If a population decreases by 11%, that means that 89% (100% - 11% = 89%) of cheetahs remains each number. 89% can be expressed as 0.89, so to calculate the change of the population, we must each year multiply the number of cheetahs by 0.89.

After 1 year: 1750 * 0.89 ≈ 1558

After 2 years: 1558 * 0.89 ≈ 1387

After 3 years: 1387 * 0.89 ≈ 1234

After 4 years: 1234 * 0.89 ≈ 1098

After 5 years: 1098 * 0.89 ≈ 977

After 6 years: 977 * 0.89 ≈ 870

After 7 years: 870 * 0.89 ≈ 774

After 8 years: 774 * 0.89 ≈ 689

After 9 years: 689 * 0.89 ≈ 613

After 10 years: 613 * 0.89 ≈ 546

Step-by-step explanation:

4 0

3 years ago

33. Suppose that the scores on a statewide standardized test are normally distributed with a mean of 78 and a standard deviation

tino4ka555 [31]

Given the scores on a statewide standardized test are normally distributed

Mean = μ = 78

Standard deviation = σ = 3

Normalize the data using the z-score by using the following formula and chart:

$z=\frac{x-\mu}{\sigma}$

Estimate the percentage of scores of the following cases:

(a) between 75 and 81

so, the z-score for the given numbers will be:

$\begin{gathered} 75\rightarrow z=\frac{75-78}{3}=\frac{-3}{3}=-1 \\ 81\rightarrow z=\frac{81-78}{3}=\frac{3}{3}=1 \end{gathered}$

As shown, the percentage when (-1 < z < 1) = 68%

(b) above 87

$87\rightarrow z=\frac{87-78}{3}=\frac{9}{3}=3$

The percentage when (z > 3) = 0.5%

(c) below 72

$72\rightarrow z=\frac{72-78}{3}=\frac{-6}{3}=-2$

The percentage when (z < -2) = 0.5 + 2 = 2.5%

(d) between 75 and 84

$\begin{gathered} 75\rightarrow z=\frac{75-78}{3}=-\frac{3}{3}=-1 \\ 84\rightarrow z=\frac{84-78}{3}=\frac{6}{3}=2 \end{gathered}$

The percentage when ( -1 < z < 2 ) = 68 + 13.5 = 81.5%

3 0

1 year ago

Write each phrase as an algebraic expression b plus 14?

s344n2d4d5 [400]

B+14 would be the algebraic expression

6 0

3 years ago