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andreyandreev [35.5K]

3 years ago

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You are an urban planner assessing the growth of a city. Ten years ago, the city's population was 250,823. Its current populatio

n is 325,823. By about what percentage has the city grown over the past ten years? Round to the nearest percent.

1 answer:

Tanzania [10]3 years ago

6 0

Answer:

I just answered it

Step-by-step explanation:

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Answer:

<h3><u>Mean</u></h3>

<u />

$\textsf{Mean}\:\overline{X}=\sf \dfrac{\textsf{sum of all the data values}}{\textsf{total number of data values}}$

$\implies \sf Mean\:(Nilo)=\dfrac{5+6+14+15}{4}=\dfrac{40}{4}=10$

$\implies \sf Mean\:(Lisa)=\dfrac{8+9+11+12}{4}=\dfrac{40}{4}=10$

<h3><u>Standard Deviation</u></h3>

$\displaystyle \textsf{Standard Deviation }s=\sqrt{\dfrac{\sum X^2-\dfrac{(\sum X)^2}{n}}{n-1}}$

$\begin{aligned}\displaystyle \textsf{Standard Deviation (Nilo)} & =\sqrt{\dfrac{(5^2+6^2+14^2+15^2)-\dfrac{(5+6+14+15)^2}{4}}{4-1}}\\\\& = \sqrt{\dfrac{482-\dfrac{40^2}{4}}{3}}\\\\& = \sqrt{\dfrac{82}{3}}\\\\& = 5.23\end{aligned}$

$\begin{aligned}\displaystyle \textsf{Standard Deviation (Lisa)} & =\sqrt{\dfrac{(8^2+9^2+11^2+12^2)-\dfrac{(8+9+11+12)^2}{4}}{4-1}}\\\\& = \sqrt{\dfrac{410-\dfrac{40^2}{4}}{3}}\\\\& = \sqrt{\dfrac{10}{3}}\\\\& = 1.83\end{aligned}$

<h3><u>Summary</u></h3>

Nilo has a mean score of 10 and a standard deviation of 5.23.

Lisa has a mean score of 10 and a standard deviation of 1.83.

The <u>mean</u> scores are the <u>same</u>.

Nilo's standard deviation is higher than Lisa's. Therefore, Nilo's test scores are more <u>spread out</u> that Lisa's, which means Lisa's test scores are more <u>consistent</u>.

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