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3 years ago

What numbers should i put in the drop down menus? the numbers i can choose are 0-9.

Mathematics

1 answer:

ElenaW [278]3 years ago

6 0

Answer:

1x+4 should be the answer

Step-by-step explanation:

6x-3x-2x=1x

4=4

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Name the property illustrated.<br><br> 6 + 4 + 3 = 6 + 3 + 4

Iteru [2.4K]

Commutative property states that order does not matter. Multiplication and addition are commutative. Related Links: Properties. Associative, Distributive and commutative properties.

5 0

3 years ago

How many computers must the AB Computer Company sell to break even? Let x be the number of computers.

Eduardwww [97]

Answer:

The Company AB must sell 116 computers to break even

Step-by-step explanation:

we have

$R(x)=1.5x$ -----> equation A

$C(x)=145+\frac{1}{4}x$ ----> equation B

where

C is the cost function

R is the revenue function

x is the number of computers sold

we know that

Break even is when the cost is equal to the revenue

equate equation A and equation B

$1.5x=145+\frac{1}{4}x$

Solve for x

subtract 1/4x both sides

$1.5x-\frac{1}{4}x=145$

$1.5x-0.25x=145$

$1.25x=145$

Divide by 1.25 both sides

$x=116$

therefore

The Company AB must sell 116 computers to break even

7 0

3 years ago

Need an explanation and answer for -25/28 ÷ (5/4)

lyudmila [28]

Answer:

-5/7

Step-by-step explanation:

When you divide fractions, you flip the second fraction, then multiply.

So it becomes -25/28 x 4/5.

This equals -100/140 which reduces to -5/7.

5 0

3 years ago

3(4a + 2) = 2(a - 2)?

Likurg_2 [28]

Answer:

A= -1

Step-by-step explanation:

8 0

3 years ago

The joint probability density function of X and Y is given by fX,Y (x, y) = ( 6 7 x 2 + xy 2 if 0 < x < 1, 0 < y < 2

fredd [130]

I'm going to assume the joint density function is

$f_{X,Y}(x,y)=\begin{cases}\frac67(x^2+\frac{xy}2\right)&\text{for }0$

a. In order for $f_{X,Y}$ to be a proper probability density function, the integral over its support must be 1.

$\displaystyle\int_0^2\int_0^1\frac67\left(x^2+\frac{xy}2\right)\,\mathrm dx\,\mathrm dy=\frac67\int_0^2\left(\frac13+\frac y4\right)\,\mathrm dy=1$

b. You get the marginal density $f_X$ by integrating the joint density over all possible values of $Y$ :

$f_X(x)=\displaystyle\int_0^2f_{X,Y}(x,y)\,\mathrm dy=\boxed{\begin{cases}\frac67(2x^2+x)&\text{for }0$

c. We have

$P(X>Y)=\displaystyle\int_0^1\int_0^xf_{X,Y}(x,y)\,\mathrm dy\,\mathrm dx=\int_0^1\frac{15}{14}x^3\,\mathrm dx=\boxed{\frac{15}{56}}$

d. We have

$\displaystyle P\left(X$

and by definition of conditional probability,

$P\left(Y>\dfrac12\mid X\frac12\text{ and }X$

$\displaystyle=\dfrac{28}5\int_{1/2}^2\int_0^{1/2}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\boxed{\frac{69}{80}}$

e. We can find the expectation of $X$ using the marginal distribution found earlier.

$E[X]=\displaystyle\int_0^1xf_X(x)\,\mathrm dx=\frac67\int_0^1(2x^2+x)\,\mathrm dx=\boxed{\frac57}$

f. This part is cut off, but if you're supposed to find the expectation of $Y$ , there are several ways to do so.

Compute the marginal density of $Y$ , then directly compute the expected value.

$f_Y(y)=\displaystyle\int_0^1f_{X,Y}(x,y)\,\mathrm dx=\begin{cases}\frac1{14}(4+3y)&\text{for }0$

$\implies E[Y]=\displaystyle\int_0^2yf_Y(y)\,\mathrm dy=\frac87$

Compute the conditional density of $Y$ given $X=x$ , then use the law of total expectation.

$f_{Y\mid X}(y\mid x)=\dfrac{f_{X,Y}(x,y)}{f_X(x)}=\begin{cases}\frac{2x+y}{4x+2}&\text{for }0$

The law of total expectation says

$E[Y]=E[E[Y\mid X]]$

We have

$E[Y\mid X=x]=\displaystyle\int_0^2yf_{Y\mid X}(y\mid x)\,\mathrm dy=\frac{6x+4}{6x+3}=1+\frac1{6x+3}$

$\implies E[Y\mid X]=1+\dfrac1{6X+3}$

This random variable is undefined only when $X=-\frac12$ which is outside the support of $f_X$ , so we have

$E[Y]=E\left[1+\dfrac1{6X+3}\right]=\displaystyle\int_0^1\left(1+\frac1{6x+3}\right)f_X(x)\,\mathrm dx=\frac87$

5 0

3 years ago