Below is the graph of the boa constrictor population. The x-axis is t, the time in years, and the y-axis is P(t), the snake popu

Question

2 months ago

9

Below is the graph of the boa constrictor population. The x-axis is t, the time in years, and the y-axis is P(t), the snake popu

lation for any given year. In this case, t = 0 indicates the year they started keeping track of the boa population. Scientists want to know how the snake population is changing every two years so they can make predictions about the next two years.
Students' Conjectures:

Two students, Sarah and Ernest, agree that the graph is exponential but disagree on how the rate changes between successive time periods.

1. Complete the table below to summarize each classmate's conjecture. (2 points: 1 point for each row of the chart)
Classmate Conjecture
Ernest

Sarah

2. Using your knowledge of exponential functions, who do you think is correct? (1 point)

Analyzing the Data:

3. Is this graph increasing or decreasing? Does this imply that the boa constrictor population is growing or diminishing? (1 point)

4. Looking at this graph, would you agree that the boa constrictor population could be a problem? (1 point)

5. What is the horizontal asymptote of this graph? (1 point)

6. What is the range? Explain. (1 point)

7. What is the y-intercept? What does this mean in terms of the snake population? (1 point)

Here is the function for this graph, where t is the time in years: P(t) = 5e0.89t.

8. Use this formula to estimate the boa constrictor population in 2 years, 4 years, and 6 years. (6 points: 2 points for each year, including 1 point for showing your work and 1 point for the answer)
t: Time in years P(t): Estimated snake population at time t
2

4

6

9. Find the rate of change for the snake population from year 0 to year 2. (2 points)

10. Apply the rate of change you found in question 9 in order to estimate the snake population for year 4. How does your answer compare with the calculated value? (1 point)

11. If there are approximately 36,660 snakes (rounded to the nearest 10) in year 10, how many will there be in year 12? (1 point)

Making a Decision:

12. Given your calculations above, is Ernest's or Sarah's conjecture correct? (2 points)

Mathematics

1 answer:

zzz [12.3K] · Answer 1 · 2026-03-17T09:24:21+03:00

I will address question number 3 since questions 1 and 2 are missing. 3. The graph is on an upward trend, indicating that P(t) values rise as time progresses. This leads us to conclude that the boa constrictor population is indeed increasing. 4. Based on the graph, it's reasonable to say that the growing boa constrictor population might pose significant issues, particularly since it grew from P(t)=7 at t=0 to P(t)=176 at t=4 years. 5. The graph's horizontal asymptote stands at P(t)=7. 6. Given that P(t)=7 at t=0 and the population continues to rise, the range of the graph is represented as [7,∞). 7. The y-intercept occurs at t=0, thus P(t)=7 at that point, which translates to a y-intercept of (0,7). 8. According to the provided formula, the estimated boa population intervals will be as follows: i. At t=2 years, using the formula P(t)=5e^(0.89t), and substituting t=2, we find P(2)=5e^(1.78)≈30. ii. For t=4 years, substituting t=4 gives P(4)=5e^(3.56)≈176. iii. At t=6 years, substituting t=6 yields P(6)=5e^(5.34)≈1043. The rate of population change from t=0 to t=2 is found using the formula [P(2)-P(0)]/(2-0). Here, P(0)=5e^(0)=5, and P(2)≈30, so (30-5)/2=12.5 results in a rate of change of 12.5. 10. Applying this rate, at t=4 years, we'd estimate P=12.5*4=50, which is lower than the calculated result because our assumption models linear growth rather than the actual exponential increase. 11. Assuming there is about 36,660 snakes in year 10, for year 12 we use P(12)=5e^(10.68), yielding approximately P(12)=217,388, rounded to the nearest 10.