-15) = 27\\\\-3 + 30 = 27\\\\27 = 27$