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spin [16.1K]
3 years ago
11

PLZ HELP! A scientist is studying wildlife. she estimates the population of bats in her state to be 270,000. she predicts the po

pulation to grow at an average annual rate of 2.9%. using the scientist's prediction, create an equation that models the population of bats, y, after x years.
Mathematics
2 answers:
algol133 years ago
6 0
I think we need to use the growth or decay formula.

A = Pe^rt

A = ending amount
P = beginning amount
r = growth or decay rate 
t = time or term
e = <span>approximately constant 2.71828

Given:
P = 270,000
r = 2.9% growth
y = ending population
x = time in years

</span><span>create an equation that models the population of bats, y, after x years.

y = Pe^rx

y = 270,000 * 2.71828^0.029x</span>
kaheart [24]3 years ago
5 0
2.9% times y = x

I do hope that this correct.
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Answer:

a. We fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

b. The 95% confidence interval for the true difference of means is -2.2468 and 36.2468. There is a probability of 95% that the true difference of means \mu_{1}-\mu_{2} is between -2.2468 and 36.2468. This confidence interval contains the number 0, this is consistent with the results we got in a. Because we fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

Step-by-step explanation:

Let \mu_{1}-\mu_{2} be the true difference between the average credit score for an adult in Virginia and the average credit score for an adult in North Carolina. We have the large sample sizes n_{1} = 40 and n_{2} = 35, the unbiased point estimate for \mu_{1}-\mu_{2} is \bar{x}_{1} - \bar{x}_{2}, i.e., 699-682 = 17.

The standard error is given by \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}, i.e.,

\sqrt{\frac{(44)^{2}}{40}+\frac{(41)^{2}}{35}} = 9.8198.

a. We want to test H_{0}: \mu_{1}-\mu_{2} = 0 vs H_{1}: \mu_{1}-\mu_{2} \neq 0 (two-tailed alternative). The rejection region is given by RR = {z | z < -1.96 or z > 1.96} where -1.96 and 1.96 are the 2.5th and 97.5th quantiles of the standard normal distribution respectively. The test statistic is Z = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}} and the observed value is z_{0} = \frac{17}{9.8198} = 1.7312. Because 1.7312 does not fall inside RR, we fail to reject the null hypothesis.

b. The endpoints for a 95% confidence interval for \mu_{1}-\mu_{2} is given by 17\pm (z_{0.05/2})9.8198, i.e., 17\pm (z_{0.025})9.8198 where z_{0.025} is the 2.5th quantile of the standard normal distribution, i.e., -1.96, so, we have 17-(1.96)(9.8198) and 17+(1.96)(9.8198), i.e., -2.2468 and 36.2468. There is a probability of 95% that the true difference of means \mu_{1}-\mu_{2} is between -2.2468 and 36.2468. This confidence interval contains the number 0, this is consistent with the results we got in a. Because we fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

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