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4 years ago

Emergency!!!! Please help Geometry

Mathematics

2 answers:

Klio2033 [76]4 years ago

7 0

The answer is A because the letter at the actual angle is always the letter in the middle.

mash [69]4 years ago

5 0

I dont do geometry but I feel like it would <L But dont quote me on it

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Jason and his family travel 160 miles in 3.2 hours. If they continue at this constant speed, how long will it take them to trave

Inga [223]

It will take them 6 hours to travel 300 miles. How I got my answer: 160 divided by 3.2 = 50 miles per hour, 300 divided by 50 = 6 hours. Hope I could help! :D

7 0

3 years ago

Read 2 more answers

Help giving brainliest

klasskru [66]

Answer:

15 square units thank you

4 0

3 years ago

Need help with this question, thanks.

Kazeer [188]

Answer:

Step-by-step explanation:

Times 11 for all sides get

11<x+13<33

then 11<x+13

-2<x

x+13<33

x<20

-2<x<20

7 0

3 years ago

What are the functions of money?

Dovator [93]

It’s b for sure if I’m wrong then it’s a

6 0

3 years ago

Suppose X and Y are random variables with joint density function. f(x, y) = 0.1e−(0.5x + 0.2y) if x ≥ 0, y ≥ 0 0 otherwise (a) I

Hatshy [7]

a. $f_{X,Y}$ is a joint density function if its integral over the given support is 1:

$\displaystyle\int_{-\infty}^\infty\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\frac1{10}\int_0^\infty\int_0^\infty e^{-x/2-y/5}\,\mathrm dx\,\mathrm dy$

$=\displaystyle\frac1{10}\left(\int_0^\infty e^{-x/2}\,\mathrm dx\right)\left(\int_0^\infty e^{-y/5}\,\mathrm dy\right)=\frac1{10}\cdot2\cdot5=1$

so the answer is yes.

b. We should first find the density of the marginal distribution, $f_Y(y)$ :

$f_Y(y)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dx=\frac1{10}\int_0^\infty e^{-x/2-y/5}\,\mathrm dy$

$f_Y(y)=\begin{cases}\dfrac15e^{-y/5}&\text{for }y\ge0\\\\0&\text{otherwise}\end{cases}$

Then

$P(Y\ge8)=\displaystyle\int_8^\infty f_Y(y)\,\mathrm dy=e^{-8/5}$

or about 0.2019.

For the other probability, we can use the joint PDF directly:

$P(X\le5,Y\le8)=\displaystyle\int_0^5\int_0^8f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=1+e^{-41/10}-e^{-5/2}-e^{-8/5}$

which is about 0.7326.

c. We already know the PDF for $Y$ , so we just integrate:

$E[Y]=\displaystyle\int_{-\infty}^\infty y\,f_Y(y)\,\mathrm dy=\frac15\int_0^\infty ye^{-y/5}\,\mathrm dy=\boxed5$

5 0

3 years ago