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3 years ago

PLEASE HELP ASAP!!!!!Test Grade!!!!!!!!please incude step by step explination

Mathematics

2 answers:

AleksandrR [38]3 years ago

6 0

<h2> the answer is </h2>

D. 10

hope it helps

<h3>step-by-step explanation</h3>

Mariulka [41]3 years ago

4 0

The answer is D. 10. Let me know if you want me to show my work

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Use the given data to find the best predicted value of the response variable x. Ten pairs of data yield r=0.003 and the regressi

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Answer:

Step-by-step explanation:

Given the regression equation :

y=2+3x

Mean of y values ; y = 5.0

Where y is the predicted variable ; x = predictor variable

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Null: H0 = 5

Alternative H1 : ≠ 5

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grigory [225]

Answer:

(1,1)(3,-2)

Step-by-step explanation:

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2 years ago

Read 2 more answers

Suppose that an airline uses a seat width of 16.5 in. Assume men have hip breadths that are normally distributed with a mean of

Alexxx [7]

Answer:

a) 0.018

b) 0

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 14.4 in

Standard Deviation, σ = 1 in

We are given that the distribution of breadths is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(breadth will be greater than 16.5 in)

P(x > 16.5)

$P( x > 16.5) = P( z > \displaystyle\frac{16.5 - 14.4}{1}) = P(z > 2.1)$

$= 1 - P(z \leq 2.1)$

Calculation the value from standard normal z table, we have,

$P(x > 16.5) = 1 - 0.982 = 0.018 = 1.8\%$

0.018 is the probability that if an individual man is randomly selected, his hip breadth will be greater than 16.5 in.

b) P( with 123 randomly selected men, these men have a mean hip breadth greater than 16.5 in)

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

P(x > 16.5)

$P( x > 16.5) = P( z > \displaystyle\frac{16.5-14.4}{\frac{1}{\sqrt{123}}}) = P(z > 23.29)$

$= 1 - P(z \leq 23.29)$

Calculation the value from standard normal z table, we have,

$P(x > 16.5) = 1 - 1 = 0$

There is 0 probability that 123 randomly selected men have a mean hip breadth greater than 16.5 in

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3 years ago