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insens350 [35]
2 years ago
12

A number cube has faces numbered 1 to 6. What is true about rolling the number cube one time? Select three options. S = {1, 2, 3

, 4, 5, 6} A. If A is a subset of S, A could be {0, 1, 2}. B. If A is a subset of S, A could be {5, 6}. C. If a subset A represents the complement of rolling a 5, then A = {1, 2, 3, 4, 6}. D. If a subset A represents the complement of rolling an even number, then A = {1, 3}.
Mathematics
2 answers:
Marat540 [252]2 years ago
5 0

Answer:

The correct answers are the options B. ; C. ; D.

Step-by-step explanation:

A numbered cube has face from 1 to 6.

The set S = {1, 2, 3, 4, 5, 6}.

A. If A is a subset of S, A could be {0, 1, 2}. This statement is false as A is a subset of B and A contains the element {0} which is not in S.

B. If A is a subset of S and A is given by {5, 6}. This statement is true as A elements of A are also in S.

C. If a subset A represents the complement of rolling a 5, then A = {1, 2, 3, 4, 6}. This statement is even true as A should contain all numbers from 1 to 6 except 5.

D. If a subset A represents the complement of rolling an even number, then A = {1, 3}. This statement is true as the numbers complement of an even number are 1, 3 and 5.

cupoosta [38]2 years ago
3 0

Answer:

its actually C and D

NOT B

Step-by-step explanation:

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Answer:

1

p(b) =  \frac{1}{6}

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p(k) =  \frac{1}{3}

3

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Step-by-step explanation:

Generally when two fair 6-sided dice is rolled the doubles are

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The total outcome of doubles is N = 6

The total outcome of the rolling the two fair 6-sided dice is

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Generally the probability that doubles (i.e., having an equal number on the two dice) were rolled is mathematically evaluated as

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p(b) =  \frac{6}{36}

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Generally when two fair 6-sided dice is rolled the outcome whose sum is 4 or less is

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Looking at this outcome we see that there are two doubles present

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The conditional probability that doubles were rolled is mathematically represented as

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p(k) =  \frac{1}{3}

Generally when two fair 6-sided dice is rolled the number of outcomes that would land on different numbers is L = 30

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=> P(a) =  \frac{10}{30}

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Answer:

As per the given statement:

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