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NNADVOKAT [17]
3 years ago
15

A game of "Doubles-Doubles" is played with two dice. If a player rolls doubles, the player earns 3 points and gets another roll.

If the player rolls doubles again, the player earns 9 more points. How many points should the player lose for not rolling doubles in order to make this a fair game?
Mathematics
2 answers:
LenKa [72]3 years ago
8 0

Answer:

27/35 points

Step-by-step explanation:

The probability of getting one double;

P(getting one double) = 5/36

Probability of getting two doubles is;

P(getting two doubles) = 1/36

Probability of getting no doubles is;

P(getting no doubles) = 1/36

For us to find the Number of points that the player should lose for not rolling doubles, we have;

(12/36) + (5(3 - x)/36) - (5x/6) = 0

(12/36) + (15 - 5x)/36 - 5x/6 = 0

Multiply through by 36 to get;

12 + 15 - 5x - 30x = 0

35x = 27

x = 27/35

sasho [114]3 years ago
4 0

Answer:

D). 1

Step-by-step explanation:

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Travka [436]
Below are the choices that can be found from other sources:

A) 36 pi 
<span>B) 6 pi </span>
<span>C) 27 pi </span>
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A sphere's volume is (4/3)*pi*r^3. 
<span>The diameter of the balloon is 6 inches. The diameter is twice the radius. The radius is 3 inches. </span>

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<span>But the balloon is only 3/4 full. So </span>

<span>36 * (3/4) * pi = 27 * pi. </span>

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Shkiper50 [21]

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Answer:

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Step-by-step explanation:

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Hope this helped :D

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Step-by-step explanation:

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