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tatyana61 [14]
2 years ago
9

There are 50 runners in a race. How many ways can the runners finish first, second, and third?

Mathematics
1 answer:
Dmitrij [34]2 years ago
8 0

Number of ways can the runners finish first, second, and third is 1,17,600 .

<u>Step-by-step explanation:</u>

Permutation is the act of arranging the members of a set into a sequence or order, or, if the set is already ordered, rearranging (reordering) its elements—a process called permuting. For example, written as tuples, there are six permutations of the set {1,2,3}, namely: (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), and (3,2,1). Formula of Permutation is : P(n,r) = \frac{n!}{(n-r)!} where n is the number of things to choose from,  and we choose r of them,  no repetitions,  order matters. Here , n = 50 , r= 3

⇒ P(n,r) = \frac{n!}{(n-r)!}

⇒ P(50,3) = \frac{50!}{(50-3)!} = \frac{50(49)(48)47!}{47!}

⇒ P(50,3) = 50(49)(48)

⇒ P(50,3) = 1,17,600

∴ Number of ways can the runners finish first, second, and third is 1,17,600 .

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<h3>Answer with explanation:</h3>

It is given that:

Circle 1 has center (−4, −7) and a radius of 12 cm.

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a. -1.60377

b. 0.25451

c. 0.344

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Step-by-step explanation:

The number of pieces in a regular bag of Skittles is approximately normally distributed with a mean of 38.4 and a standard deviation of 2.12.

a)What is the z-score value of a randomly selected bag of Skittles that has 35 Skittles? a) 1.62 b) -1.62 c) 3.40 d) -3.40 e)1.303.

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Option b) -1.62 is correct

b) What is the probability that a randomly selected bag of Skittles has at least 37 Skittles? a) .152 b) .247 c) .253 d).747e).7534. .

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Mean of 38.4 and a standard deviation of 2.12.

z = (37 - 38.4)/2.12

= -0.66038

P-value from Z-Table:

P(x<37) = 0.25451

The probability that a randomly selected bag of Skittles has at least 37 Skittles is 0.25451

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c) What is the probability that a randomly selected bag of Skittles has between 39 and 42 Skittles? a) .112 b) .232 c) .344 d).457 e).6125.

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Mean of 38.4 and a standard deviation of 2.12.

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z = (42 - 38.4)/2.12

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Probability value from Z-Table:

P(x = 42) = 0.95526

The probability that a randomly selected bag of Skittles has between 39 and 42 Skittles is:

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d) What is the percentile rank of a randomly selected bag of Skittles that has 40 Skittles in it? a)82nd b) 78th c) 75th d)25th e)22nd

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P-value from Z-Table:

P(x = 40) = 0.77479

Converting to percentage = 0.77479× 100

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