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BigorU [14]
3 years ago
8

In the casino game of roulette, a gambler can bet on which of 38 numbers the ball will land when the roulette wheel is spun. On

a $2 bet, a gambler gains $70 (so a net profit of $68) if he or she chooses the winning number, but loses the $2 otherwise. The probability of choosing the winning number is 1/38 and the probability of not choosing the winning number is 37/38. Let X denote the net gain from a roulette spin. What is the expected winnings for one spin?
Mathematics
1 answer:
ollegr [7]3 years ago
4 0

Answer:

Expected winnings for one spin = 0.1

Step-by-step explanation:

Expected Value = Sum of [(Probability of X)(Value of X)]

E(X) = Σ [P(X).X]

Two Events : X = +68 , P(X) = 1/38 and X = -2 , P(X) = 37/38

E(X) = [(68) (1/38)] + [(-2)(37/38)]

[(68)(0.03)] + [(-2)(0.97)]

= 2.04 - 1.94

= 0.1

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{(-5,64), (2,1)}

linear equation: y = -9x + 19
quadratic equation: y = x² - 6x + 9

Substitute the y in the quadratic equation by the its value in the linear equation.

-9x + 19 = x² - 6x + 9
      -  19               - 19    *subtract 19 to both sides
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Set each factor = 0 and solve
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Find the corresponding value of y using the linear equation.
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x = -5                              x = 2
y = -9(-5) + 19               y = -9(2) + 19
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(-5,64)                              (2,1)

Check each value on each equation.
y = x² - 6x + 9
(-5,64)                                                  (2,1)
64 = (-5)² - 6(-5) + 9                          1 = 2² - 6(2) + 9
64 = 25 + 30 + 9                                1 = 4 - 12 + 9
64 = 64                                               1 = 1

y = -9x + 19
64 = -9(-5) + 19                                  1 = -9(2) + 19
64 = 45 + 19                                      1 = -18 + 19
64 = 64                                              1 = 1

{(-5,64), (2,1)}

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