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2 years ago

أحتاج إلى مساعدة كيف تقدم كوزي Tell me what this say for free branleist no transition script

Mathematics

2 answers:

telo118 [61]2 years ago

8 0

Answer: Hope This Helps! :D

Step-by-step explanation:

I need help how to progress cozy

likoan [24]2 years ago

6 0

Answer:

you need help to progress cozy im guessing?

Step-by-step explanation:

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Evaluate 2(L+W)<br> for L=10 and W= 2

arsen [322]

Answer:

Step-by-step explanation:

2(10+2)

turns into 2×12

so, that would be 24

5 0

1 year ago

Find the distance between the points (-1,-5) and (-10,7).

Paladinen [302]

Answer:

15 units

Step-by-step explanation:

Calculate the distance d using the distance formula

d = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 }$

with (x₁, y₁ ) = (- 1, - 5) and (x₂, y₂ ) = (- 10, 7)

d = $\sqrt{(-10+1)^2+(7+5)^2}$

= $\sqrt{(-9)^2+12^2}$

= $\sqrt{81+144}$

= $\sqrt{225}$

= 15 units

5 0

2 years ago

consider the line shown on the graph. enter the equation of the line in the form y=mx where m is the slope.

Amanda [17]

Answer:

y= 3x

Step-by-step explanation:

6 0

2 years ago

The stadium can seat 15,000 fans. If the stadium is 85 percent full, how many tickets were sold?

Vera_Pavlovna [14]

12,750 tickets were sold. 0.85*15,000

8 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