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pav-90 [236]
2 years ago
6

Find the slope of the line that contains the points (-8, -2) and (-17, 4).

Mathematics
2 answers:
amid [387]2 years ago
6 0

Answer:

<h2>-2/3 </h2>

Step-by-step explanation:

<h3>m=y2-y1/x2-x1
m=4-(-2)/-17-(-8)
m=4+2/-17+8
m=6/-9
m=-2/3 </h3>
IRINA_888 [86]2 years ago
6 0
<h2><u>S</u>O<u>L</u>V<u>I</u>N<u>G</u></h2>

\Large\maltese\underline{\textsf{A. \space What is Asked}}

Find the slope of the line that contains the points (-8,-2) and (-17,4)

\Large\maltese\underline{\textsf{B.\space This problem has been solved!}}

If we are given two points, we can easily find the slope of the line that passes through the points. With points (y₂, y₁) and (x₂, x₁), the slope is \bf{\dfrac{y2-y1}{x2-x1}.

Similarly, with points (-8,-2) and (-17,4) the slope is.

\bf{\dfrac{4-(-2)}{-17-(-8)}} | as we can see, it is possible to simplify this further

\bf{\dfrac{4+2}{-17+8}} | simplify

\bf{\dfrac{6}{-9} | reduce

\bf{-\dfrac{2}{3}}

\rule{300}{1.7}

\bf{Result:}

                    \bf{=-\dfrac{2}{3}}

\boxed{\bf{aesthetic\not101}}

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natali 33 [55]

Answer:

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Step-by-step explanation:

If you're looking for..

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4 0
3 years ago
If Y has a geometric distribution with probability of success p, show that the moment-generating function for Y is
butalik [34]

Answer:

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Step-by-step explanation:

Let the random variable Y have a geometric distribution g (y;p) = pq y-¹

The m.g.f of the geometric distribution is derived as below

By definition , M₀ (t) = E (e^ ty) = ∑ (e^ ty )( q ^ y-1)p    ( for ∑ , y varies 1 to infinity)

                   = pe^t ∑(e^tq)^y-1

                    = pe^t/1- qe^t,     where qe^t <1

In order to differentiate the m.g.f we write it as

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 M₀` (t)  = pe^-t (e^-t -q) ^ -2  and

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Hence

E (y) = p (1-q)-² = 1/p

E (y²) =2 p (1-q)-³ - p (1-q)-²

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5 0
2 years ago
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Step by step:

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(because we changed side square root becomes exponent ²)

x=49+4

x=53

3 0
3 years ago
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denis23 [38]
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