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3 years ago

15

A population of 1,750 cheetahs decreases by 11% per year. How many cheetahs will there be in the population after 10 years?

2 answers:

Sever21 [200]3 years ago

7 0

First, write an equation.

1750 = number of cheetahs
.89 = Rate (1-.11)
x = Number of years

1750(.89)^x

1750(.89)^10 = 545.68 (you cant have half a cheetah) so around 546

Cerrena [4.2K]3 years ago

3 0

It would be a population of 546 cheetahs

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2 years ago

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Given

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$\bar x_1 = 13.6$ $\bar x_2 = 11.6$

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This is calculated as: $\bar x_1 - \bar x_2$

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$1 - \alpha = 0.90$

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$z_{\alpha/2} = z_{0.05}$

The z score is:

$z_{\alpha/2} = z_{0.05} =1.645$

The endpoints of the confidence level is:

$(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$

$2.0 \± 1.645 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}$

$2.0 \± 1.645 * \sqrt{\frac{4.41}{60}+\frac{9}{35}}$

$2.0 \± 1.645 * \sqrt{0.0735+0.2571}$

$2.0 \± 1.645 * \sqrt{0.3306}$

$2.0 \± 0.9458$

Split

$(2.0 - 0.9458) \to (2.0 + 0.9458)$

$(1.0542) \to (2.9458)$

Hence, the 90% confidence interval is:

$CI =(1.0542,2.9458)$

Solving (c): 95% confidence interval

We have:

$c = 95\%$

$c = 0.95$

Confidence level is: $1 - \alpha$

$1 - \alpha = c$

$1 - \alpha = 0.95$

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$z_{\alpha/2} = z_{0.05/2}$

$z_{\alpha/2} = z_{0.025}$

The z score is:

$z_{\alpha/2} = z_{0.025} =1.96$

The endpoints of the confidence level is:

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$2.0 \± 1.96 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}$

$2.0 \± 1.96* \sqrt{\frac{4.41}{60}+\frac{9}{35}}$

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$(0.8730) \to (2.1270)$

Hence, the 95% confidence interval is:

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