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guajiro [1.7K]
4 years ago
9

On Monday Mrs. Wise bought 3 lbs of apples at $4 per lb. On Tuesday, apples were on sale for 20% off so she bought another 5 lbs

.
What was the average price for all apples Mrs. Wise bought?
Mathematics
1 answer:
Vesna [10]4 years ago
7 0
Average price=total price/total weight, let p be the average price:

p=(3*4+5*4*(100-20)/100)/8

p=(12+16)/8

p=$3.50/lb
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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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