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2 years ago

Can somone give me the answer to 2 x 0

Mathematics

2 answers:

marshall27 [118]2 years ago

8 0

Any number times 0 is going to equal 0

Rzqust [24]2 years ago

6 0

Ur answer is 0...!!!!

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What is the slope of the line through the points (-3, 5) and (4, 5)?

Natasha_Volkova [10]

Answer:

Slope: 0

Step-by-step explanation:

(-3,5) (4,5)

y2-y1/x2-x1

5-5/4+3=0/7

0/7 simplifies to 0.

So the slope of the line (-3,5) (4,5) is 0.

Hope this helps! ;)

5 0

2 years ago

Read 2 more answers

2x 2 (9x 2 −16)

xz_007 [3.2K]

Is the equation2x2(9x+2-16) if so you would do -16+2 first which is -14

Then multiply 2x2 equals 4

4(9x-14)

Multiply the 4 to 9x and -14

You get 36x-56

7 0

2 years ago

What is equal to 6,530,000,000?

dsp73

Gee wiz man I don't know. that is a really big number. have you tried google? :/

5 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

A teacher asks students to list the math lessons they learned on nine random school days in the past month. She evaluates the st

N76 [4]

<span>Hope this helps: C. Convenience sampling</span>

8 0

3 years ago