A certain type of bacteria increases at a steady rate of 20% per hour, if there were 100 bacteria at t = 0, how many bacteria would there after seven hours??

The growth of bacteria follows the formula for exponential growth:

\[
N(t) = N_0 \times (1 + r)^t
\]

Where:
- \( N(t) \) is the population after time \( t \),
- \( N_0 \) is the initial population,
- \( r \) is the growth rate (as a decimal),
- \( t \) is the time in hours.

In this case:
- \( N_0 = 100 \),
- \( r = 0.20 \) (20% growth rate per hour),
- \( t = 7 \) hours.

Now, plugging the values into the formula:

\[
N(7) = 100 \times (1 + 0.20)^7
\]

Let's calculate that.

\[
N(7) = 100 \times (1.20)^7
\]

I'll compute the value.After 7 hours, the bacteria population will have grown to approximately 358.

Would you like further details or have any questions?

Here are 5 related questions to expand on this topic:
1. How would the population change if the growth rate were 30% instead of 20%?
2. What would be the population after 10 hours at the same growth rate?
3. How can you calculate the time it takes for the population to double?
4. What happens if the growth rate decreases over time?
5. How would you model bacteria growth if the rate changes each hour?

**Tip:** In exponential growth, small changes in the growth rate can lead to significant differences over time!

A certain type of bacteria increases at a steady rate of 20% per hour, if there were 100 bacteria at t = 0, how many bacteria would there be after seven hours?

Learn how to calculate the population of bacteria growing at a steady rate of 20% per hour after 7 hours, using the exponential growth formula N(t) = N_0 × (1 + r)^t. This problem covers key concepts in algebra and exponential functions, suitable for students in Grades 9-12.

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