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3 years ago

Quadrilateral ABCD with A(2,2), B(-3,5), C(-4,0), D(2, -2) translated 1 unit up and 2 units right

Mathematics

1 answer:

wlad13 [49]3 years ago

8 0

A(3,4) B(-2.7) C(-3,2) D(3,0)
Add 1 to the 1st # of each
Add 2 to the 2nd #of each

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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

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 ...

If I agree to pay you $n^2 if heads comes up first on the nth toss.

this implies that , you need to be paid $\sum \limits ^{n}_{i=1} n^2 P(X=n)$

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

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

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

The value of y is 20 when x = 15 <em> i think</em>

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Step-by-step explanation:

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