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Sergio [31]
3 years ago
14

Simplify (8m2+m+7m2)-(3m-4m3+2m2)

Mathematics
1 answer:
kondaur [170]3 years ago
4 0

(8m^2+m+7m^2)-(3m-4m^3+2m^2)\\\\=8m^2+m+7m^2-3m+4m^3-2m^2\qquad|\text{combine like terms}\\\\=(4m^3)+(8m^2+7m^2-2m^2)+(m-3m)\\\\=4m^3+13m^2-2m

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If Sarah drink 2 cups of water and then drink one more how many cups of water did Sarah drink?
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Answer:

A

Step-by-step explanation:

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2+1=3 cups of water?

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Please Need Help. Thank You :)<br><br><br>Please find the Volume of these cylinders:
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Answer:

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Step-by-step explanation:

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You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confide
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Question:

You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If​ convenient, use technology to construct the confidence intervals. A random sample of 45 home theater systems has a mean price of ​$114.00. Assume the population standard deviation is ​$15.30. Construct a​ 90% confidence interval for the population mean.

Answer:

At the 90% confidence level, confidence interval = 110.2484 < μ < 117.7516

At the 95% confidence level, confidence interval = 109.53 < μ < 118.48

The 95% confidence interval is wider

Step-by-step explanation:

Here, we have

Sample size, n = 45

Sample mean, \bar x = $114.00

Population standard deviation, σ = $15.30

The formula for Confidence Interval, CI is given by the following relation;

CI=\bar{x}\pm z\frac{\sigma}{\sqrt{n}}

Where, z is found for the 90% confidence level as ±1.645

Plugging in the values, we have;

CI=114\pm 1.645 \times \frac{15.3}{\sqrt{45}}

or CI: 110.2484 < μ < 117.7516

At 95% confidence level, we have our z value given as z = ±1.96

From which we have CI=114\pm 1.96 \times \frac{15.3}{\sqrt{45}}

Hence CI: 109.53 < μ < 118.48

To find the wider interval, we subtract their minimum from the maximum as follows;

90% Confidence level: 117.7516 - 110.2484 = 7.5

95% Confidence level: 118.47503 - 109.5297 = 8.94

Therefore, the 95% confidence interval is wider.

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