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Sagot :
To estimate the population in 2010, we use the exponential growth formula:
[tex]\[ P = A e^{kt} \][/tex]
Step-by-step solution:
1. Identify the initial values:
- Initial population in 1995 ([tex]\( P_{1995} \)[/tex]) = 228 million
- Population in 2001 ([tex]\( P_{2001} \)[/tex]) = 230 million
- Time difference from 1995 to 2001 ([tex]\( t_{1995 \to 2001} \)[/tex]) = 2001 - 1995 = 6 years
2. Determine the growth rate constant [tex]\( k \)[/tex]:
We know that in 2001 the population is given by:
[tex]\[ P_{2001} = P_{1995} e^{k \cdot t_{1995 \to 2001}} \][/tex]
Plugging in the known values:
[tex]\[ 230 = 228 e^{6k} \][/tex]
Solving for [tex]\( k \)[/tex]:
[tex]\[ \frac{230}{228} = e^{6k} \][/tex]
[tex]\[ \ln \left(\frac{230}{228}\right) = 6k \][/tex]
[tex]\[ k = \frac{\ln \left(\frac{230}{228}\right)}{6} \][/tex]
Evaluating this, we get:
[tex]\[ k \approx 0.001456 \][/tex]
3. Estimate the population in 2010:
Time difference from 1995 to 2010 ([tex]\( t_{1995 \to 2010} \)[/tex]) = 2010 - 1995 = 15 years
Using the exponential growth formula again:
[tex]\[ P_{2010} = P_{1995} e^{k \cdot t_{1995 \to 2010}} \][/tex]
Plugging in the values:
[tex]\[ P_{2010} = 228 e^{0.001456 \cdot 15} \][/tex]
Evaluating the exponent:
[tex]\[ P_{2010} \approx 233.033 \][/tex]
4. Round the population to the nearest million:
[tex]\[ \text{Population in 2010} \approx 233 \text{ million} \][/tex]
So, the estimated population in 2010 is 233 million.
[tex]\[ P = A e^{kt} \][/tex]
Step-by-step solution:
1. Identify the initial values:
- Initial population in 1995 ([tex]\( P_{1995} \)[/tex]) = 228 million
- Population in 2001 ([tex]\( P_{2001} \)[/tex]) = 230 million
- Time difference from 1995 to 2001 ([tex]\( t_{1995 \to 2001} \)[/tex]) = 2001 - 1995 = 6 years
2. Determine the growth rate constant [tex]\( k \)[/tex]:
We know that in 2001 the population is given by:
[tex]\[ P_{2001} = P_{1995} e^{k \cdot t_{1995 \to 2001}} \][/tex]
Plugging in the known values:
[tex]\[ 230 = 228 e^{6k} \][/tex]
Solving for [tex]\( k \)[/tex]:
[tex]\[ \frac{230}{228} = e^{6k} \][/tex]
[tex]\[ \ln \left(\frac{230}{228}\right) = 6k \][/tex]
[tex]\[ k = \frac{\ln \left(\frac{230}{228}\right)}{6} \][/tex]
Evaluating this, we get:
[tex]\[ k \approx 0.001456 \][/tex]
3. Estimate the population in 2010:
Time difference from 1995 to 2010 ([tex]\( t_{1995 \to 2010} \)[/tex]) = 2010 - 1995 = 15 years
Using the exponential growth formula again:
[tex]\[ P_{2010} = P_{1995} e^{k \cdot t_{1995 \to 2010}} \][/tex]
Plugging in the values:
[tex]\[ P_{2010} = 228 e^{0.001456 \cdot 15} \][/tex]
Evaluating the exponent:
[tex]\[ P_{2010} \approx 233.033 \][/tex]
4. Round the population to the nearest million:
[tex]\[ \text{Population in 2010} \approx 233 \text{ million} \][/tex]
So, the estimated population in 2010 is 233 million.
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