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Leokris [45]
3 years ago
6

Are the triangles similar? If yes, write a similarity statement and give a scale factor (k).

Mathematics
1 answer:
Irina-Kira [14]3 years ago
5 0

Answer:

choice 2) Yes

Triangle ACB is congruent to triangle XZY because of the Side-Side-Side theorem

k = 1.5

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Let X be a continuous random variable with density functionf(x)={1−|x|0 for −1
BlackZzzverrR [31]

Answer:

f_Y (y) =\frac{d}{dy} = 2 \frac{1}{2 |\sqrt{y}|} -1 = \frac{1}{|\sqrt{y}|} -1,0 \leq Y\leq 1

Step-by-step explanation:

For this case we want to find the density function for Y=X^2

And we have the following density function for the random variable X:

f(X) = 1- |X|,-1 \leq X \leq 1

So we can do this replacing Y=X^2

F_Y (Y \leq y) = P(X^2 \leq y)

If we apply square root on both sides we got:

P(-\sqrt{y} \leq X \leq \sqrt{y}) = \int_{-\sqrt{y}}^0 1+t dt +\int_{0}^{\sqrt{y}} 1-t dt

And if we integrate we got this:

F_Y (y) = [t+ \frac{t^2}{2}] \Big|_{-\sqrt{y}}^0+ [t -\frac{t^2}{2}] \Big|_{0}^{\sqrt{y}}

And replacing we got:

F_Y (y) = [0 -(-\sqrt{y} +\frac{y}{2})] + [\sqrt{y} -\frac{y}{2}]

F_Y (y) = 2 |\sqrt{y}| -y

And if we want to find the density function we just need to derivate the pdf like this:

f_Y (y) =\frac{d}{dy} = 2 \frac{1}{2 |\sqrt{y}|} -1 = \frac{1}{|\sqrt{y}|} -1,0 \leq Y\leq 1

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