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RUDIKE [14]
3 years ago
9

Cindy has a job on a orchard.

Mathematics
1 answer:
Ulleksa [173]3 years ago
6 0

Answer:

$18.60

$16.00

_________

$34.60

Step-by-step explanation:

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I’m stuck on this question. any help would be greatly appreciated
Mademuasel [1]

Using the Pythagorean theorem we get:

y^2+8^2=12^2.

Simplifying the above result we get:

y^2+64=144.

Subtracting 64 from the above equation we get:

\begin{gathered} y^2+64-64=144-64, \\ y^2=80. \end{gathered}

Therefore:

y=\sqrt{80}\approx8.9.

Answer:

y=8.9.

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2 years ago
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vova2212 [387]

Answer:

60%

Step-by-step explanation:

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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
how many decimal places are in the product of a number with decimal places to the hundredths multiplied by a number with decimal
ExtremeBDS [4]
3 decimal places. You start with the number with the two decimal places and then you add a decimal place because the number you're multiplying it by has one decimal place.
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3 years ago
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- :-) heyoo my name is cole
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