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brilliants [131]
3 years ago
10

Line A is perpendicular to line B. The slope of line A is -1/9 what is the slope of line b?

Mathematics
2 answers:
fredd [130]3 years ago
8 0
The slope of the line is 9
Pani-rosa [81]3 years ago
4 0
\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}}
{\stackrel{slope}{-\cfrac{1}{9}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{9}{1}}\qquad \stackrel{negative~reciprocal}{+\cfrac{9}{1}}\implies 9}
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Artyom0805 [142]

Answer:

In the long run, ou expect to  lose $4 per game

Step-by-step explanation:

Suppose we play the following game based on tosses of a fair coin. You pay me $10, and I agree to pay you $n^2 if heads comes up first on the nth toss.

Assuming X be the toss on which the first head appears.

then the geometric distribution of X is:

X \sim geom(p = 1/2)

the probability function P can be computed as:

P (X = n) = p(1-p)^{n-1}

where

n = 1,2,3 ...

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\sum \limits ^{n}_{i=1} n^2 P(X=n) = E(X^2)

\sum \limits ^{n}_{i=1} n^2 P(X=n) =Var (X) + [E(X)]^2

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p}{p^2}+(\dfrac{1}{p})^2        ∵  X \sim geom(p = 1/2)

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p}{p^2}+\dfrac{1}{p^2}

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p+1}{p^2}

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{2-p}{p^2}

\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{2-\dfrac{1}{2}}{(\dfrac{1}{2})^2}

\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ \dfrac{4-1}{2} }{{\dfrac{1}{4}}}

\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ \dfrac{3}{2} }{{\dfrac{1}{4}}}

\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ 1.5}{{0.25}}

\sum \limits ^{n}_{i=1} n^2 P(X=n) =6

Given that during the game play, You pay me $10 , the calculated expected loss = $10 - $6

= $4

∴

In the long run, you expect to  lose $4 per game

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