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The given exponential model is:

\[
A = 72.9 \cdot e^{0.008t}
\]

where \( A \) is the population in millions, and \( t \) represents the number of years after 2003. You are tasked with determining the population in 2003, which corresponds to \( t = 0 \) (since it's the reference year).

Substitute \( t = 0 \) into the equation:

\[
A = 72.9 \cdot e^{0.008 \cdot 0}
\]

This simplifies to:

\[
A = 72.9 \cdot e^0 = 72.9 \cdot 1 = 72.9
\]

Thus, the population of the country in 2003 was **72.9 million**.

Would you like further clarification on this? Here are some related questions:

1. How does changing the growth rate \( 0.008 \) affect the population over time?
2. What would the population be in 2010, seven years after 2003?
3. How can you use logarithms to solve for \( t \) when the population reaches a certain value?
4. How does the base \( e \) function relate to real-world exponential growth models?
5. What assumptions are made when using an exponential growth model?

**Tip:** In exponential models, the rate of growth or decay is constant in relative terms (percentage-wise), not in absolute terms. This is why populations modeled by exponential equations grow faster over time.

The exponential model A = 72.9e^{0.008t} describes the population, A, of a country in millions, t years after 2003. Use the model to determine the population of the country in 2003.

Use the exponential growth model A = 72.9e^{0.008t} to calculate the population of a country in 2003. This problem demonstrates the use of exponential functions to model real-world population growth scenarios. Suitable for Grades 9-12.

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