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3 years ago

In the diagram of XYZ, which term describes point ?

Mathematics

1 answer:

spin [16.1K]3 years ago

4 0

The answer should be B. Incenter

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If f(x) = k(x - 2)^2, where k is positive, what is the effect on the graph of f(x) as k increases

pashok25 [27]

The answer is B. Lana Rhoades

4 0

3 years ago

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What is the answer?

rosijanka [135]

Answer: 60

Step-by-step explanation:

It’s an equilateral triangle based on the congruent sides.

4 0

3 years ago

HELP

Otrada [13]

Answer:

(x + 2)² + (y + 9)² = 49

Step-by-step explanation:

Equation:

(x - h)² + (y - k)² = r²

(x - -2)² + (y - -9)² = 7²

(x + 2)² + (y + 9)² = 49

5 0

3 years ago

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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

Based on the diagram, what is sinA ?<br><br> Enter your answer as a fraction in the boxes.

Leya [2.2K]

Answer:

sinA = $\frac{a}{c}$

Step-by-step explanation:

sinA = $\frac{opposite}{hypotenuse}$ = $\frac{BC}{AB}$ = $\frac{a}{c}$

3 0

4 years ago

Read 2 more answers