To find out the population of the bacteria after 978 minutes, we can use the concept of exponential growth due to doubling.
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Determine the number of doubling periods in 978 minutes: The time taken for one doubling is 163 minutes. To find out how many times the bacteria will double in 978 minutes, we divide 978 by 163:
\[ \text{Number of doublings} = \frac{978}{163} \approx 6 \]
We can confirm this is approximately 6 as:
\[ 163 \times 6 = 978 \quad \text{(exactly)} \]
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Calculate the population after 6 doublings: The population doubles every time, so the formula for the population after \( n \) doublings is:
\[ P = P_0 \times 2^n \]
where:
- \( P_0 = 55,700 \) (initial population),
- \( n = 6 \) (number of doublings).
Now substituting these values:
\[ P = 55,700 \times 2^6 \]
We first calculate \( 2^6 \):
\[ 2^6 = 64 \]
Now, substitute this back into the equation:
\[ P = 55,700 \times 64 \]
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Perform the multiplication:
\[ P = 55,700 \times 64 = 3,564,800 \]
Therefore, the population of the bacteria after 978 minutes will be 3,564,800.
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