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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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First of all, we would assign a variable to the different amount of money as follows:

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Translating the word problem into an algebraic expression, we have;

For a bill of less than $100, the entire amount is due:

0 ≤ x < $100, therefore, y = x

For a bill of at least $100 but less than $500, the minimum due is $100: $100 ≤ x < $500, therefore, y = 100

For a bill of at least $500 but less than $1,000, the minimum due is $300:

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Since the credit limit is equal to $2000, we have $1000 ≤ x < $2000 and y = $500

In conclusion, the graph that correctly shows the minimum amount due for a credit amount (x)  is graph B.

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