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makkiz [27]
2 years ago
12

The population of Detroit, Michigan, decreased from 1,027,974 in 1990 to 688,701 in 2013 (Source: U.S. Census Bureau). Find the

average rate of change in the population of Detroit, Michigan, over the 23-year period.
Mathematics
1 answer:
AleksandrR [38]2 years ago
5 0

the slope goes by several names

• average rate of change

• rate of change

• deltaY over deltaX

• Δy over Δx

• rise over run

• gradient

• constant of proportionality

however, is the same cat wearing different costumes.

to get the slope of any straight line, we simply need two points off of it, so let's use the ones provided in that

\begin{array}{|cc|ll} \cline{1-2} \stackrel{years}{x}&\stackrel{population}{y}\\ \cline{1-2} 1990&1027974\\ 2013&688701\\ \cline{1-2} \end{array}\hspace{5em} (\stackrel{x_1}{1990}~,~\stackrel{y_1}{1027974})\qquad (\stackrel{x_2}{2013}~,~\stackrel{y_2}{688701}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{688701}-\stackrel{y1}{1027974}}}{\underset{run} {\underset{x_2}{2013}-\underset{x_1}{1990}}} \implies \cfrac{-339273}{23}\implies -14751\qquad \stackrel{rate~of}{change}

let's notice, is negative since the population decreased.

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Rus_ich [418]

Answer:

\hat p = \frac{14}{105}= 0.133

And that represent the proportion of failures.

The confidence interval for the mean is given by the following formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

If we replace the values obtained we got:

0.133 - 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.0475

0.133 + 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.2185

The 99% confidence interval would be given by (0.0475;0.2185)

Step-by-step explanation:

Previous concept

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 population proportion have the following distribution

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by \alpha=1-0.99=0.01 and \alpha/2 =0.005. And the critical value would be given by:

z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58

The proportion estimated would be:

\hat p = \frac{14}{105}= 0.133

And that represent the proportion of failures.

The confidence interval for the mean is given by the following formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

If we replace the values obtained we got:

0.133 - 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.0475

0.133 + 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.2185

The 99% confidence interval would be given by (0.0475;0.2185)

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Step-by-step explanation:

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Answer:

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