9. Suppose an experimental population of amoeba increases according to the law of exponential growth. There were 100 amoeba after the second day of the experiment and 300 amoeba after the fourth day. Approximately how many amoeba were in the original sample?
A. 5
B. 33
C. 71
D. 10
E. Not enough information to determine
Since the population tripled in 2 days, the population after t days is
p = (100/3)*3^(t/2)
To answer this question, we can use the formula for exponential growth:
N = N0 * (1 + r)^t
where:
- N is the final population size
- N0 is the initial population size
- r is the growth rate (expressed as a decimal)
- t is the time period
We know that after the second day (t = 2), the population is 100, and after the fourth day (t = 4), the population is 300. We need to determine N0, the initial population size.
Let's start by plugging in the values we have:
100 = N0 * (1 + r)^2
300 = N0 * (1 + r)^4
Divide the second equation by the first equation to eliminate N0:
300/100 = [(N0 * (1 + r)^4)] / [(N0 * (1 + r)^2)]
3 = (1 + r)^2
√3 = 1 + r
Solve for r:
√3 - 1 = r
Now, substitute the value of r back into any of the original equations. Let's use the first equation:
100 = N0 * (√3)^2
100 = N0 * 3
N0 = 100/3
So, the approximate number of amoeba in the original sample is 100/3, which is approximately 33.
Therefore, the answer is B. 33.