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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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Step-by-step explanation:

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Stels [109]

Answer:

See below.

Step-by-step explanation:

First, we can see that \lim_{x \to 2}  (f(x))= -1.

Thus, for the question, we can just plug -1 in:

\lim_{x \to 2} (\frac{x}{f(x)+1})=\frac{(2)}{-1+1}  =und.

Saying undefined (or unbounded) will be correct.

However, note that as x approaches 2, the values of y decrease in order to get to -1. In other words, f(x) will always be greater or equal to -1 (you can also see this from the graph). This means that as x approaches 2, f(x) will approach -.99 then -.999 then -.9999 until it reaches -1 and then go back up. What is important is that because of this, we can determine that:

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4 0
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____ [38]

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Answer:

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