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3 years ago

9

ms.fords class start art at 9:30 and finish at 10:15.they spend twice as much time in math class.If they start math at 1:10,What

time do they finish math?

2 answers:

hram777 [196]3 years ago

6 0

They finish math class at 2:45

julia-pushkina [17]3 years ago

6 0

If art class is only you 45 minutes long and math is twice as long then math is 90 minutes causing math to get outof class at 2:40.

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I will give brainliest to any correct answer. Question is the image attached, please answer asap.

AlekseyPX

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An item is regularly priced at $97. Lucy bought it at a discount of 55% off the regular price.

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

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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

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3 years ago

Lynn paid $17.25 for ribbon. The cost of the ribbon was $0.75 per foot. How many feet of ribbon did

snow_tiger [21]

Answer: 23

Step-by-step explanation:

1. Divide the amount Lynn paid for the ribbon, 17.25, by the amount the ribbon cost per foot, 0.75, to get how many feet of ribbon she bought.

17.25 / 0.75 = 23

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Answer:

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Step-by-step explanation:

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