Answered

At Westonci.ca, we provide reliable answers to your questions from a community of experts. Start exploring today! Join our platform to connect with experts ready to provide precise answers to your questions in various areas. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.

a. In 2000, the population of a country was approximately 6.32 million, and by 2050 it is projected to grow to 10 million. Use the exponential growth model [tex]A = A_0 e^{kt}[/tex], in which [tex]t[/tex] is the number of years after 2000 and [tex]A_0[/tex] is in millions, to find an exponential growth function that models the data.

The exponential growth function that models the data is [tex]A = \square[/tex]
(Simplify your answer. Use integers or decimals for any numbers in the expression. Round to two decimal places as needed.)

b. By which year will the population be 12 million?

Sagot :

GinnyAnswer
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]
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.