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3 years ago

Brainliest + 30 points

Mathematics

2 answers:

Evgesh-ka [11]3 years ago

5 0

Answer:

Step-by-step explanation:

Plug the numbers into the distance formula:

$d = \sqrt{(x2 - x1)^2 + (y2-y1)^2} \\$

$d = \sqrt{(11 - 5)^2 + (16-8)^2} \\$

Then solve:

You get 10

Luden [163]3 years ago

4 0

From Geocity to Shapetown is

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I need help..... please answer ASAP if you know how to solve this

kvv77 [185]

For part A, The answer is that the car gets better gas mileage. We can see it from the graph that the number of gallons used is on the X axis, and the distance traveled using those number of gallons is on the Y axis. The easiest way to compare would be to look at the 1 gallon of gas. You can see that you can travel 25 miles on 1 gallon of gas. The truck on the other hand will get you 18 miles per gallon. Imagine putting 1 in for X, the Y value would be 18 if you did this. The graph just shows us a visual way of saying the same thing. To determine how much farther the car with a girl on 8 gallons of gas, you would just multiply 8 by 25 for the number of miles traveled by the car. You would multiply 8 by 18 to find the number of miles traveled for the truck. The answers are 200 miles for the car and 144 miles for the truck. 200-144=56 miles farther for the car.

6 0

3 years ago

If X and Y are independent continuous positive random

Leni [432]

a) $Z=\frac XY$ has CDF

$F_Z(z)=P(Z\le z)=P(X\le Yz)=\displaystyle\int_{\mathrm{supp}(Y)}P(X\le yz\mid Y=y)P(Y=y)\,\mathrm dy$

$F_Z(z)\displaystyle=\int_{\mathrm{supp}(Y)}P(X\le yz)P(Y=y)\,\mathrm dy$

where the last equality follows from independence of $X,Y$ . In terms of the distribution and density functions of $X,Y$ , this is

$F_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy$

Then the density is obtained by differentiating with respect to $z$ ,

$f_Z(z)=\displaystyle\frac{\mathrm d}{\mathrm dz}\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy=\int_{\mathrm{supp}(Y)}yf_X(yz)f_Y(y)\,\mathrm dy$

b) $Z=XY$ can be computed in the same way; it has CDF

$F_Z(z)=P\left(X\le\dfrac zY\right)=\displaystyle\int_{\mathrm{supp}(Y)}P\left(X\le\frac zy\right)P(Y=y)\,\mathrm dy$

$F_Z(z)\displaystyle=\int_{\mathrm{supp}(Y)}F_X\left(\frac zy\right)f_Y(y)\,\mathrm dy$

Differentiating gives the associated PDF,

$f_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}\frac1yf_X\left(\frac zy\right)f_Y(y)\,\mathrm dy$

Assuming $X\sim\mathrm{Exp}(\lambda_x)$ and $Y\sim\mathrm{Exp}(\lambda_y)$ , we have

$f_{Z=\frac XY}(z)=\displaystyle\int_0^\infty y(\lambda_xe^{-\lambda_xyz})(\lambda_ye^{\lambda_yz})\,\mathrm dy$

$\implies f_{Z=\frac XY}(z)=\begin{cases}\frac{\lambda_x\lambda_y}{(\lambda_xz+\lambda_y)^2}&\text{for }z\ge0\\0&\text{otherwise}\end{cases}$

and

$f_{Z=XY}(z)=\displaystyle\int_0^\infty\frac1y(\lambda_xe^{-\lambda_xyz})(\lambda_ye^{\lambda_yz})\,\mathrm dy$

$\implies f_{Z=XY}(z)=\lambda_x\lambda_y\displaystyle\int_0^\infty\frac{e^{-\lambda_x\frac zy-\lambda_yy}}y\,\mathrm dy$

I wouldn't worry about evaluating this integral any further unless you know about the Bessel functions.

6 0

3 years ago

3х + Зу = -x+5y ordered pair

likoan [24]

Answer:

2x-y=0

x=1/2y

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

What is the solution to −6 − (−2)?

fomenos

- 6 - (-2)

- (-2) = 2
* two negatives = one positive

-6 + 2 = -4

-4 is your answer

hope this helps

7 0

3 years ago

Read 2 more answers

I will mark brainliest and 20 pts

vekshin1

I believe the Surface Area is 37 inches^2 and the Volume is 15 inches^2

6 0

3 years ago