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gavmur [86]
2 years ago
9

A graph shows the horizontal axis numbered 1 to 5 and the vertical axis numbered 1 to 5. points and a line show a downward trend

. which is most likely the correlation coefficient for the set of data shown? –0.83 –0.21 0.21 0.83
Mathematics
1 answer:
andrezito [222]2 years ago
6 0

The  most likely correlation coefficient for the set of data shown is: -0.21.

<h3>Correlation coefficient</h3>

We would be using CORREL in excel to determine the correlation coefficient by inputting the set data below

X Y

A (1,4)

B (2, 1.5)

C (3,3)

D (4,4)

E (5,2)

Hence:

Correlation coefficient=-0.20801

Correlation coefficient=-0.21 (Approximately)

Therefore the  most likely correlation coefficient for the set of data shown is: -0.21.

Learn more about correlation coefficient here:brainly.com/question/2735094

#SPJ4

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Answer:1 2/3

Step-by-step explanation:

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The is thinking of a number which he calls n. He finds 1/3 of the number and subtracts 5
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Step-by-step explanation:

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For an outdoor concert by the city​ orchestra, concert organizers estimate that 16 comma 000 people will attend if it is not rai
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Answer:  13300

========================================

Work Shown:

A = event that it rains

B = event that it does not rain

P(A) = 0.30

P(B) = 1-P(A) = 1-0.30 = 0.70

Multiply the attendance figures with their corresponding probabilities

  • if it rains, then 7000*P(A) = 7000*0.30 = 2100
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Add up the results: 2100+11200 = 13300

This is the expected value. This is basically the average based on the probabilities. The average is more tilted toward the higher end of the spectrum (closer to 16000 than it is to 7000) because there is a higher chance that it does not rain.

3 0
3 years ago
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Vedmedyk [2.9K]

Answer:

a) 151lb.

b) 6.25 lb

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

In this problem, we have that:

\mu = 151, \sigma = 25, n = 16

So

a) The expected value of the sample mean of the weights is 151 lb.

(b) What is the standard deviation of the sampling distribution of the sample mean weight?

This is s = \frac{\sigma}{\sqrt{n}} = \frac{25}{\sqrt{16}} = 6.25

8 0
3 years ago
The ratio of men to women working for a company is 5 to 7. If there are 126 women working for the company, what is the total num
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Answer:

There are 216 employees

Step-by-step explanation:

First you divide 126 by 7 (number of women) and you get 18 then you times that by 5 (total number of men) and you get 90 finally you add 90 (men) and 126 (women) to get 216 employees in total.

8 0
3 years ago
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