Statesville's population in 2010 was about 24,800, and was growing by about 1.1% each year. If this continues, what will Statesville's population be in 2019? [Round to the nearest person.] people Calculator
Statesville's population in 2010 was about 24,800, and was growing by about 1.1% each year. If this continues, what will Statesville's population be in 2019? [Round to the nearest person.] people Calculator
Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE:
1. Give the measures of the complement and the supplement of an angle measuring 35°.
Related questions
Question
![### Statesville Population Growth Calculation
#### Problem Statement:
Statesville's population in 2010 was about 24,800 and was growing by about 1.1% each year. If this growth rate continues, what will Statesville's population be in 2019? [Round to the nearest person.]
#### Input Fields:
- **Population:** *(Text Field - User input required)*
#### Tools Available:
- **Calculator:** *(Button - To assist with calculations)*
#### Action:
- **Submit Question:** *(Button - To submit the answer for verification)*
##### Please enter your calculated population estimate for Statesville in 2019 in the text field and press the "Submit Question" button to check your answer.
Using the formula for exponential growth:
\[ P(t) = P_0 \times (1 + r)^t \]
where:
- \( P(t) \) is the population at time \( t \),
- \( P_0 \) is the initial population,
- \( r \) is the growth rate per period,
- \( t \) is the number of periods (years in this case).
You can calculate the population as follows:
\[ t = 2019 - 2010 = 9 \text{ years} \]
\[ P_0 = 24,800 \]
\[ r = 0.011 \]
\[ P(9) = 24,800 \times (1 + 0.011)^9 \]
Perform this calculation using your calculator and enter the result in the provided text field rounded to the nearest person.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F38801349-e7ca-44ac-b05b-a57835e16b3c%2F547297e7-0fc9-49f7-ba31-f99c93ebe044%2Fn7rg7mw_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Statesville Population Growth Calculation
#### Problem Statement:
Statesville's population in 2010 was about 24,800 and was growing by about 1.1% each year. If this growth rate continues, what will Statesville's population be in 2019? [Round to the nearest person.]
#### Input Fields:
- **Population:** *(Text Field - User input required)*
#### Tools Available:
- **Calculator:** *(Button - To assist with calculations)*
#### Action:
- **Submit Question:** *(Button - To submit the answer for verification)*
##### Please enter your calculated population estimate for Statesville in 2019 in the text field and press the "Submit Question" button to check your answer.
Using the formula for exponential growth:
\[ P(t) = P_0 \times (1 + r)^t \]
where:
- \( P(t) \) is the population at time \( t \),
- \( P_0 \) is the initial population,
- \( r \) is the growth rate per period,
- \( t \) is the number of periods (years in this case).
You can calculate the population as follows:
\[ t = 2019 - 2010 = 9 \text{ years} \]
\[ P_0 = 24,800 \]
\[ r = 0.011 \]
\[ P(9) = 24,800 \times (1 + 0.011)^9 \]
Perform this calculation using your calculator and enter the result in the provided text field rounded to the nearest person.
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