Question
A certain population of beetles is increasing at a rate of 22% per month if they are currently if there are currently 350 Beatles the equation f of x = 350 * 1.22x can be used to model the situation finish the chart and plot the points to determine which graph correctly displays the situation
Answers
GPT-4o mini
To model the beetle population with the equation \( f(x) = 350 \times 1.22^x \), where \( x \) is the number of months and \( f(x) \) is the population at that time, we can calculate the population for a series of months.
Here is a chart for the population over the first few months:
| Month (x) | Population (f(x)) |
|-----------|----------------------------|
| 0 | \( 350 \times 1.22^0 = 350 \) |
| 1 | \( 350 \times 1.22^1 \approx 427 \) |
| 2 | \( 350 \times 1.22^2 \approx 520 \) |
| 3 | \( 350 \times 1.22^3 \approx 635 \) |
| 4 | \( 350 \times 1.22^4 \approx 774 \) |
| 5 | \( 350 \times 1.22^5 \approx 946 \) |
Now we compute each population value:
1. For \( x = 0 \):
\[
f(0) = 350 \times 1.22^0 = 350
\]
2. For \( x = 1 \):
\[
f(1) = 350 \times 1.22^1 \approx 350 \times 1.22 = 427
\]
3. For \( x = 2 \):
\[
f(2) = 350 \times 1.22^2 \approx 350 \times 1.4884 \approx 520
\]
4. For \( x = 3 \):
\[
f(3) = 350 \times 1.22^3 \approx 350 \times 1.8158 \approx 635
\]
5. For \( x = 4 \):
\[
f(4) = 350 \times 1.22^4 \approx 350 \times 2.2196 \approx 774
\]
6. For \( x = 5 \):
\[
f(5) = 350 \times 1.22^5 \approx 350 \times 2.8490 \approx 996
\]
Now, plotting these points on a graph:
- (0, 350)
- (1, 427)
- (2, 520)
- (3, 635)
- (4, 774)
- (5, 946)
To determine which graph correctly displays this situation, look for a graph that shows an exponential increase starting from the point (0, 350) and rising steeply as \( x \) increases. The growth rate should show a consistent upward curve indicating rapid growth due to the 22% increase per month.
When you have the options, select the graph that matches this exponential growth pattern!
Here is a chart for the population over the first few months:
| Month (x) | Population (f(x)) |
|-----------|----------------------------|
| 0 | \( 350 \times 1.22^0 = 350 \) |
| 1 | \( 350 \times 1.22^1 \approx 427 \) |
| 2 | \( 350 \times 1.22^2 \approx 520 \) |
| 3 | \( 350 \times 1.22^3 \approx 635 \) |
| 4 | \( 350 \times 1.22^4 \approx 774 \) |
| 5 | \( 350 \times 1.22^5 \approx 946 \) |
Now we compute each population value:
1. For \( x = 0 \):
\[
f(0) = 350 \times 1.22^0 = 350
\]
2. For \( x = 1 \):
\[
f(1) = 350 \times 1.22^1 \approx 350 \times 1.22 = 427
\]
3. For \( x = 2 \):
\[
f(2) = 350 \times 1.22^2 \approx 350 \times 1.4884 \approx 520
\]
4. For \( x = 3 \):
\[
f(3) = 350 \times 1.22^3 \approx 350 \times 1.8158 \approx 635
\]
5. For \( x = 4 \):
\[
f(4) = 350 \times 1.22^4 \approx 350 \times 2.2196 \approx 774
\]
6. For \( x = 5 \):
\[
f(5) = 350 \times 1.22^5 \approx 350 \times 2.8490 \approx 996
\]
Now, plotting these points on a graph:
- (0, 350)
- (1, 427)
- (2, 520)
- (3, 635)
- (4, 774)
- (5, 946)
To determine which graph correctly displays this situation, look for a graph that shows an exponential increase starting from the point (0, 350) and rising steeply as \( x \) increases. The growth rate should show a consistent upward curve indicating rapid growth due to the 22% increase per month.
When you have the options, select the graph that matches this exponential growth pattern!