Answer:
7432 bacteria
Step-by-step explanation:
The given equation is an exponential function that describes continuous growth at a rate of k per unit time. To use this equation, we must do some computation to find the value of k.
Alternatively, we can use an exponential function that will let us use directly the values given in the problem.
P = A·b^(t/p)
where A is the initial population, b is the growth factor in period p.
For the values given here, the equation is ...
P = 4000(4400/4000)^(t/2) = 4000(1.1^(t/2))
For t=13, the predicted population is ...
P = 4000(1.1^(13/2)) ≈ 7432.12
The predicted population after 13 hours is 7432 bacteria.
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Additional comment
The value of k in P=Ae^(kt) is ...
k = (1/p)ln(b) = 1/2·ln(1.1) ≈ 0.047655 . . . . . 'b' and 'p' described above
The predicted population has 4 significant figures, so obtaining that value accurately requires a value of k with at least 5 significant figures.
We chose the formula we did so as to (a) obtain better prediction accuracy with digits entered into the calculator, (b) avoid the need to describe and calculate k.
The value of k tells us the equivalent continuous growth rate is about 4.766% per hour.
