Welcome to Westonci.ca, the place where your questions find answers from a community of knowledgeable experts. Get immediate and reliable solutions to your questions from a knowledgeable community of professionals on our platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
Sure! Let's solve this problem step by step using the exponential growth formula [tex]\( P = A e^{kt} \)[/tex].
### Step 1: Understand the given information
- Population in 1995 [tex]\((P_{1995})\)[/tex]: 228 million
- Population in 2001 [tex]\((P_{2001})\)[/tex]: 230 million
- Year difference from 1995 to 2001 [tex]\((\Delta t_{1995-2001})\)[/tex]: [tex]\(2001 - 1995 = 6\)[/tex] years
- Year difference from 1995 to 2010 [tex]\((\Delta t_{1995-2010})\)[/tex]: [tex]\(2010 - 1995 = 15\)[/tex] years
### Step 2: Use the exponential growth formula
The exponential growth formula is:
[tex]\[ P = A e^{kt} \][/tex]
We need to find the growth rate [tex]\( k \)[/tex]. To do this, we use the populations in 1995 and 2001:
[tex]\[ P_{2001} = P_{1995} e^{k \cdot \Delta t_{1995-2001}} \][/tex]
[tex]\[ 230 = 228 e^{6k} \][/tex]
### Step 3: Solve for [tex]\( k \)[/tex]
First, divide both sides by 228:
[tex]\[ \frac{230}{228} = e^{6k} \][/tex]
Take the natural logarithm of both sides:
[tex]\[ \ln \left( \frac{230}{228} \right) = 6k \][/tex]
[tex]\[ k = \frac{\ln \left( \frac{230}{228} \right)}{6} \][/tex]
Using the natural logarithm:
[tex]\[ k \approx \frac{0.0088}{6} \approx 0.0015 \][/tex]
### Step 4: Use [tex]\( k \)[/tex] to estimate the population in 2010
We use the value of [tex]\( k \)[/tex] we found to estimate the population in 2010:
[tex]\[ P_{2010} = P_{1995} e^{k \cdot \Delta t_{1995-2010}} \][/tex]
[tex]\[ P_{2010} = 228 e^{0.0015 \cdot 15} \][/tex]
Calculate the exponent:
[tex]\[ 0.0015 \cdot 15 = 0.0225 \][/tex]
So,
[tex]\[ P_{2010} = 228 e^{0.0225} \][/tex]
Calculate the exponential term:
[tex]\[ e^{0.0225} \approx 1.0228 \][/tex]
Finally, calculate [tex]\( P_{2010} \)[/tex]:
[tex]\[ P_{2010} \approx 228 \cdot 1.0228 \approx 233.224 \][/tex]
Rounded to the nearest million, the population in 2010 is [tex]\( 233 \)[/tex] million.
### Conclusion
The estimated population in 2010 is:
[tex]\[ \boxed{233} \][/tex]
### Step 1: Understand the given information
- Population in 1995 [tex]\((P_{1995})\)[/tex]: 228 million
- Population in 2001 [tex]\((P_{2001})\)[/tex]: 230 million
- Year difference from 1995 to 2001 [tex]\((\Delta t_{1995-2001})\)[/tex]: [tex]\(2001 - 1995 = 6\)[/tex] years
- Year difference from 1995 to 2010 [tex]\((\Delta t_{1995-2010})\)[/tex]: [tex]\(2010 - 1995 = 15\)[/tex] years
### Step 2: Use the exponential growth formula
The exponential growth formula is:
[tex]\[ P = A e^{kt} \][/tex]
We need to find the growth rate [tex]\( k \)[/tex]. To do this, we use the populations in 1995 and 2001:
[tex]\[ P_{2001} = P_{1995} e^{k \cdot \Delta t_{1995-2001}} \][/tex]
[tex]\[ 230 = 228 e^{6k} \][/tex]
### Step 3: Solve for [tex]\( k \)[/tex]
First, divide both sides by 228:
[tex]\[ \frac{230}{228} = e^{6k} \][/tex]
Take the natural logarithm of both sides:
[tex]\[ \ln \left( \frac{230}{228} \right) = 6k \][/tex]
[tex]\[ k = \frac{\ln \left( \frac{230}{228} \right)}{6} \][/tex]
Using the natural logarithm:
[tex]\[ k \approx \frac{0.0088}{6} \approx 0.0015 \][/tex]
### Step 4: Use [tex]\( k \)[/tex] to estimate the population in 2010
We use the value of [tex]\( k \)[/tex] we found to estimate the population in 2010:
[tex]\[ P_{2010} = P_{1995} e^{k \cdot \Delta t_{1995-2010}} \][/tex]
[tex]\[ P_{2010} = 228 e^{0.0015 \cdot 15} \][/tex]
Calculate the exponent:
[tex]\[ 0.0015 \cdot 15 = 0.0225 \][/tex]
So,
[tex]\[ P_{2010} = 228 e^{0.0225} \][/tex]
Calculate the exponential term:
[tex]\[ e^{0.0225} \approx 1.0228 \][/tex]
Finally, calculate [tex]\( P_{2010} \)[/tex]:
[tex]\[ P_{2010} \approx 228 \cdot 1.0228 \approx 233.224 \][/tex]
Rounded to the nearest million, the population in 2010 is [tex]\( 233 \)[/tex] million.
### Conclusion
The estimated population in 2010 is:
[tex]\[ \boxed{233} \][/tex]
We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.