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3 years ago

Through the points (-10,-20) and (5,10)

Mathematics

1 answer:

pychu [463]3 years ago

7 0

Do you need to find the m?
If yes so the solution will be like that
(10- -20) / (5- - 10) = (10+20)/(5+10) = 30/15= 2

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I'm assuming $\alpha$ is the shape parameter and $\beta$ is the scale parameter. Then the PDF is

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a. The expectation is

$E[X]=\displaystyle\int_{-\infty}^\infty xf_X(x)\,\mathrm dx=\frac29\int_0^\infty x^2e^{-x^2/9}\,\mathrm dx$

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$E[X]=\displaystyle-xe^{-x^2/9}\bigg|_0^\infty+\int_0^\infty e^{-x^2/9}\,\mathrm x$

$E[X]=\displaystyle\int_0^\infty e^{-x^2/9}\,\mathrm dx$

Substitute $x=3y^{1/2}$ , so that $\mathrm dx=\dfrac32y^{-1/2}\,\mathrm dy$ :

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$\boxed{E[X]=\dfrac32\Gamma\left(\dfrac12\right)=\dfrac{3\sqrt\pi}2\approx2.659}$

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$\mathrm{Var}[X]=E[(X-E[X])^2]=E[X^2-2XE[X]+E[X]^2]=E[X^2]-E[X]^2$

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$u=x^2\implies\mathrm du=2x\,\mathrm dx$

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$E[X^2]=\displaystyle-x^2e^{-x^2/9}\bigg|_0^\infty+2\int_0^\infty xe^{-x^2/9}\,\mathrm dx$

$E[X^2]=\displaystyle2\int_0^\infty xe^{-x^2/9}\,\mathrm dx$

Substitute $x=3y^{1/2}$ again to get

$E[X^2]=\displaystyle9\int_0^\infty e^{-y}\,\mathrm dy=9$

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b. The probability that $X\le3$ is

$P(X\le 3)=\displaystyle\int_{-\infty}^3f_X(x)\,\mathrm dx=\frac29\int_0^3xe^{-x^2/9}\,\mathrm dx$

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$\boxed{P(1\le X\le3)=\dfrac{e^{8/9}-1}e\approx0.527}$

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2 years ago

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Answer:

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We know that Jane has 6 more shoes than 1/2 of what Mercedes has. We can use this information to make an equation.

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$\displaystyle \\ 
\texttt{Notations: } \\ 
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