Use the table to answer the question.

0 350
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A certain population of beetles is increasing at a rate of 22% per month. If there are currently 350 beetles, the equation f(x)=350(1.22)x
can be used to model the situation. Finish the chart and plot the points to determine which graph correctly displays the situation.(1 point)
Responses

A decreasing curve with an arrow at the end is plotted on a coordinate plane. The x-axis for months ranges from 0 to 10 in increments of 1. The y-axis for Beetles ranges from 0 to 350 in increments of 50.
Image with alt text: A decreasing curve with an arrow at the end is plotted on a coordinate plane. The x-axis for months ranges from 0 to 10 in increments of 1. The y-axis for Beetles ranges from 0 to 350 in increments of 50.

An increasing curve with an arrow on the top is plotted on a coordinate plane. The x-axis for months ranges from 0 to 10 in increments of 1. The y-axis for Beetles ranges from 0 to 2000 in increments of 250.
Image with alt text: An increasing curve with an arrow on the top is plotted on a coordinate plane. The x-axis for months ranges from 0 to 10 in increments of 1. The y-axis for Beetles ranges from 0 to 2000 in increments of 250.

A decreasing curve with an arrow at the end is plotted on a coordinate plane. The x-axis for months ranges from 0 to 10 in increments of 1. The y-axis for Beetles ranges from 0 to 350 in increments of 50.
Image with alt text: A decreasing curve with an arrow at the end is plotted on a coordinate plane. The x-axis for months ranges from 0 to 10 in increments of 1. The y-axis for Beetles ranges from 0 to 350 in increments of 50.

An increasing curve with an arrow on the top is plotted on a coordinate plane. The x-axis for months ranges from 0 to 2,000 in increments of 200. The y-axis for Beetles ranges from 0 to 10 in increments of 1.
Image with alt text: An increasing curve with an arrow on the top is plotted on a coordinate plane. The x-axis for months ranges from 0 to 2,000 in increments of 200. The y-axis for Beetles ranges from 0 to 10 in increments of 1.

1 answer

To complete the table and determine the correct graph for the beetle population modeled by the equation \( f(x) = 350(1.22)^x \), we first need to calculate the number of beetles for each month from 0 to 5.

  1. At \( x = 0 \): \[ f(0) = 350(1.22)^0 = 350(1) = 350 \]

  2. At \( x = 1 \): \[ f(1) = 350(1.22)^1 = 350 \times 1.22 \approx 427 \]

  3. At \( x = 2 \): \[ f(2) = 350(1.22)^2 = 350 \times 1.4884 \approx 521 \]

  4. At \( x = 3 \): \[ f(3) = 350(1.22)^3 = 350 \times 1.8155 \approx 636 \]

  5. At \( x = 4 \): \[ f(4) = 350(1.22)^4 = 350 \times 2.2136 \approx 775 \]

  6. At \( x = 5 \): \[ f(5) = 350(1.22)^5 = 350 \times 2.6950 \approx 943 \]

Now, let's summarize the calculated values in the table format:

| Month (x) | Number of Beetles (f(x)) | |-----------|---------------------------| | 0 | 350 | | 1 | 427 | | 2 | 521 | | 3 | 636 | | 4 | 775 | | 5 | 943 |

Next, we have the following points to plot:

  • (0, 350)
  • (1, 427)
  • (2, 521)
  • (3, 636)
  • (4, 775)
  • (5, 943)

Since the population of beetles is increasing exponentially (due to the 22% increase every month), we look for a graph that represents an increasing curve. The only correct options for this that fit the description provided would be the "increasing curve" graphs.

Therefore, the correct graph will be the one that shows an increasing curve that corresponds to values increasing well over 350 and moving towards values like 943 and generally beyond that within the range laid out.

Given the descriptions you provided, the appropriate graph is:

  • An increasing curve with an arrow on the top is plotted on a coordinate plane. The x-axis for months ranges from 0 to 10 in increments of 1. The y-axis for Beetles ranges from 0 to 2000 in increments of 250.

This correctly reflects the exponential growth of the beetle population over time as per the model provided.

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