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3 years ago

⚠️HURRY WHORTH 30 points ⚠️Charlie bikes 9 miles everyday which equation models the situation

Mathematics

2 answers:

astraxan [27]3 years ago

5 0

Answer:

C.9d=m

Step-by-step explanation:

Dima020 [189]3 years ago

4 0

Answer:

The answer is C im sure i just took the thest and got 100% good luck!!!

Step-by-step explanation:

I just guess n got it right now i wanna help yhu out have a bless day!

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Find the area of the trapezoid

katovenus [111]

Answer: 49 in^2

Step-by-step explanation:

Trapezoid area formula = (a+b/2) x h

So...

(5+9/2) x 7

= 49 in^2

4 0

3 years ago

Read 2 more answers

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

What is the area of this figure

dusya [7]

Insert a picture so I can help you out :)

3 0

3 years ago

Read 2 more answers

zelly works 30 hours a week at a food market for $8.50 an hour she takes home $7.65 an hour after deductions. what is her rate f

morpeh [17]

Answer:

Her deduction is 10%, or 0.85

Step-by-step explanation:

5 0

3 years ago

While shopping at the school bookstore Ric purchased 4 book covers for $1.25 each and a pen that cost 2/5 as much as a book cove

Anvisha [2.4K]

Answer: Ric will get $4.50 back in change

Step-by-step explanation: $1.25 each cover times 4 covers is $5. 2/5 of 1.25 is .50. add $5 and $.50 is $5.50. if he pays with a ten, subtract, and he will get $4.50. hope this helpedd❤️

4 0

3 years ago