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Consider the following data.

\begin{tabular}{|l|c|c|}
\hline \multicolumn{3}{|l|}{ Rates for Cheetah Population from [tex]$2011$[/tex] to [tex]$2012$[/tex]} \\
\hline Cause of change & [tex]$2011$[/tex] & [tex]$2012$[/tex] \\
\hline Deaths & 2 & 1 \\
\hline Births & 5 & 4 \\
\hline Immigration & 6 & 1 \\
\hline Emigration & 8 & 3 \\
\hline \hline
\end{tabular}

Which would have to happen to keep the population growth of cheetahs in 2013 the same as in the previous years?

A. 4 deaths, 7 births, 2 immigration, 6 emigration
B. 3 deaths, 6 births, 5 immigration, 7 emigration
C. 5 deaths, 2 births, 8 immigration, 3 emigration
D. 1 death, 5 births, 3 immigration, 7 emigration


Sagot :

To solve this question, let's break it down into a step-by-step process, following what we need to calculate and consider:

1. Calculate the population growth for each year:
- For [tex]\(2011\)[/tex]:
[tex]\[ \text{Growth}_{2011} = \text{Births}_{2011} + \text{Immigration}_{2011} - \text{Deaths}_{2011} - \text{Emigration}_{2011} \][/tex]
Substituting the given values:
[tex]\[ \text{Growth}_{2011} = 5 + 6 - 2 - 8 = 5 + 6 - 10 = 1 \][/tex]

- For [tex]\(2012\)[/tex]:
[tex]\[ \text{Growth}_{2012} = \text{Births}_{2012} + \text{Immigration}_{2012} - \text{Deaths}_{2012} - \text{Emigration}_{2012} \][/tex]
Substituting the given values:
[tex]\[ \text{Growth}_{2012} = 4 + 1 - 1 - 3 = 4 + 1 - 4 = 1 \][/tex]

2. Assume the average population growth needed for 2013 is the same as the previous years:
- Calculate the average growth over the two years:
[tex]\[ \text{Average Growth} = \frac{\text{Growth}_{2011} + \text{Growth}_{2012}}{2} \][/tex]
[tex]\[ \text{Average Growth} = \frac{1 + 1}{2} = \frac{2}{2} = 1 \][/tex]

3. Test each option to find which one maintains this average population growth for 2013:

- For Option 1: 4 deaths, 7 births, 2 immigration, 6 emigration
[tex]\[ \text{Growth}_{2013} = 7 + 2 - 4 - 6 = 7 + 2 - 10 = -1 \][/tex]
This does not match the average growth of 1.

- For Option 2: 3 deaths, 6 births, 5 immigration, 7 emigration
[tex]\[ \text{Growth}_{2013} = 6 + 5 - 3 - 7 = 6 + 5 - 10 = 1 \][/tex]
This matches the average growth of 1.

- For Option 3: 5 deaths, 2 births, 8 immigration, 3 emigration
[tex]\[ \text{Growth}_{2013} = 2 + 8 - 5 - 3 = 2 + 8 - 8 = 2 \][/tex]
This does not match the average growth of 1.

- For Option 4: 1 death, 5 births, 3 immigration, 7 emigration
[tex]\[ \text{Growth}_{2013} = 5 + 3 - 1 - 7 = 5 + 3 - 8 = 0 \][/tex]
This does not match the average growth of 1.

So, the option that maintains the population growth rate of cheetahs in 2013 the same as the previous years is:

[tex]\[ \boxed{3 \text{ deaths}, 6 \text{ births}, 5 \text{ immigration}, 7 \text{ emigration}} \][/tex]
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