The population of a town can be modeled by the equation P, left parenthesis, t, right parenthesis, equals, 14, comma, 400, e, start superscript, minus, 0, point, 0, 5, 1, t, end superscriptP(t)=14,400e 
−0.051t
  where tt is measured in years since 2010.

a. Is the population increasing or decreasing?

b. What was the population in 2010?

c. Predict the population of the town in 20 years. (Round to the nearest person.)

Let's analyze the given population model:

The population \( P(t) \) is given by the equation:
\[
P(t) = 14,400 \, e^{-0.051t}
\]
where:
- \( t \) is the time in years since 2010.
- \( e \) is the base of the natural logarithm (approximately 2.718).
- The exponent \( -0.051t \) determines the rate of change.

### a. Is the population increasing or decreasing?

To determine whether the population is increasing or decreasing, we need to look at the exponent in the function. The exponent is \( -0.051t \). Since the coefficient of \( t \) is negative (-0.051), the exponent is negative, and as time \( t \) increases, \( e^{-0.051t} \) decreases because raising \( e \) to a negative exponent results in a fraction less than 1.

Thus, the population is **decreasing** over time.

### b. What was the population in 2010?

To find the population in 2010, we substitute \( t = 0 \) (since 2010 is the starting point).

\[
P(0) = 14,400 \, e^{-0.051 \times 0} = 14,400 \, e^0
\]

Since \( e^0 = 1 \):

\[
P(0) = 14,400 \times 1 = 14,400
\]

So, the population in 2010 was **14,400**.

### c. Predict the population of the town in 20 years.

To find the population 20 years after 2010 (i.e., in 2030), we substitute \( t = 20 \) into the equation:

\[
P(20) = 14,400 \, e^{-0.051 \times 20}
\]

First, calculate the exponent:

\[
-0.051 \times 20 = -1.02
\]

Now, substitute this back into the equation:

\[
P(20) = 14,400 \, e^{-1.02}
\]

Next, let's calculate \( e^{-1.02} \) and find \( P(20) \).The predicted population of the town in 20 years (in 2030) is **5,193** people (rounded to the nearest person).

### Summary of Results:
- **a.** The population is decreasing.
- **b.** The population in 2010 was **14,400**.
- **c.** The predicted population in 2030 is **5,193**.

Would you like more details or have any questions?

### Related Questions:
1. How do exponential growth and decay models differ mathematically?
2. What is the significance of the base \( e \) in natural exponential functions?
3. How would the population change if the rate were positive instead of negative?
4. Can we determine when the population will be halved from its initial value?
5. How do you compute the rate of change for exponential functions?

**Tip:** Remember that the sign of the exponent in an exponential model determines whether it represents growth or decay.

The population of a town can be modeled by the equation P(t) = 14,400e^(-0.051t) where t is measured in years since 2010. a. Is the population increasing or decreasing? b. What was the population in 2010? c. Predict the population of the town in 20 years. (Round to the nearest person.)

Explore how to analyze population changes using the exponential decay model P(t) = 14,400e^(-0.051t). Learn to determine if the population is increasing or decreasing, calculate the population in 2010, and predict the population in 20 years. Suitable for students in Grades 10-12.

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