https://file.aiquickdraw.com/mathbot/file/1730224711051cvk4i1vl.jpg

The equation given is:

\[
P = 0.0089t^2 + 1.1149t + 78.4491
\]

where \( P \) is the population in millions, and \( t \) is the number of years after 1900.

To estimate the population in 2025, we calculate \( t \), which is the number of years since 1900:

\[
t = 2025 - 1900 = 125
\]

Now, substitute \( t = 125 \) into the equation for \( P \):

\[
P = 0.0089(125)^2 + 1.1149(125) + 78.4491
\]

First, calculate \( 125^2 \):

\[
125^2 = 15625
\]

Now substitute:

\[
P = 0.0089(15625) + 1.1149(125) + 78.4491
\]

Calculate each term:

\[
0.0089 \times 15625 = 139.0625
\]
\[
1.1149 \times 125 = 139.3625
\]

Now, sum these terms with the constant:

\[
P = 139.0625 + 139.3625 + 78.4491 = 356.8741
\]

So, the estimated population in 2025 is approximately **356.9 million**.

Would you like further details or have any questions?

Here are five related questions to consider:
1. How would the population estimate change if we used a different year, say 2050?
2. What effect does the \( t^2 \) term have on the population growth over time?
3. How would the model change if we considered a more recent equation with different coefficients?
4. Could you explain how to graph this quadratic equation and what its curve might represent?
5. How accurate do you think this model is for predicting population, and what factors might affect its precision?

**Tip:** When using models, always check if they apply to the range you're interested in, as projections beyond the model's dataset might lead to inaccurate results.

The equation P = 0.0089t^2 + 1.1149t + 78.4491 models the United States population, P, in millions since 1900. If t represents the number of years after 1900, then what is the estimated population in 2025 to the nearest tenth of a million?

This problem involves estimating the United States population in the year 2025 using a quadratic model, P = 0.0089t^2 + 1.1149t + 78.4491, where P is the population in millions and t represents years since 1900. Detailed solution shows how to substitute and solve for P.

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