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3 years ago

In the polynomial below.

Mathematics

2 answers:

Kaylis [27]3 years ago

6 0

Answer: 2

Step-by-step explanation:

2 terms

algol [13]3 years ago

3 0

Answer:

Step-by-step explanation:uywegfywieflwef

efeerewer

werwer

wer

werwer

tew

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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:

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

3 0

4 years ago

Graph the line passing through the given point and having the given slope. Give the slope intercept form of the equation of the

shutvik [7]

Equation of the line is y=mx+b
the line passing through point (1;-2) so x=1 y=-2
m=-0.5 so put all these values in equetion and found value of b
-2=-0.5*1+b
-2=-0.5+b
b=-1.5
so equation of the line is y=-0.5x-1.5
Answer:
equation of the line is y=-0.5x-1.5

6 0

3 years ago

Read 2 more answers

There are 17 girls and 23 boys in P.E class.Of these students,10 girls and 14 boys will try out for basketball.What percentage o

insens350 [35]

Answer:

$60%$ %