Question
A certain population of beetles is increasing at a rate of 22% per month. If there are currently
350 beetles, the equation f() = 350(1.22)= can be used to model the situation. Finish the
chart and plot the points to determine which graph correctly displays the situation. (1 point)
350 beetles, the equation f() = 350(1.22)= can be used to model the situation. Finish the
chart and plot the points to determine which graph correctly displays the situation. (1 point)
Answers
GPT-4o mini
To model the growth of the population of beetles, we can use the equation for exponential growth. Given that the population increases at a rate of 22% per month, the equation should actually be:
\[
f(t) = 350(1.22)^t
\]
where \( f(t) \) represents the population after \( t \) months and 350 is the initial population.
Now, let's calculate the population of beetles for several months. We can fill in a chart with values for \( t \) (number of months) and \( f(t) \) (population of beetles).
| Month (t) | f(t) = 350(1.22)^t |
|-----------|--------------------|
| 0 | 350 * (1.22)^0 = 350 |
| 1 | 350 * (1.22)^1 = 350 * 1.22 ≈ 427 |
| 2 | 350 * (1.22)^2 = 350 * 1.4884 ≈ 520 |
| 3 | 350 * (1.22)^3 = 350 * 1.8105 ≈ 633 |
| 4 | 350 * (1.22)^4 = 350 * 2.2046 ≈ 771 |
| 5 | 350 * (1.22)^5 = 350 * 2.6885 ≈ 941 |
Now, here are the approximate values:
- For \( t = 0 \) : \( f(0) \approx 350 \)
- For \( t = 1 \) : \( f(1) \approx 427 \)
- For \( t = 2 \) : \( f(2) \approx 520 \)
- For \( t = 3 \) : \( f(3) \approx 633 \)
- For \( t = 4 \) : \( f(4) \approx 771 \)
- For \( t = 5 \) : \( f(5) \approx 941 \)
To plot these points on a graph, the \( x \)-axis will represent the time in months (0 to 5), and the \( y \)-axis will represent the population of beetles. You would plot the following points:
1. (0, 350)
2. (1, 427)
3. (2, 520)
4. (3, 633)
5. (4, 771)
6. (5, 941)
After plotting these points, you will observe an exponential curve indicative of the 22% growth rate. Look for a graph that reflects this upward, accelerating trend. If it’s available, choose the graph that best matches this growth behavior.
\[
f(t) = 350(1.22)^t
\]
where \( f(t) \) represents the population after \( t \) months and 350 is the initial population.
Now, let's calculate the population of beetles for several months. We can fill in a chart with values for \( t \) (number of months) and \( f(t) \) (population of beetles).
| Month (t) | f(t) = 350(1.22)^t |
|-----------|--------------------|
| 0 | 350 * (1.22)^0 = 350 |
| 1 | 350 * (1.22)^1 = 350 * 1.22 ≈ 427 |
| 2 | 350 * (1.22)^2 = 350 * 1.4884 ≈ 520 |
| 3 | 350 * (1.22)^3 = 350 * 1.8105 ≈ 633 |
| 4 | 350 * (1.22)^4 = 350 * 2.2046 ≈ 771 |
| 5 | 350 * (1.22)^5 = 350 * 2.6885 ≈ 941 |
Now, here are the approximate values:
- For \( t = 0 \) : \( f(0) \approx 350 \)
- For \( t = 1 \) : \( f(1) \approx 427 \)
- For \( t = 2 \) : \( f(2) \approx 520 \)
- For \( t = 3 \) : \( f(3) \approx 633 \)
- For \( t = 4 \) : \( f(4) \approx 771 \)
- For \( t = 5 \) : \( f(5) \approx 941 \)
To plot these points on a graph, the \( x \)-axis will represent the time in months (0 to 5), and the \( y \)-axis will represent the population of beetles. You would plot the following points:
1. (0, 350)
2. (1, 427)
3. (2, 520)
4. (3, 633)
5. (4, 771)
6. (5, 941)
After plotting these points, you will observe an exponential curve indicative of the 22% growth rate. Look for a graph that reflects this upward, accelerating trend. If it’s available, choose the graph that best matches this growth behavior.