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atroni [7]
3 years ago
6

How many 5-member chess teams can be chosen from 15 interested players? Consider only the members selected, not their board posi

tions. 3,003 120 360,360
Mathematics
1 answer:
ryzh [129]3 years ago
7 0

<u>Answer</u>:

3003 number of 5-member chess teams can be chosen from 15 interested players.

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

Given:

Number of the interested players =  15

To Find:

Number of 5-member chess teams that can be chosen = ?

Solution:

Combinations are a way to calculate the total outcomes of an event where order of the outcomes does not matter. To calculate combinations, we will use the formulanCr = \frac{n!}{r!(n - r)!}

where

n represents the total number of items,

r represents the number of items being chosen at a time.

Now  we have n = 15 and r = 5

Substituting the values,

15C_5 = \frac{15!}{5!(15- 5)!}

15C_5 = \frac{15!}{5!(10)!}

15C_5 = \frac{15!}{5!(10)!}

15C_5 = \frac{15\times \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 }{5 \times 4 \times 3 \times 2 \times 1(10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)}

15C_5 = \frac{15\times \times 14 \times 13 \times 12 \times 11}{(5 \times 4 \times 3 \times 2 \times 1)}

15C_5 = \frac{360360}{120}

15C_5 = 3003

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irina1246 [14]

Answer:

We conclude that commute time in her city is less than the amount reported by the survey which is 25.4 minutes.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 25.4 minutes

Sample mean, \bar{x} = 22.1 minutes

Sample size, n = 15

Alpha, α = 0.10

Sample standard deviation, s = 5.3 minutes

First, we design the null and the alternate hypothesis

H_{0}: \mu = 25.4\text{ minutes}\\H_A: \mu < 25.4\text{ minutes}

We use one-tailed t(left) test to perform this hypothesis.

Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }

Putting all the values, we have

t_{stat} = \displaystyle\frac{22.1 - 25.4}{\frac{5.3}{\sqrt{15}} } = -2.411

Now, t_{critical} \text{ at 0.10 level of significance, 14 degree of freedom } = -1.345

Since,                    

t_{stat} < t_{critical}

We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

Thus, there is enough evidence to conclude that commute time in her city is less than the amount reported by the survey which is 25.4 minutes.

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Answer:

<em>The building is 61.5 m tall</em>

Step-by-step explanation:

The image below is a diagram where all the given distances and angles are shown. We have additionally added some variables:

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x  = base of each triangle

The angles a and b can be easily found by subtracting the given angles from 90° since they are complementary angles, thus:

a = 90° - 37° = 53°

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Now we apply the tangent ratio on both triangles separately:

\displaystyle \tan a=\frac{\text{opposite leg}}{\text{adjacent leg}}

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From the last equation:

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Substituting into the first equation:

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Operating on the right side:

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Rearranging:

\displaystyle \tan 53^\circ+\tan 48^\circ=\frac{150}{h}

Solving for h:

\displaystyle h=\frac{150}{\tan 53^\circ+\tan 48^\circ}

Calculating:

h = 61.5 m

The building is 61.5 m tall

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