Having completed your first few tasks, you move now to a portion of the lab where important bacteria are being grown. Based on the data collected, you have determined that the population of this bacteria in the range measured can be modeled as A(t)=185(1.11)tAt=1851.11t, where tt is given in days. To the nearest whole number, what will the bacteria population be after 66 days?

The given function for the bacteria population is:

\[
A(t) = 185(1.11)^t
\]

where \( t \) is the time in days. We are asked to find the population after 66 days, so we substitute \( t = 66 \) into the equation:

\[
A(66) = 185(1.11)^{66}
\]

Now, let's calculate that.The bacteria population after 66 days will be approximately **181,338**.

Do you have any questions or need further details?

Here are 5 related questions to explore further:
1. What would the bacteria population be after 30 days?
2. How long will it take for the bacteria population to double?
3. If the growth rate changes to 1.15, what will the population be after 50 days?
4. What is the continuous growth rate equivalent to a discrete rate of 1.11?
5. How would the population change if the initial population was 250 instead of 185?

**Tip:** Exponential growth quickly escalates, so even a small increase in the growth rate can have a significant impact over time.

Having completed your first few tasks, you move now to a portion of the lab where important bacteria are being grown. Based on the data collected, you have determined that the population of this bacteria in the range measured can be modeled as A(t)=185(1.11)^t, where t is given in days. To the nearest whole number, what will the bacteria population be after 66 days?

Calculate the bacteria population after 66 days using the exponential growth model A(t) = 185(1.11)^t. Learn about exponential growth and its impact on population modeling. Suitable for high school students in Grades 10-12.

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