Please answer the questions before 12 am

you have an annual salary of 82,338 your monthly expenes include a 1,153 mortagage payment a 380 car lease payment, 131 in minum

Marysya12 [62]

the answer is:

80.503

7 0

3 years ago

Brandon and his friend Adam both mow lawns. Brandon earns $ per week mowing lawns while Adam earns $ per week mowing lawns. Part

olga55 [171]

Answer:

Step-by-step explanation:

3 0

2 years ago

if each cube inside the rectangular prism has an edge length of three-forth inch then what is the volume

sergeinik [125]

Base×hight×width you are just doing 3/4×3/4×3/4 and what ever your answer is put a little 3 above it so it almost looks like an exponent. Pretend that ABC is the answer it would look like this format = ABCinches3 but the three is tiny

3 0

2 years ago

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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

2 years ago

According to the table, which ordered pair is a local minimum of the function, f(x)?

sveticcg [70]

Answer:

Option D

Step-by-step explanation:

Given is a table which gives the order pair of x and y for a function f(x)

We have to find the local minimum of the function f(x)

We have from the table the values of different f(x)

Of all we find the least value is -15 and for this x value is -2 or +2

Hence f(x) has minimum at two points

(-2,-15) and (2,-15)

Out of 4 options given we find that (2,-15) appears in IV option

Hence option D is right answer

7 0

3 years ago

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