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3 years ago

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M is the midpoint of segment CS. C is at (-8, 2) M is at the origin. What are the coordinates of S?

1 answer:

ahrayia [7]3 years ago

4 0

C(-8,2) and M(0,0) , since M is at the origin. Let x₁ and y₁ be the
coordinates of S →s(x₁ , y₁)

C(-8,2) and S(x₁ , y₁)
The coordinates of M, the midpoint of CS are M(x₂ , y₂)

a) x₂ = (-8 + x₁)/2 , but x₂ = 0, then :
0 = -4+x₁/2 and x₁ = 8

b) y₂ = (2+y₁)/2 , but y₂ = o, then:
0 = 2+ y₁/2 and y₂ = -2

Then the coordinates of S are S(8 , -2)

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By definition of conditional probability,

P(<em>Y</em>₁ > <em>Y</em>₂ | <em>Y</em>₁ < 2 <em>Y</em>₂) = P((<em>Y</em>₁ > <em>Y</em>₂) and (<em>Y</em>₁ < 2 <em>Y</em>₂)) / P(<em>Y</em>₁ < 2 <em>Y</em>₂)

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$=\displaystyle\frac1{49}\int_0^\infty\int_{\frac{y_1}2}^{y_1}e^{-\frac{y_1+y_2}7}\,\mathrm dy_2\,\mathrm dy_1$

$=\displaystyle-\frac17\int_0^\infty\int_{-\frac{3y_1}{14}}^{-\frac{2y_1}7}e^u\,\mathrm du\,\mathrm dy_1$

$=\displaystyle-\frac17\int_0^\infty\left(e^{-\frac{2y_1}7} - e^{-\frac{3y_1}{14}}\right)\,\mathrm dy_1$

$=\displaystyle-\frac17\left(-\frac72e^{-\frac{2y_1}7} + \frac{14}3 e^{-\frac{3y_1}{14}}\right)\bigg|_0^\infty$

$=\displaystyle-\frac17\left(\frac72 - \frac{14}3\right)=\frac16$

• denominator:

$P(Y_1$

(I leave the details of the second integral to you)

Then you should end up with

P(<em>Y</em>₁ > <em>Y</em>₂ | <em>Y</em>₁ < 2 <em>Y</em>₂) = (1/6) / (2/3) = 1/4

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