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LenKa [72]
3 years ago
13

Some one please help me ​

Mathematics
2 answers:
Ksju [112]3 years ago
6 0

Answer:

b>k c>L  ac>JL  bc>KL

Step-by-step explanation:

poggers

Nana76 [90]3 years ago
3 0

Answer:

B = K

C = L

AC = JL

BC = KL

hope this helps!

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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

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= $4

∴

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

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4 years ago
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