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3 years ago

What is the slope of a line parallel to a line with slope of 9

Mathematics

1 answer:

marishachu [46]3 years ago

7 0

Answer:

slope of 9

Step-by-step explanation:

parallel means never touching, so it would have the same slope.

Hope this helps!

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Mary bought a 7lb bag of flour. She used 1/3 of the bag to bake wedding treats.

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Answer:

she is left with 4.67 LB of flour.

Step-by-step explanation:

7 divided by 3 = 2.3333333

2.3333333 x 2 = 4.67

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3 years ago

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3 years ago

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3 years ago

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Which of the following figures are rhombuses?

Tpy6a [65]

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3 years ago

In 2018, Mike Krzyewski and John Calipari topped the list of highest paid college basketball coaches (Sports Illustrated website

expeople1 [14]

From the data given, we estimate the population mean and population standard deviation. Then, we use this estimate to find a 95% confidence interval for the population variance and the population standard deviation.

Sample:

Salaries in millions of dollars: 2.2, 1.5, 0.5, 1.3, 2.4, 1.5, 2.7, 0.3, 2.0, 0.3

Question a:

The mean is the sum of all values divided by the number of values. So

$\overline{x} = \frac{2.2 + 1.5 + 0.5 + 1.3 + 2.4 + 1.5 + 2.7 + 0.3 + 2.0 + 0.3}{10} = 1.42$

The sample mean salary is of 1.42 million.

Question b:

The standard deviation is the square root of the difference squared between each value and the mean, divided by one less than the number of values.

$s = \sqrt{\frac{(2.2-1.42)^2 + (1.5-1.42)^2 + (0.5-1.42)^2 + (1.3-1.42)^2 + (2.4-1.42)^2 + (1.5-1.42)^2 + (2.7-1.42)^2 + ...}{9}} = 0.8772$

Thus, the estimate for the population standard deviation is of 0.8772 million.

Question c:

The sample size is $n = 10$

The significance level is $\alpha = 1 - 0.05 = 0.95$

The estimate, which is the sample standard deviation, is of $s = 0.8772$ .

Now, we have to find the critical values for the Pearson distribution. They are:

$\chi^2_{\frac{\alpha}{2},n-1} = \chi^2_{0.025,9} = 19.0228$

$\chi^2_{1-\frac{\alpha}{2},n-1} = \chi^2_{0.975,9} = 2.7004$

The confidence interval for the population variance is:

$\frac{(n-1)s^2}{\chi^2_{\frac{\alpha}{2},n-1}} < \sigma^2 < \frac{(n-1)s^2}{\chi^2_{1-\frac{\alpha}{2},n-1}}$

$\frac{9*0.8772^2}{19.0228} < \sigma^2 < \frac{9*0.8772^2}{2.7004}$

$0.3641 < \sigma^2 < 2.5646$

Thus, the 95% confidence interval for the population variance is (0.3641, 2.5646)

Question d:

Standard deviation is the square root of variance, so:

$\sqrt{0.3641} = 0.6034$

$\sqrt{2.5646} = 1.6014$

The 95% confidence interval for the population standard deviation is (0.6034, 1.6014).

For more on confidence intervals for population mean/standard deviation, you can check brainly.com/question/13807706

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2 years ago