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Westkost [7]
2 years ago
15

If a number is chosen at random from the set {1, 2, 3, 4, . . ., 18}, what is the probability that the number chosen is a factor

of 17?
Mathematics
1 answer:
choli [55]2 years ago
5 0
Looking at the set, we are given 18 elements. 17 is prime; it has only two factors: 1 and 17, since 1•17=17. So, the question is really asking what is the probability the numbers 1 or 17 is chosen. As mentioned earlier, 17 is prime, so there are two possible choices: 1 and 17.

P (probability) = possible outcomes / total outcomes

It is important to note that these events are “or” events, meaning that the probability can only be determined by choosing a 1 or a 17; you can’t randomly chose a 1 and 17 at the same time. So, the formula is:

P(A or B) = P(A) + P(B)

All this is saying is that given two possible outcomes, the probability occurs independent of each event; they don’t occur at the same time.

P(1 or 17) = P(1)/18 + P(1)/18

P(1 or 17) = 2/18

Since 17 is prime, it’s two and only factors are 1 and 17. The probability of randomly choosing a 1 or 17 is 2/18, meaning that there are 2 elements in the set out of a possible 18 elements that can be randomly chosen.

2/18 simplifies to 1/9


So, your answer is 1/9
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Answer:

Marissa should have used 30 as the common denominator.

Step-by-step explanation:

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Looking for the LCM of 6, 3, 7/5, 1/5 and 3 we get 30.

Since the numbers are to be listed from least to greatest, then ordering from left to right is the correct form of listing since the numbers on the number line increase going from left to right.

This implies that the only mistake she made was she should've used 30 as the common denominator.

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3 years ago
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Answer:

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Step-by-step explanation:

Given that:

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