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The average tempture for -7 , 2 , -9 , -4 ,and -7

jekas [21]

To find the average, combine all the numbers together and divide by the amount there is.

-7 + 2 - 9 - 4 - 7 = -25

-25/5 = -5

-5 is your average temperature

hope this helps

5 0

3 years ago

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The width-to-height ratio of a certain computer monitor

Mnenie [13.5K]

<h2>24 inches</h2>

Step-by-step explanation:

It is given that $\frac{\textbf{width of computer monitor}}{\textbf{height of computer monitor}}=\frac{16}{9}$

It is given that $\text{height of computer monitor}=13.5$

$\text{width of computer monitor} =\text{height of computer monitor}\times \frac{16}{9}$

$\text{width of computer monitor}=13.5\times \frac{16}{9} =\frac{216}{9} =24$

Hence the required width in inches is 24.

5 0

3 years ago

and oil storage tank is a cylinder with a height of 50 feet in a diameter of 20 feet. What is the volume of the tank? Use 3.14 f

anyanavicka [17]

Answer: 7850 ft

Step-by-step explanation:

4 0

3 years ago

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Help plz...

tino4ka555 [31]

Answer:

A

Step-by-step explanation:

4^3+7=4^10

5 0

3 years ago

The following data were collected from 12 rain gauges in a park. Build a 95% CI for the mean rainfall at the park.

dybincka [34]

Answer:

Critical values: $t_{\alpha/2}=-2.201$ $t_{1-\alpha/2}=2.201$

95% confidence interval would be given by (3.646;4.472)

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The data is:

4.65 3.89 2.73 4.35 3.80 4.86 4.33 4.37 4.76 4.05 3.05 3.87

2) Compute the sample mean and sample standard deviation.

In order to calculate the mean and the sample deviation we need to have on mind the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$

=AVERAGE(4.65,3.89,2.73, 4.35, 3.8, 4.86, 4.33, 4.37, 4.76, 4.05, 3.05, 3.87)

On this case the average is $\bar X= 4.059$

=STDEV.S(4.65,3.89,2.73, 4.35, 3.8, 4.86, 4.33, 4.37, 4.76, 4.05, 3.05, 3.87)

The sample standard deviation obtained was s=0.6503

3) Find the critical value t* Use the formula for a CI to find upper and lower endpoints

In order to find the critical value we need to take in count that our sample size n =12 <30 and on this case we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . The degrees of freedom are given by: