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3 years ago

7

Count from 582 using a place value

1 answer:

Yuri [45]3 years ago

8 0

5 hundreds

8 tens

2 ones

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Which fraction is equivalent to <br> 1/3

Umnica [9.8K]

Answer:

there's many! three examples of equivalent fractions to 1/3 are: 3/9 9/27. and 27/81

Hope that helps!

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3 years ago

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Mitchell jogged 2.2 miles on Friday. He jogged 0.7 miles less on Saturday than on Friday. What was the total distance he jogged

Nookie1986 [14]

You jogged 3.7 total miles! 2.2 - 0/7 = 1.5 + 2.2 = 3.7! Hope I helped!

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3 years ago

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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.

So

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

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

4 0

3 years ago

Simplify: –4(2x – 3y) + 4x – 2(x + 6)

kodGreya [7K]

Multiply the first bracket by -4. Multiply the second bracket by -2

-8x+12y+4x-2x-12

Negative and Negative = Positive

Negative and Positive = Negative

Then do -8x+4x-2x

= -6x+12y-12

Answer is D. -6x+12y-12

4 0

3 years ago

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X^2+2x+2=x^2+2<br> What is the answer to this equation?

Whitepunk [10]

Follow the steps above

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3 years ago