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Aleksandr [31]
2 years ago
12

The population of a small town in Ohio was 3,800 people in I. It has slowly been decreasing at a rate of 25% per year. What will

the population of the town be in 2025?
Mathematics
1 answer:
MrRa [10]2 years ago
3 0

Answer:

676 assuming the population is 3800 in 2019. If it's a different year then change t.

Step-by-step explanation:

P(t) = P_{0} (1+r)^t \\\\P(6) = 3800(1-.25)^6 \\=P(6) = 3800(.75)^6\\=676


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A certain change in a process for manufacturing component parts is being considered. Samples are taken under both the existing a
mr_godi [17]

Answer:

And the 90% confidence interval for the difference would be given by:(-0.022;0.0017).  

Step-by-step explanation:

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".

p_A represent the real population proportion for brand A  

\hat p_A =\frac{80}{2000}=0.04 represent the estimated proportion for the new process

n_A=2000 is the sample size required for Brand A

p_B represent the real population proportion for brand b  

\hat p_B =\frac{75}{1500}=0.05 represent the estimated proportion for the before process

n_B=1500 is the sample size required for before process

z represent the critical value for the margin of error  

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})  

The confidence interval for the difference of two proportions would be given by this formula  

(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}  

For the 90% confidence interval the value of \alpha=1-0.90=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=1.64  

And replacing into the confidence interval formula we got:  

 (0.04 -0.05) - 1.64 \sqrt{\frac{0.04(1-0.04)}{2000} +\frac{0.05(1-0.05)}{1500}}=-0.022  

(0.04 -0.05) + 1.64 \sqrt{\frac{0.04(1-0.04)}{2000} +\frac{0.05(1-0.05)}{1500}}=0.0017  

And the 90% confidence interval would be given (-0.022;0.0017).  

3 0
2 years ago
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