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Sagot :
Let's address the problem step-by-step, considering the answers are:
1. The initial population in 2000, [tex]\( A_0 \)[/tex], is 6.32 million.
2. The population in 2050, [tex]\( A_{2050} \)[/tex], is projected to be 10 million.
3. The number of years after 2000 until 2050, [tex]\( t_{2050} \)[/tex], is 50 years.
### Part a: Finding the Exponential Growth Function
We use the exponential growth model:
[tex]\[ A = A_0 e^{kt} \][/tex]
Given:
[tex]\[ A_0 = 6.32 \][/tex]
[tex]\[ A_{2050} = 10 \][/tex]
[tex]\[ t_{2050} = 50 \][/tex]
First, we need to find the growth rate constant [tex]\( k \)[/tex]. The population in 2050 allows us to set up the equation:
[tex]\[ 10 = 6.32 e^{50k} \][/tex]
We solve for [tex]\( k \)[/tex] as follows:
[tex]\[ e^{50k} = \frac{10}{6.32} \][/tex]
[tex]\[ 50k = \ln\left(\frac{10}{6.32}\right) \][/tex]
[tex]\[ k = \frac{\ln\left(\frac{10}{6.32}\right)}{50} \][/tex]
The computed value of [tex]\( k \)[/tex] is approximately 0.009177317696705593. Rounding [tex]\( k \)[/tex] to two decimal places, we get [tex]\( k \approx 0.01 \)[/tex].
Thus, the exponential growth function that models the data is:
[tex]\[ A = 6.32 e^{0.01t} \][/tex]
### Part b: Finding the Year When the Population Will Be 12 Million
We use the exponential growth function to find the year when the population will be 12 million.
Set [tex]\( A = 12 \)[/tex]:
[tex]\[ 12 = 6.32 e^{0.01t} \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ e^{0.01t} = \frac{12}{6.32} \][/tex]
[tex]\[ 0.01t = \ln\left(\frac{12}{6.32}\right) \][/tex]
[tex]\[ t = \frac{\ln\left(\frac{12}{6.32}\right)}{0.01} \][/tex]
From the computation, we find [tex]\( t \)[/tex] is approximately 69.86654083680715.
To find the year:
[tex]\[ \text{Year} = 2000 + t \][/tex]
[tex]\[ \text{Year} \approx 2000 + 69.87 = 2069.87 \][/tex]
Rounding to the nearest year, the population will reach 12 million around the year 2070.
### Final Answers
a. The exponential growth function that models the data is:
[tex]\[ A = 6.32 e^{0.01t} \][/tex]
b. The population will be 12 million by approximately the year:
[tex]\[ 2070 \][/tex]
1. The initial population in 2000, [tex]\( A_0 \)[/tex], is 6.32 million.
2. The population in 2050, [tex]\( A_{2050} \)[/tex], is projected to be 10 million.
3. The number of years after 2000 until 2050, [tex]\( t_{2050} \)[/tex], is 50 years.
### Part a: Finding the Exponential Growth Function
We use the exponential growth model:
[tex]\[ A = A_0 e^{kt} \][/tex]
Given:
[tex]\[ A_0 = 6.32 \][/tex]
[tex]\[ A_{2050} = 10 \][/tex]
[tex]\[ t_{2050} = 50 \][/tex]
First, we need to find the growth rate constant [tex]\( k \)[/tex]. The population in 2050 allows us to set up the equation:
[tex]\[ 10 = 6.32 e^{50k} \][/tex]
We solve for [tex]\( k \)[/tex] as follows:
[tex]\[ e^{50k} = \frac{10}{6.32} \][/tex]
[tex]\[ 50k = \ln\left(\frac{10}{6.32}\right) \][/tex]
[tex]\[ k = \frac{\ln\left(\frac{10}{6.32}\right)}{50} \][/tex]
The computed value of [tex]\( k \)[/tex] is approximately 0.009177317696705593. Rounding [tex]\( k \)[/tex] to two decimal places, we get [tex]\( k \approx 0.01 \)[/tex].
Thus, the exponential growth function that models the data is:
[tex]\[ A = 6.32 e^{0.01t} \][/tex]
### Part b: Finding the Year When the Population Will Be 12 Million
We use the exponential growth function to find the year when the population will be 12 million.
Set [tex]\( A = 12 \)[/tex]:
[tex]\[ 12 = 6.32 e^{0.01t} \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ e^{0.01t} = \frac{12}{6.32} \][/tex]
[tex]\[ 0.01t = \ln\left(\frac{12}{6.32}\right) \][/tex]
[tex]\[ t = \frac{\ln\left(\frac{12}{6.32}\right)}{0.01} \][/tex]
From the computation, we find [tex]\( t \)[/tex] is approximately 69.86654083680715.
To find the year:
[tex]\[ \text{Year} = 2000 + t \][/tex]
[tex]\[ \text{Year} \approx 2000 + 69.87 = 2069.87 \][/tex]
Rounding to the nearest year, the population will reach 12 million around the year 2070.
### Final Answers
a. The exponential growth function that models the data is:
[tex]\[ A = 6.32 e^{0.01t} \][/tex]
b. The population will be 12 million by approximately the year:
[tex]\[ 2070 \][/tex]
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