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Elena L [17]
3 years ago
14

In the American version of the Game Roulette, a wheel has 18 black slots, 18 red slots and 2 green slots. All slots are the same

size. A person can wager on either red or black. Green is reserved for the house. If a player wagers $5 on either red or black and that color comes up, they win $10 otherwise they lose their wager. What is the expected value of playing the game once
Mathematics
1 answer:
m_a_m_a [10]3 years ago
5 0

Answer:

-$0.26

Step-by-step explanation:

Calculation to determine the expected value of playing the game once

Expected value= [18/(18+18+2) x $5)]- [20/(18+18+2) x $5]

Expected value= ($18/38 x $5) - (20/38 x $5)

Expected value= ($2.37-$2.63)

Expected value= -$0.26

Therefore the expected value of playing the game once is -$0.26

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How many solutions are in this equation?
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Please help me in this questions....​
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Part (i)

I'm going to use the notation T(n) instead of T_n

To find the first term, we plug in n = 1

T(n) = 2 - 3n

T(1) = 2 - 3(1)

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The first term is -1

Repeat for n = 2 to find the second term

T(n) = 2 - 3n

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<h3>Answers: -1, -4</h3>

==============================================

Part (ii)

Plug in T(n) = -61 and solve for n

T(n) = 2 - 3n

-61 = 2 - 3n

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n = -63/(-3)

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Note that plugging in n = 21 leads to T(21) = -61, similar to how we computed the items back in part (i).

<h3>Answer:  21st term</h3>

===============================================

Part (iii)

We're given that T(n) = 2 - 3n

Let's compute T(2n). We do so by replacing every copy of n with 2n like so

T(n) = 2 - 3n

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This means 2n = 2*8 = 16. So subtracting T(8) - T(16) will get us 24.

<h3>Answer: 8</h3>
4 0
3 years ago
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