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2 years ago

10

PLZ HELP ASAP THANKS IN ADVANCED!! :)

2 answers:

umka2103 [35]2 years ago

7 0

Every bottle costs $1.50

x = Number of bottles sold

x(1.50) = x times 1.50

hodyreva [135]2 years ago

3 0

$\underline{\ \ x\ \ |\ \ y}\\\underline{\ \ 0\ \ |\ \ 0}\\\underline{\ \ 1\ \ |\ 1.50}\\\underline{\ \ 2\ \ |\ 2\cdot1.5=3}\\\underline{\ \ 3\ |\ 3\cdot1.5=4.5}\\\underline{\ \ 4\ |\ 4\cdot1.5=6}\\\\\boxed{y=1.5x}$

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beks73 [17]

Answer:

.

Step-by-step explanation:

8 0

2 years ago

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

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

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3 years ago

Solve <br>60000000+63636733-7373

Yuri [45]

Answer:

your answer is 123629360

Step-by-step explanation:

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i hope this helped

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7 0

2 years ago

Pls help thanks in advance

mixer [17]

I give u the answer or did u already do this

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2 years ago

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KATRIN_1 [288]

Answer:

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Step-by-step explanation:

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2 years ago