Solve and show work. The population of New England was 4,000 in the year 1750. The population continues to increase at a rate of 28%. per year, a write an equation that models the population of New England where + is the number of years since 1750, b) Determine the population of New England in the year 1823 using the model.
Solve and show work. The population of New England was 4,000 in the year 1750. The population continues to increase at a rate of 28%. per year, a write an equation that models the population of New England where + is the number of years since 1750, b) Determine the population of New England in the year 1823 using the model.
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Question
![**Title: Population Growth Model for New England**
---
**Problem: Exponential Growth Calculation**
**Goal:** Solve and show work for the given population growth problem.
### Given Information:
- The population of New England in the year 1750 was 4,000.
- The population continues to increase at a rate of 2.8% per year.
### Tasks:
1. **Write an equation that models the population of New England where \( t \) is the number of years since 1750.**
2. **Determine the population of New England in the year 1823 using the model.**
### Solution:
1. **Creating the Equation:**
To model the population growth, we can use the exponential growth formula:
\[
P(t) = P_0 \times (1 + r)^t
\]
Where:
- \( P(t) \) is the population at year \( t \).
- \( P_0 \) is the initial population (4,000 in this case).
- \( r \) is the growth rate (2.8% or 0.028 as a decimal).
- \( t \) is the number of years since the initial year (1750).
Plugging in the given values:
\[
P(t) = 4000 \times (1 + 0.028)^t
\]
This is the model of the population of New England based on the given growth rate.
2. **Determining the Population in 1823:**
To find the population in the year 1823, we need to determine how many years have passed since 1750:
\[
t = 1823 - 1750 = 73 \text{ years}
\]
Now, substitute \( t = 73 \) into the equation:
\[
P(73) = 4000 \times (1 + 0.028)^{73}
\]
Calculate the result:
\[
P(73) = 4000 \times (1.028)^{73}
\]
Using a calculator to evaluate the exponential expression:
\[
(1.028)^{73} \approx 7.161
\]
Now, multiply by the initial population:
\[
P(73) = 4000 \times](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F76555388-6a8d-4d01-aabe-a551666898f4%2Fcc0ed370-8526-4622-95bb-081bd35e1ace%2F65zniu_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Population Growth Model for New England**
---
**Problem: Exponential Growth Calculation**
**Goal:** Solve and show work for the given population growth problem.
### Given Information:
- The population of New England in the year 1750 was 4,000.
- The population continues to increase at a rate of 2.8% per year.
### Tasks:
1. **Write an equation that models the population of New England where \( t \) is the number of years since 1750.**
2. **Determine the population of New England in the year 1823 using the model.**
### Solution:
1. **Creating the Equation:**
To model the population growth, we can use the exponential growth formula:
\[
P(t) = P_0 \times (1 + r)^t
\]
Where:
- \( P(t) \) is the population at year \( t \).
- \( P_0 \) is the initial population (4,000 in this case).
- \( r \) is the growth rate (2.8% or 0.028 as a decimal).
- \( t \) is the number of years since the initial year (1750).
Plugging in the given values:
\[
P(t) = 4000 \times (1 + 0.028)^t
\]
This is the model of the population of New England based on the given growth rate.
2. **Determining the Population in 1823:**
To find the population in the year 1823, we need to determine how many years have passed since 1750:
\[
t = 1823 - 1750 = 73 \text{ years}
\]
Now, substitute \( t = 73 \) into the equation:
\[
P(73) = 4000 \times (1 + 0.028)^{73}
\]
Calculate the result:
\[
P(73) = 4000 \times (1.028)^{73}
\]
Using a calculator to evaluate the exponential expression:
\[
(1.028)^{73} \approx 7.161
\]
Now, multiply by the initial population:
\[
P(73) = 4000 \times
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