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The problem states that the initial bacterial population is 5000 at time \( t = 0 \), and the rate of growth is given by \( 1,000 \cdot 9^t \) bacteria per hour, where \( t \) is in hours. The task is to determine the population after one hour, rounding the result to the nearest whole number.

### Solution:

The population after one hour can be found by integrating the rate of growth from \( t = 0 \) to \( t = 1 \), since the growth rate is not constant but changes over time.

1. Initial population: \( P(0) = 5000 \)
2. Rate of growth: \( \frac{dP}{dt} = 1,000 \cdot 9^t \)

The total population after 1 hour can be found by:
\[
P(1) = P(0) + \int_0^1 1,000 \cdot 9^t \, dt
\]

Let's calculate this integral.The population of the bacteria after one hour is approximately 8,641 when rounded to the nearest whole number.

Would you like further details or explanations on how this was computed? 

Here are 5 related questions you might be interested in:
1. What is the exact formula for the population at any time \( t \)?
2. How does the growth rate change over time for this bacterial population?
3. What happens to the population after two hours?
4. Can you explain the steps involved in integrating the growth rate function?
5. How would the population change if the initial growth rate was different?

**Tip:** When dealing with growth problems, identifying the rate function and initial conditions is key to setting up and solving the differential equation.

A bacterial population is 5000 at time t = 0, and its rate of growth is 1,000 * 9^t bacteria per hour at time t hours. What is the population after one hour? (Round your answer to the nearest whole number.)

Learn how to calculate the bacterial population after 1 hour given an initial population of 5000 and a growth rate of 1,000 * 9^t bacteria per hour. This problem uses exponential growth and integration techniques suitable for high school students.

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