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tatyana61 [14]
3 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]3 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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Saul decides to use the IQR to measure the spread of the data. Saul calculates the IQR of the data set to be 27.
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Answer:

78

Step-by-step explanation:

The IQR is quartile 3 - quartile 1. First, we need to find the mean of the dataset, which is 50 -- the fifth number. Then, the median of the dataset on the left of the fifth number is the first quartile, and the median of the dataset on the right is the third. We get Q1 to then be halfway between 30 and 50, or 40, and Q3 to be halfway between 56 and n, so Q3 = (56+n)/2. Since we need Q3-Q1=27, or (56+n)/2-40=27, we have (56+n)/2=67 and 56+n=134, so n = 134-56=78

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What is the simplified form of 4 X ? 3 + 7?
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Step-by-step explanation:

add 3+7=10

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Simplify 6(x + 2) + 7.<br><br> 6x + 9<br> 6x + 15<br> 6x + 19
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i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.
Natali [406]

Answer:

sin\ \varnothing = \frac{-1}{10}\sqrt 2

cos\ \varnothing = \frac{-7}{10}\sqrt 2

tan\ \varnothing = \frac{1}{7}

cot\ \varnothing = 7

sec\ \varnothing = \frac{-5}{7}\sqrt 2

csc\ \varnothing = -5\sqrt 2

Step-by-step explanation:

Given

N = (-7,-1) --- terminal side of \varnothing

Required

Determine the values of trigonometric functions of \varnothing.

For \varnothing, the trigonometry ratios are:

sin\ \varnothing = \frac{y}{r}       cos\ \varnothing = \frac{x}{r}       tan\ \varnothing = \frac{y}{x}

cot\ \varnothing = \frac{x}{y}       sec\ \varnothing = \frac{r}{x}       csc\ \varnothing = \frac{r}{y}

Where:

r^2 = x^2 + y^2

r = \sqrt{x^2 + y^2

In N = (-7,-1)

x = -7 and y = -1

So:

r = \sqrt{(-7)^2 + (-1)^2

r = \sqrt{50

r = \sqrt{25 * 2

r = \sqrt{25} * \sqrt 2

r = 5 * \sqrt 2

r = 5 \sqrt 2

<u>Solving the trigonometry functions</u>

sin\ \varnothing = \frac{y}{r}

sin\ \varnothing = \frac{-1}{5\sqrt 2}

Rationalize:

sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}

sin\ \varnothing = \frac{-\sqrt 2}{5*2}

sin\ \varnothing = \frac{-\sqrt 2}{10}

sin\ \varnothing = \frac{-1}{10}\sqrt 2

cos\ \varnothing = \frac{x}{r}

cos\ \varnothing = \frac{-7}{5\sqrt 2}

Rationalize

cos\ \varnothing = \frac{-7}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}

cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}

cos\ \varnothing = \frac{-7\sqrt 2}{10}

cos\ \varnothing = \frac{-7}{10}\sqrt 2

tan\ \varnothing = \frac{y}{x}

tan\ \varnothing = \frac{-1}{-7}

tan\ \varnothing = \frac{1}{7}

cot\ \varnothing = \frac{x}{y}

cot\ \varnothing = \frac{-7}{-1}

cot\ \varnothing = 7

sec\ \varnothing = \frac{r}{x}

sec\ \varnothing = \frac{5\sqrt 2}{-7}

sec\ \varnothing = \frac{-5}{7}\sqrt 2

csc\ \varnothing = \frac{r}{y}

csc\ \varnothing = \frac{5\sqrt 2}{-1}

csc\ \varnothing = -5\sqrt 2

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2/7 because both 8 and 28 are divisible by 4
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