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sweet [91]
3 years ago
7

I’ll give brainliest

Mathematics
1 answer:
Nimfa-mama [501]3 years ago
4 0

Answer:

Step-by-step explanation:

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Please do thank you so much, brainliest, and thanks and points
erik [133]

Answer:

ok

Step-by-step explanation:

5 0
3 years ago
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Write each equation in slope-intercept form of the equation of a line.
Fynjy0 [20]

The x-intercept is 4

The y-intercept is 3

The slope is -3/4

8 0
3 years ago
Find the volume of the cone shown below
Vadim26 [7]

Answer:

3.14*9^2*12/3=

Step-by-step explanation:

V=π*r^2*h/3

3

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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
What is 12 times twelve minus one + 1
notsponge [240]

Answer:

144

Step-by-step explanation:

12x12=144-1=143+1=144

Hope this helps!

8 0
3 years ago
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