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3 years ago

Can someone please help me

Mathematics

1 answer:

N76 [4]3 years ago

5 0

Answer:

A = 7 x 9.3

A = 65.1 yd²

Step-by-step explanation:

(I am assuming that either 7 or 9.3 refers to the height. It is unclear.)

A = bh

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Find the next two numbers in a pattern? 7, - 28,112, - 448 __, __, . . .

Igoryamba

Answer:

1,792 -7168

the pattern is to multiply by -4

Step-by-step explanation:

7 · -4 = -28

-28 · -4 = 112

112 · -4 = -448

-448 · -4 = 1,792

1,792 · -4 = -7168

4 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

Help

earnstyle [38]

Aas is not d triangle buddy

8 0

3 years ago

A+b=95 4a+1.50b=267.5

Papessa [141]

9514 1404 393

Answer:

a=50
b=45

Step-by-step explanation:

We can subtract the second equation from 4 times the first:

4(a +b) -(4a +1.5b) = 4(95) -(267.5)

2.5b = 112.5 . . . . . simplify

b = 45 . . . . . . . . . . divide by 2.5

a = 95 -b = 50 . . . find 'a'

The solution is (a, b) = (50, 45).

5 0

2 years ago

Charlie ate 2/3 of his chocolate bar. Kelsey ate

Contact [7]

I think he’s because 3/6 is half and 2/3 is half

3 0

3 years ago

Read 2 more answers

Can someone please help me​

Can someone please help me