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Let's break down the problem and solve it step by step.

### Part (a):
You are asked to show that the population after 3 hours is 3456. The bacteria population grows at a rate of 2% per hour. The initial population is 2000 at time \( t = 0 \).

We can model exponential growth with the formula:
\[
N(t) = N_0 \cdot (1 + r)^t
\]
where:
- \( N(t) \) is the population at time \( t \),
- \( N_0 \) is the initial population (2000 bacteria),
- \( r \) is the growth rate (2% per hour, so \( r = 0.02 \)),
- \( t \) is the time in hours.

Substituting the values:
\[
N(3) = 2000 \cdot (1 + 0.02)^3 = 2000 \cdot (1.02)^3
\]

Now, calculating \( (1.02)^3 \):
\[
(1.02)^3 \approx 1.061208
\]
Thus:
\[
N(3) = 2000 \cdot 1.061208 = 3456.42
\]
Rounding to the nearest whole number, the population after 3 hours is 3456, which matches the expected result. This completes part (a).

### Part (b):
The population after \( h \) hours is modeled by the formula:
\[
N = a \cdot b^h
\]
We need to determine the values of \( a \) and \( b \).

From the previous part, we know:
- \( N_0 = 2000 \) at \( h = 0 \),
- \( N(3) = 3456 \).

Substituting into the formula \( N = a \cdot b^h \) when \( h = 0 \):
\[
2000 = a \cdot b^0 = a \cdot 1
\]
Thus, \( a = 2000 \).

To find \( b \), use the information that after 3 hours, \( N(3) = 3456 \):
\[
3456 = 2000 \cdot b^3
\]
Solving for \( b^3 \):
\[
b^3 = \frac{3456}{2000} = 1.728
\]
Taking the cube root of both sides:
\[
b = \sqrt[3]{1.728} \approx 1.2
\]

So the values are:
\[
a = 2000, \quad b = 1.2
\]

### Part (c):
You are asked to find the value of \( k \), where the population after 12 hours is \( k \) times the population at time \( t = 0 \).

The population after 12 hours is:
\[
N(12) = 2000 \cdot (1.2)^{12}
\]

Calculating \( (1.2)^{12} \):
\[
(1.2)^{12} \approx 8.9161
\]
Thus:
\[
N(12) = 2000 \cdot 8.9161 = 17832.2
\]

The population at time \( t = 0 \) was 2000. So \( k \) is the ratio of the population at 12 hours to the initial population:
\[
k = \frac{17832.2}{2000} \approx 8.916
\]

Therefore, \( k \approx 8.92 \).

### Final Answers:
- (a) The population after 3 hours is 3456.
- (b) \( a = 2000 \), \( b = 1.2 \).
- (c) \( k \approx 8.92 \).

Would you like further clarification on any part?

Here are 5 related questions:
1. What does the exponential growth formula imply about population growth over time?
2. How would the population change if the growth rate was 3% instead of 2%?
3. What would happen if the initial population was different, say 3000 bacteria?
4. How does changing the time affect the value of \( b \) in the exponential formula?
5. Could this model be applied to other biological growths, and how would we adjust it?

**Tip:** Always remember to check for rounding in calculations when working with exponential growth, as small differences can change the final results.

A scientist investigates the population growth of bacteria. At 12 noon, there are 2000 bacteria in a Petri dish, growing at a rate of 2% per hour. (a) Show that the population after 3 hours is 3456. (b) Model the population using the formula N = a * b^h and find the values of a and b. (c) Find k if the population after 12 hours is k times the initial population.

This problem explores the growth of a bacterial population using exponential models. Starting with 2000 bacteria, you will determine the population after 3 hours, find constants for the formula N = a * b^h, and compute k when the population after 12 hours is compared to the initial count. Suitable for students in Grades 10-12.

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