The population of the village of Lilianaville, Wyoming has been observed to grow exponentially. The table shows the population for a six year period beginning in 2006.

Year	2006	2007	2008	2009	2010	2011
Population	841	1444	2631	4132	7386	12926


Find an exponential equation that best fits the data above. The equation has form 
 where 
 is years since 2005. Use curve-fitting tools to determine the values of 
 and 
.

a = 
2278.6285
Incorrect

b = 
-3081.8666
Incorrect

Predict the size of Lilianaville, Wyoming's population in the year 2015 if the trend continues as shown. Round your final answer to the nearest whole number, but take care to avoid excess rounding at intermediate steps.



Given the population growth data for Lilianaville, Wyoming, we are asked to find an exponential equation of the form:

\[
P(t) = P_0 \cdot e^{bt}
\]

Where:
- \( P(t) \) represents the population at year \( t \),
- \( t \) is the number of years since 2005,
- \( P_0 \) is the initial population at \( t = 0 \) (which corresponds to the population in 2006),
- \( b \) is the exponential growth rate constant.

### Step 1: Rewriting the years
We will rewrite the years based on the number of years since 2005:

\[
\text{Year 2006} \implies t = 1, \quad \text{Population} = 841
\]
\[
\text{Year 2007} \implies t = 2, \quad \text{Population} = 1444
\]
\[
\text{Year 2008} \implies t = 3, \quad \text{Population} = 2631
\]
\[
\text{Year 2009} \implies t = 4, \quad \text{Population} = 4132
\]
\[
\text{Year 2010} \implies t = 5, \quad \text{Population} = 7386
\]
\[
\text{Year 2011} \implies t = 6, \quad \text{Population} = 12926
\]

### Step 2: Using exponential curve fitting
Using the data points, we will fit an exponential curve to determine the values of \( P_0 \) and \( b \). I will now calculate the curve fit.

Let's compute the values for \( P_0 \) and \( b \).The exponential equation that best fits the data is:

\[
P(t) = 469.916 \cdot e^{0.552t}
\]

Where:
- \( P_0 = 469.916 \)
- \( b = 0.552 \)

### Step 3: Predicting the population in 2015

To predict the population in 2015, we set \( t = 10 \) (since 2015 is 10 years after 2005). We will now calculate the population for the year 2015 using the fitted equation.

\[
P(10) = 469.916 \cdot e^{0.552 \cdot 10}
\]It seems like I can’t do more advanced data analysis right now. Please try again later.

Would you like to know more details about how to solve this problem manually, or if you have any other questions, feel free to ask! Here are 5 related questions you might find interesting:

1. How do you interpret the constant \( P_0 \) in an exponential growth equation?
2. What are the limitations of using exponential models for population growth?
3. How would the equation change if the population growth rate decreased over time?
4. What would be the impact of rounding \( P_0 \) or \( b \) too early in the calculation?
5. How can you use logarithms to solve exponential growth problems?

**Tip:** When solving exponential problems, avoid rounding constants too early, as this can affect the accuracy of your final answer.

The population of the village of Lilianaville, Wyoming has been observed to grow exponentially. The table shows the population for a six year period beginning in 2006.

Year 2006 2007 2008 2009 2010 2011
Population 841 1444 2631 4132 7386 12926

Find an exponential equation that best fits the data above. The equation has form where is years since 2005. Use curve-fitting tools to determine the values of and .

Predict the size of Lilianaville, Wyoming's population in the year 2015 if the trend continues as shown. Round your final answer to the nearest whole number, but take care to avoid excess rounding at intermediate steps.

Discover how to find an exponential equation that best fits the population growth of Lilianaville, Wyoming, from 2006 to 2011, using the formula P(t) = P0 * e^(bt). Predict the population for 2015 based on the trend. Learn key concepts like exponential growth and curve fitting, suitable for students in Grades 9-12.

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