Question
a certain population of beetles is increasing at a rate of 22% per month. if there are are currently 350 beetles, the equation f(x) = 350(1.22)x can be used to model the situation. finish the chart and plot points to determine which graph correctly displays the situatio
Answers
GPT-4o mini
Given the information, we can calculate the population of beetles over several months using the equation \( f(x) = 350(1.22)^x \), where \( x \) is the number of months and \( f(x) \) represents the population of beetles after \( x \) months.
Let's calculate the population for the first few months (from 0 to 5 months):
- **Month 0**:
\[
f(0) = 350(1.22)^0 = 350 \cdot 1 = 350
\]
- **Month 1**:
\[
f(1) = 350(1.22)^1 = 350 \cdot 1.22 \approx 427
\]
- **Month 2**:
\[
f(2) = 350(1.22)^2 = 350 \cdot 1.4884 \approx 520
\]
- **Month 3**:
\[
f(3) = 350(1.22)^3 = 350 \cdot 1.810248 \approx 633
\]
- **Month 4**:
\[
f(4) = 350(1.22)^4 = 350 \cdot 2.209 \approx 772
\]
- **Month 5**:
\[
f(5) = 350(1.22)^5 = 350 \cdot 2.6926 \approx 943
\]
Now we have the following data points:
| Month (x) | Population (f(x)) |
|-----------|--------------------|
| 0 | 350 |
| 1 | 427 |
| 2 | 520 |
| 3 | 633 |
| 4 | 772 |
| 5 | 943 |
### Plotting the Points
You can plot these points on a graph with:
- The x-axis representing the number of months (0 to 5).
- The y-axis representing the population of beetles.
### Graphing Guidance
- At \( x = 0 \), \( f(x) = 350 \)
- At \( x = 1 \), \( f(x) \approx 427 \)
- At \( x = 2 \), \( f(x) \approx 520 \)
- At \( x = 3 \), \( f(x) \approx 633 \)
- At \( x = 4 \), \( f(x) \approx 772 \)
- At \( x = 5 \), \( f(x) \approx 943 \)
### Expected Graph Shape
The graph of this data will show exponential growth because the beetle population is increasing at a constant percentage (22% per month). The curve will rise more steeply as time goes on due to the nature of compound interest effects.
You can use graphing software or graph paper to connect the points smoothly, which should create a steeply rising curve indicative of exponential growth. This will help you visually confirm which graph displays the situation correctly. If you have specific graph options to analyze, compare their shapes to the expected trajectory outlined here.
Let's calculate the population for the first few months (from 0 to 5 months):
- **Month 0**:
\[
f(0) = 350(1.22)^0 = 350 \cdot 1 = 350
\]
- **Month 1**:
\[
f(1) = 350(1.22)^1 = 350 \cdot 1.22 \approx 427
\]
- **Month 2**:
\[
f(2) = 350(1.22)^2 = 350 \cdot 1.4884 \approx 520
\]
- **Month 3**:
\[
f(3) = 350(1.22)^3 = 350 \cdot 1.810248 \approx 633
\]
- **Month 4**:
\[
f(4) = 350(1.22)^4 = 350 \cdot 2.209 \approx 772
\]
- **Month 5**:
\[
f(5) = 350(1.22)^5 = 350 \cdot 2.6926 \approx 943
\]
Now we have the following data points:
| Month (x) | Population (f(x)) |
|-----------|--------------------|
| 0 | 350 |
| 1 | 427 |
| 2 | 520 |
| 3 | 633 |
| 4 | 772 |
| 5 | 943 |
### Plotting the Points
You can plot these points on a graph with:
- The x-axis representing the number of months (0 to 5).
- The y-axis representing the population of beetles.
### Graphing Guidance
- At \( x = 0 \), \( f(x) = 350 \)
- At \( x = 1 \), \( f(x) \approx 427 \)
- At \( x = 2 \), \( f(x) \approx 520 \)
- At \( x = 3 \), \( f(x) \approx 633 \)
- At \( x = 4 \), \( f(x) \approx 772 \)
- At \( x = 5 \), \( f(x) \approx 943 \)
### Expected Graph Shape
The graph of this data will show exponential growth because the beetle population is increasing at a constant percentage (22% per month). The curve will rise more steeply as time goes on due to the nature of compound interest effects.
You can use graphing software or graph paper to connect the points smoothly, which should create a steeply rising curve indicative of exponential growth. This will help you visually confirm which graph displays the situation correctly. If you have specific graph options to analyze, compare their shapes to the expected trajectory outlined here.