A strain of bacteria growing on your desktop doubles every five minutes. Assuming that you start with only one bacterium, how many bacteria could be present at the end of 96 minutes?

To solve this problem, we need to understand that the bacteria double every 5 minutes, starting with one bacterium. This type of growth is exponential, and the formula for exponential growth can be written as:

\[
N(t) = N_0 \times 2^{t / d}
\]

Where:
- \(N(t)\) is the number of bacteria at time \(t\),
- \(N_0\) is the initial number of bacteria,
- \(t\) is the total time in minutes,
- \(d\) is the doubling time (in this case, 5 minutes),
- \(2^{t / d}\) represents the number of doublings.

### Given:
- \(N_0 = 1\) (starting with one bacterium),
- \(t = 96\) minutes,
- \(d = 5\) minutes.

Now, let's calculate how many times the bacteria double in 96 minutes:

\[
\text{Number of doublings} = \frac{96}{5} = 19.2
\]

Since the number of doublings must be a whole number, we round it down to 19 (as you can't have a fraction of a doubling).

Now apply the formula:

\[
N(96) = 1 \times 2^{19} = 524,288 \text{ bacteria}.
\]

So, at the end of 96 minutes, there could be **524,288 bacteria**.

Do you want further details on exponential growth? 

Here are 5 related questions:
1. What would happen if the doubling time was changed to 3 minutes?
2. How does exponential growth differ from linear growth?
3. If the bacteria died at a certain rate, how would that affect the calculation?
4. How would you model this growth if the initial bacteria count was 100 instead of 1?
5. What is the total number of doublings after 120 minutes?

**Tip:** In exponential growth, small changes in time can lead to massive changes in population.

Explore how a strain of bacteria doubles every 5 minutes, starting from one bacterium. Learn to calculate the number of bacteria after 96 minutes using exponential growth. The solution uses the formula N(t) = N_0 * 2^(t/d), suitable for students in Grades 8-10.

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