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Anastaziya [24]

3 years ago

15

Rita has a loan of 45,580.This loan has a simple interest rate of 4% per year. What is the amount of interest rhat rita will be

charged on this loan at the end of one year?

1 answer:

krek1111 [17]3 years ago

8 0

Total Interest charged
=principle×interest rate×time
I=prt
I=45,580×0.04×1
I=1,823.2

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To compute this integral, recall the definition of the Gamma function,

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$u=x\implies\mathrm du=\mathrm dx$

$\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}$

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

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

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

$E[X]=\displaystyle\frac32\int_0^\infty y^{-1/2}e^{-y}\,\mathrm dy$

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

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

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

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Then

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

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