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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 & 2011 & 2012 \\
\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 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 :

GinnyAnsweridn
Let's solve the problem step by step:

First, we need to understand the net population change in each year. For both 2011 and 2012, the net change can be computed using the formula:

[tex]\[ \text{Net Change} = \text{Births} + \text{Immigration} - \text{Deaths} - \text{Emigration} \][/tex]

### Calculation for 2011
- Deaths: 2
- Births: 5
- Immigration: 6
- Emigration: 8

[tex]\[ \text{Net Change 2011} = 5 \, (\text{births}) + 6 \, (\text{immigration}) - 2 \, (\text{deaths}) - 8 \, (\text{emigration}) \][/tex]
[tex]\[ \text{Net Change 2011} = 11 - 10 \][/tex]
[tex]\[ \text{Net Change 2011} = 1 \][/tex]

### Calculation for 2012
- Deaths: 1
- Births: 4
- Immigration: 1
- Emigration: 3

[tex]\[ \text{Net Change 2012} = 4 \, (\text{births}) + 1 \, (\text{immigration}) - 1 \, (\text{deaths}) - 3 \, (\text{emigration}) \][/tex]
[tex]\[ \text{Net Change 2012} = 5 - 4 \][/tex]
[tex]\[ \text{Net Change 2012} = 1 \][/tex]

### Since the net change is the same in 2011 and 2012, we need to find the option for 2013 that gives the same net change:
#### Option 1: 4 deaths, 7 births, 2 immigration, 6 emigration
[tex]\[ \text{Net Change} = 7 + 2 - 4 - 6 \][/tex]
[tex]\[ \text{Net Change} = 9 - 10 \][/tex]
[tex]\[ \text{Net Change} = -1 \][/tex]

#### Option 2: 3 deaths, 6 births, 5 immigration, 7 emigration
[tex]\[ \text{Net Change} = 6 + 5 - 3 - 7 \][/tex]
[tex]\[ \text{Net Change} = 11 - 10 \][/tex]
[tex]\[ \text{Net Change} = 1 \][/tex]

#### Option 3: 5 deaths, 2 births, 8 immigration, 3 emigration
[tex]\[ \text{Net Change} = 2 + 8 - 5 - 3 \][/tex]
[tex]\[ \text{Net Change} = 10 - 8 \][/tex]
[tex]\[ \text{Net Change} = 2 \][/tex]

#### Option 4: 1 death, 5 births, 3 immigration, 7 emigration
[tex]\[ \text{Net Change} = 5 + 3 - 1 - 7 \][/tex]
[tex]\[ \text{Net Change} = 8 - 8 \][/tex]
[tex]\[ \text{Net Change} = 0 \][/tex]

### Conclusion
The option that results in the same net population change of 1 (the same as in 2011 and 2012) is:

- Option 2: 3 deaths, 6 births, 5 immigration, 7 emigration

Therefore, to keep the population growth of cheetahs in 2013 the same as the previous years, the events should be:
[tex]\( \boxed{3 \, \text{deaths}, \, 6 \, \text{births}, \, 5 \, \text{immigration}, \, 7 \, \text{emigration}} \)[/tex]
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