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4 years ago

Hi....................

Mathematics

1 answer:

Svetradugi [14.3K]4 years ago

3 0

good evening ,

Answer:

The right answer is the first one

Step-by-step explanation:

Area = length × width

m = (1/5)×(2/3).

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Graph the solution of the inequality on a number line c<5

polet [3.4K]

C < 5

there will be an open circle on 5....it is open because there is no equal sign in the problem.....the shading will be to the left....because less then is shaded to the left

5 0

4 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

A hollow crystal ball is made of glass that is 3 cm thick. The ball has a radius to the outside surface of 8 cm. What is the app

Vadim26 [7]

Answer:

1620.2

Sorry if I'm late The correct answer is 1620.2

Have a good day!

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Answer:

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Rom4ik [11]

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