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kaheart [24]
2 years ago
12

Find the slope of the line that passes through (87, 91) and (88, -4).

Mathematics
1 answer:
Elanso [62]2 years ago
5 0
<h2>Answer:    slope = - 95</h2><h2> </h2>

Step-by-step explanation:    

      The question gives us two points, (87, 91) and (88, -4), from which we can find the slope and later the equation of the line.

<u> </u>

<u>Finding the Slope</u>  

The slope of the line (m) = (y₂ - y₁) ÷ (x₂ - x₁)      

                                        =  (91 - (- 4)) ÷ (87 - 88)    

                                        =  - 95

<em><u /></em>

<em><u /></em>

<em><u>Checking my answer:</u></em>

<em>Finding the Equation</em>  

We can now use the point-slope form (y - y₁) = m(x - x₁)) to write the equation for this line:  

                                ⇒  y - (-4) =  - 95 (x - 88)

                                       y + 4 =  - 95 (x - 88)

<em>To test my answer, I have included a Desmos Graph that I graphed using the information provided in the question and my answer.</em>

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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

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

\Gamma(x)=\displaystyle\int_0^\infty t^{x-1}e^{-t}\,\mathrm dt

For this particular integral, first integrate by parts, taking

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

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

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7 0
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(6^-8)^4
KIM [24]

Your Answer is A.

(-8) x 4

= -8 x 4

= -32

= 1/6 ^32


HOPE THIS HELPS!!

PLEASE MARK ME AS BRAINLIEST!!

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