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3 years ago

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m∠RSU=(5x+7)° and m∠UST=(9x−1)°. If ∠RSU and ∠UST are complementary, what is the measure of each angle?

1 answer:

ArbitrLikvidat [17]3 years ago

8 0

Answer:

37° and 53° respectively.

Step-by-step explanation:

We need to know that the angles are complementtary if ∠RSU + ∠UST = 90°. Then,

5x+7 + 9x-1 = 90

5x + 9x +7 -1 = 90

14x +6 = 90

14x = 90 - 6

14x = 84

x = 84/14

x = 6.

Then, the angle ∠RSU has measure 5(6)+ 7 = 30+7 = 37, and the angle ∠UST has measure 9(6) - 1 = 54-1 = 53.

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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 $\mathrm dx=\dfrac32y^{-1/2}\,\mathrm dy$ :

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

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

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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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$E[X^2]=\displaystyle\int_{-\infty}^\infty x^2f_X(x)\,\mathrm dx=\frac29\int_0^\infty x^3e^{-x^2/9}\,\mathrm dx$

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

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

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

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$\mathrm{Var}[X]=9-E[X]^2$

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3 years ago