Math Problem Statement

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?

Solution

The given function for the bacteria population is:

A(t)=185(1.11)tA(t) = 185(1.11)^t

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

A(66)=185(1.11)66A(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.

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Math Problem Analysis

Mathematical Concepts

Exponential Growth
Population Modeling
Algebra

Formulas

A(t) = 185(1.11)^t

Theorems

Exponential Growth Theorem

Suitable Grade Level

Grades 10-12

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