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An arcade manager finds that revenue, R, based on a per-game fee, ffor unlimited play can be modeled by the function R = - 480f...Asked by 54t4n45
                An arcade manager finds that revenue, R, based on a per-game fee, f, for unlimited play can be modeled by the function R=−480f2+3,120f
. Which of the following correctly interprets the average rate of change in revenue if the per-game fee increased from $1 to $3?(1 point)
Responses
Revenue would increase by an average of $1,200 per $1 increase in the per-game fee.
Revenue would increase by an average of $1,200 per $1 increase in the per-game fee.
Revenue would decrease by an average of $1,200 per $1 increase in the per-game fee.
Revenue would decrease by an average of $1,200 per $1 increase in the per-game fee.
The per-game fee would increase by an average of $1,200 per $1 increase in the revenue.
The per-game fee would increase by an average of $1,200 per $1 increase in the revenue.
The per-game fee would decrease by an average of $1,200 per $1 increase in the revenue.
            
        . Which of the following correctly interprets the average rate of change in revenue if the per-game fee increased from $1 to $3?(1 point)
Responses
Revenue would increase by an average of $1,200 per $1 increase in the per-game fee.
Revenue would increase by an average of $1,200 per $1 increase in the per-game fee.
Revenue would decrease by an average of $1,200 per $1 increase in the per-game fee.
Revenue would decrease by an average of $1,200 per $1 increase in the per-game fee.
The per-game fee would increase by an average of $1,200 per $1 increase in the revenue.
The per-game fee would increase by an average of $1,200 per $1 increase in the revenue.
The per-game fee would decrease by an average of $1,200 per $1 increase in the revenue.
Answers
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                    Answered by
            GPT-4o mini
            
    To find the average rate of change in revenue, \( R \), when the per-game fee, \( f \), increases from $1 to $3, we can evaluate the function \( R = -480f^2 + 3120f \) at these two points.
1. First, calculate \( R \) at \( f = 1 \):
\[
R(1) = -480(1^2) + 3120(1) = -480 + 3120 = 2640
\]
2. Next, calculate \( R \) at \( f = 3 \):
\[
R(3) = -480(3^2) + 3120(3) = -480(9) + 9360 = -4320 + 9360 = 5040
\]
3. Now, calculate the change in revenue as the fee changes from $1 to $3:
\[
\Delta R = R(3) - R(1) = 5040 - 2640 = 2400
\]
4. The change in the per-game fee is:
\[
\Delta f = 3 - 1 = 2
\]
5. The average rate of change in revenue per $1 increase in the per-game fee can be found by dividing the change in revenue by the change in the per-game fee. Since the change in the fee is $2, the average increase in revenue per $1 increase in fee is:
\[
\text{Average rate of change} = \frac{\Delta R}{\Delta f} = \frac{2400}{2} = 1200
\]
Thus, the average rate of change in revenue is **$1,200 per $1 increase in the per-game fee**.
Therefore, the correct interpretation is:
**Revenue would increase by an average of $1,200 per $1 increase in the per-game fee.**
    
1. First, calculate \( R \) at \( f = 1 \):
\[
R(1) = -480(1^2) + 3120(1) = -480 + 3120 = 2640
\]
2. Next, calculate \( R \) at \( f = 3 \):
\[
R(3) = -480(3^2) + 3120(3) = -480(9) + 9360 = -4320 + 9360 = 5040
\]
3. Now, calculate the change in revenue as the fee changes from $1 to $3:
\[
\Delta R = R(3) - R(1) = 5040 - 2640 = 2400
\]
4. The change in the per-game fee is:
\[
\Delta f = 3 - 1 = 2
\]
5. The average rate of change in revenue per $1 increase in the per-game fee can be found by dividing the change in revenue by the change in the per-game fee. Since the change in the fee is $2, the average increase in revenue per $1 increase in fee is:
\[
\text{Average rate of change} = \frac{\Delta R}{\Delta f} = \frac{2400}{2} = 1200
\]
Thus, the average rate of change in revenue is **$1,200 per $1 increase in the per-game fee**.
Therefore, the correct interpretation is:
**Revenue would increase by an average of $1,200 per $1 increase in the per-game fee.**
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