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4 years ago

Suppose that the dollar cost of producing x appliances is c(x)=1000+ 90x- 0.2x 2

Mathematics

1 answer:

Aleksandr [31]4 years ago

6 0

Answer:

Step-by-step explanation:

Given that the dollar cost of producing x appliances is

$c(x)=1000+ 90x- 0.2x^2$

A) Average cost can be obtained by dividing c(x) by units = x

i.e. average cost = $\frac{1000}{x} +90-0.2x$

B) Marginal cost = dC/dx = $90-0.4x$

When x=140 marginal cost $= 90-0.4(140)\\=34$

C) We calculate

c(140) = $=1000+90(140)-0.2(140^2)\\\\=9680$

c(141) = $=1000+90(141)-0.2(141^2)\\\\\\=9713.80$

Marginal cost = difference = 33.80

Same as 34 shown in marginal cost.

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An internet service provider uses the function R = 3 log(a + 2) + 15 to relate R, the sales revenue from hosting websites, to a,

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Answer:

a=10R−153−2

Step-by-step explanation:

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

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

Write the following equation in standard form:22+52 +4 24 - 220

LekaFEV [45]

Answer:2.78×10^2

Step-by-step explanation:

22+52+424-220

=278

In standard form

=2.78×10^2

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3 years ago

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Your answer is C). Pi/4

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3 years ago

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Mila [183]

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3 years ago

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