- Analyze the data and observe that \(n\) increases as \(t\) increases, suggesting an exponential model.
- Test the candidate model \(n = 600(1.2)^t\) by substituting the given values of \(t\).
- Compare calculated results with table values; the model's output closely aligns with the provided estimates.
- Conclude that the best fit is model (d).
- Final Answer: \(\boxed{d}\)
Explanation
- Data and Problem Analysis
The provided table shows the relationship between time \(t\) in days and the estimated population of organisms \(n\):
- \(t=2, n=860\)
- \(t=4, n=1250\)
- \(t=6, n=1800\)
- \(t=8, n=2600\)
- \(t=10, n=3720\)
We observe that as time \(t\) increases, the number of organisms \(n\) grows significantly. We are asked to choose the model that best fits this data from four given options:
A\(t = 600(1.2)^n\)
B\(n = 1.2(600)^t\)
C\(t = 1.2(600)^n\)
D\(n = 600(1.2)^t\)
- Testing the Proposed Models
We test the models by substituting the given values of \(t\) and checking if the resulting \(n\) matches the data provided in the table.
Consider model d: \(n = 600(1.2)^t\)
- For \(t=2\): \(n = 600(1.2)^2 = 600(1.44) = 864 \approx 860\)
- For \(t=4\): \(n = 600(1.2)^4 = 600(2.0736) = 1244.16 \approx 1250\)
- For \(t=6\): \(n = 600(1.2)^6 = 600(2.985984) = 1791.59 \approx 1800\)
- For \(t=8\): \(n = 600(1.2)^8 = 600(4.29981696) = 2579.89 \approx 2600\)
- For \(t=10\): \(n = 600(1.2)^{10} = 600(6.191736) = 3715.04 \approx 3720\)
The values calculated using model (d) are very close to the values provided in the table, confirming it is an excellent fit for the population growth over time.
- Evaluating Alternatives and Conclusion
Evaluating the other options:
- Option (b) \(n = 1.2(600)^t\) would grow extremely fast for even small values of \(t\). For example, at \(t=2\), \(n = 1.2(600^2) = 1.2(360,000) = 432,000\), which is nowhere near the recorded value of 860.
- Options (a) and (c) suggest that \(t\) is a function of \(n\) in an exponential manner, which would mean that as \(n\) increases, \(t\) increases exponentially. This is the inverse of the observed data where \(n\) increases as \(t\) increases, and the provided structures are not consistent with the data behavior.
Therefore, model (d) is the only reasonable choice among the options provided.
Examples
Exponential growth models like \(n = ab^t\) are frequently used in biology and ecology to estimate population growth of bacteria, plants, or animals under ideal conditions. For instance, if you are monitoring the growth of a yeast colony in a laboratory setting, you can use collected measurements of the count over time to find the parameters \(a\) (initial population) and \(b\) (growth factor). This allows scientists to make accurate predictions about future population sizes without having to count every single organism daily.