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zzz [600]
3 years ago
14

Gambler is deciding whether or not to take a bet. she must pay $40 to take the bet, but if she wins, she willprofit$225.theproba

bility that she wins the bet is ¼. what is the player’s expected value in this situation
Mathematics
1 answer:
Vesnalui [34]3 years ago
3 0
Expected value of the bet is
the sum of the products of value of outcome and its probability,
less the amount paid to place the bet.

Outcomes value probability
win            225    1/4
lose               0    3/4
cost of bet = 40

So expected value of bet
E[X]=225*(1/4)+0*(3/4)-40
=56.25-40
=16.25

This means that in the long run, gambler will win, since the expected value is positive.  (does NOT mean she will win in the next bet!)
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If AC = 8x - 14 and EC = 2x + 11, solve for x.
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Answer:

Step-by-step explanation:

8x-14=2x+11

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5 0
2 years ago
Which theorem would be used to prove the triangles below as congruent?
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  A:  HL

Step-by-step explanation:

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2 years ago
State the domain of the relation {(4, 5), (2, 3), (−3, 0), (2, 1)}.
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3 years ago
how many ways can 8 distinguishable books be placed in 5 distinguishable shelves so that each shelf contains one book
musickatia [10]

There are 6720 ways by 8 distinguishable books be placed in 5 shelves.

According to statement

The number of books (n) is 8

The number of shelves (r) is 5

Now, we find the ways by which the 8 books be placed in 5 distinguishable shelves

From Permutation formula

P(n,r) = n! / (n-r)!

Substitute the values then

P(n,r) = 8! / (8-5)!

P(n,r) = (8*7*6*5*4*3*2*1) / (3*2*1)

P(n,r) = 8*7*6*5*4

P(n,r) = 6720

So, there are 6720 ways by 8 distinguishable books be placed in 5 shelves.

Learn more about PERMUTATION here brainly.com/question/11732255

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