Welcome to Westonci.ca, where curiosity meets expertise. Ask any question and receive fast, accurate answers from our knowledgeable community. Get quick and reliable solutions to your questions from a community of experienced professionals on our platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
To estimate the population of a country in the year 2016, using its population data from the years 1994 and 1998, we can apply the exponential growth formula:
[tex]\[ P = A e^{kt} \][/tex]
where:
- [tex]\( P \)[/tex] is the population at time [tex]\( t \)[/tex].
- [tex]\( A \)[/tex] is the initial population.
- [tex]\( k \)[/tex] is the growth rate.
- [tex]\( t \)[/tex] is the time in years.
### Step 1: Identify the known values
- Initial population in 1994: [tex]\( A = 76 \)[/tex] million
- Population in 1998: [tex]\( P = 81 \)[/tex] million
- The time span between 1994 and 1998: [tex]\( t = 1998 - 1994 = 4 \)[/tex] years
- Target year for population estimate: 2016
### Step 2: Find the growth rate [tex]\( k \)[/tex]
First, we rearrange the exponential growth formula to solve for the growth rate [tex]\( k \)[/tex]:
[tex]\[ P = A e^{kt} \][/tex]
[tex]\[ \frac{P}{A} = e^{kt} \][/tex]
[tex]\[ \ln\left(\frac{P}{A}\right) = kt \][/tex]
[tex]\[ k = \frac{1}{t} \ln\left(\frac{P}{A}\right) \][/tex]
Substituting the known values:
[tex]\[ P = 81 \, \text{million} \][/tex]
[tex]\[ A = 76 \, \text{million} \][/tex]
[tex]\[ t = 4 \, \text{years} \][/tex]
[tex]\[ k = \frac{1}{4} \ln\left(\frac{81}{76}\right) \][/tex]
From the result, we know:
[tex]\[ k \approx 0.015928953596526945 \, \text{(per year)} \][/tex]
### Step 3: Estimate the population in 2016
Now we use the exponential growth formula again to find the population in 2016.
Let's calculate the time span from 1994 to 2016:
[tex]\[ t_{\text{target}} = 2016 - 1994 = 22 \, \text{years} \][/tex]
Using the exponential growth formula:
[tex]\[ P = A e^{kt_{\text{target}}} \][/tex]
Substitute the values:
[tex]\[ A = 76 \, \text{million} \][/tex]
[tex]\[ k \approx 0.015928953596526945 \, \text{(per year)} \][/tex]
[tex]\[ t_{\text{target}} = 22 \, \text{years} \][/tex]
[tex]\[ P = 76 \cdot e^{0.015928953596526945 \cdot 22} \][/tex]
From the result, we know:
[tex]\[ P \approx 107.89627181144573 \, \text{million} \][/tex]
Rounding to the nearest million:
[tex]\[ P = 108 \, \text{million} \][/tex]
### Answer:
The estimated population in 2016 is [tex]\(\boxed{108 \, \text{million}}\)[/tex].
[tex]\[ P = A e^{kt} \][/tex]
where:
- [tex]\( P \)[/tex] is the population at time [tex]\( t \)[/tex].
- [tex]\( A \)[/tex] is the initial population.
- [tex]\( k \)[/tex] is the growth rate.
- [tex]\( t \)[/tex] is the time in years.
### Step 1: Identify the known values
- Initial population in 1994: [tex]\( A = 76 \)[/tex] million
- Population in 1998: [tex]\( P = 81 \)[/tex] million
- The time span between 1994 and 1998: [tex]\( t = 1998 - 1994 = 4 \)[/tex] years
- Target year for population estimate: 2016
### Step 2: Find the growth rate [tex]\( k \)[/tex]
First, we rearrange the exponential growth formula to solve for the growth rate [tex]\( k \)[/tex]:
[tex]\[ P = A e^{kt} \][/tex]
[tex]\[ \frac{P}{A} = e^{kt} \][/tex]
[tex]\[ \ln\left(\frac{P}{A}\right) = kt \][/tex]
[tex]\[ k = \frac{1}{t} \ln\left(\frac{P}{A}\right) \][/tex]
Substituting the known values:
[tex]\[ P = 81 \, \text{million} \][/tex]
[tex]\[ A = 76 \, \text{million} \][/tex]
[tex]\[ t = 4 \, \text{years} \][/tex]
[tex]\[ k = \frac{1}{4} \ln\left(\frac{81}{76}\right) \][/tex]
From the result, we know:
[tex]\[ k \approx 0.015928953596526945 \, \text{(per year)} \][/tex]
### Step 3: Estimate the population in 2016
Now we use the exponential growth formula again to find the population in 2016.
Let's calculate the time span from 1994 to 2016:
[tex]\[ t_{\text{target}} = 2016 - 1994 = 22 \, \text{years} \][/tex]
Using the exponential growth formula:
[tex]\[ P = A e^{kt_{\text{target}}} \][/tex]
Substitute the values:
[tex]\[ A = 76 \, \text{million} \][/tex]
[tex]\[ k \approx 0.015928953596526945 \, \text{(per year)} \][/tex]
[tex]\[ t_{\text{target}} = 22 \, \text{years} \][/tex]
[tex]\[ P = 76 \cdot e^{0.015928953596526945 \cdot 22} \][/tex]
From the result, we know:
[tex]\[ P \approx 107.89627181144573 \, \text{million} \][/tex]
Rounding to the nearest million:
[tex]\[ P = 108 \, \text{million} \][/tex]
### Answer:
The estimated population in 2016 is [tex]\(\boxed{108 \, \text{million}}\)[/tex].
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.