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Alexandra [31]
2 years ago
14

Solve. x + 6/7 = 3^2/9 x = 23/63 x = 1^ 7/16 x = 2^ 23/63 x = 4^ 5/ 63

Mathematics
1 answer:
djverab [1.8K]2 years ago
6 0
<span><span><span><span>Exam Number 700160RR</span><span>Grade 100</span><span>Date Graded 07/06/18 *</span></span><span><span>
* If your first attempt was below passing you will need to complete your retake or wait 30 days to see the correct responses. Correct exam answers will be made available to you when your lesson grade is finalized.</span><span>Continue Learning 
</span></span></span><span><span> Question NumberYour AnswerAnswerReference</span><span> 1.CCorrect</span><span> 2.DCorrect</span><span> 3.CCorrect</span><span> 4.DCorrect</span><span> 5.DCorrect</span><span> 6.DCorrect</span><span> 7.DCorrect</span><span> 8.DCorrect</span><span> 9.BCorrect</span><span> 10.ACorrect</span><span> 11.DCorrect</span><span> 12.DCorrect</span><span> 13.ACorrect</span><span> 14.ACorrect</span><span> 15.ACorrect</span><span> 16.BCorrect</span><span> 17.ACorrect</span><span> 18.BCorrect</span><span> 19.CCorrect</span><span> 20.CCorrect</span></span></span>
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If you roll two six-faced dice together, you will get 36 possible outcomes. 4 pts 1. List all possible outcomes of the experimen
bezimeni [28]

Given:

Two dice are rolled together.

Total number of possible outcomes.

To find:

The list of total possible outcomes.

The probability of getting a sum of 11 in these outcomes.

The probability of getting a sum less than or equal to 4.

The probability of getting a sum of 13 or more.

Solution:

If two dice are rolled together, then the total number of possible outcomes is 36 and list of total possible outcomes is

S = {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

Sum of 11 in these outcomes = {(5,6),(6,5),(6,6)} = 3

The probability of getting a sum of 11 in these outcomes is

P(\text{sum=11})=\dfrac{3}{36}

P(\text{sum=11})=\dfrac{1}{12}

Therefore, the probability of getting a sum of 11 in these outcomes is \dfrac{1}{12}.

Sum less than or equal to 4 = {(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)} = 6

The probability of getting a sum less than or equal to 4 is

P({sum\leq 4})=\dfrac{6}{36}

P({sum\leq 4})=\dfrac{1}{6}

Therefore, the probability of getting a sum less than or equal to 4 is \dfrac{1}{6}.

Sum of 13 or more = empty set because maximum sum is 12.

The probability of getting a sum of 13 or more is

P(sum\geq 13)=\dfrac{0}{36}

P({sum\geq 13})=0

Therefore, the probability of getting a sum of 13 or more is 0.

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Answer:

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5 x5 =25

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Answer:

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