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Sagot :
To solve this problem, let's use the given formula to predict the ticket sales for the sixth week after the movie's release. The formula is:
[tex]\[ P_t = P_0 e^{-kt} \][/tex]
where:
- [tex]\( P_t \)[/tex] is the ticket sales at week [tex]\( t \)[/tex].
- [tex]\( P_0 \)[/tex] is the initial ticket sales (first week).
- [tex]\( e \)[/tex] is the base of the natural logarithm (approximately 2.71828).
- [tex]\( k \)[/tex] is the decay rate.
- [tex]\( t \)[/tex] is the number of weeks after the release.
Given values:
- [tex]\( P_0 = 16.3 \)[/tex] million dollars (initial ticket sales).
- [tex]\( k = 0.4 \)[/tex] (decay rate).
- [tex]\( t = 6 \)[/tex] (number of weeks after release).
We need to find [tex]\( P_t \)[/tex] for [tex]\( t = 6 \)[/tex].
Substitute the given values into the formula:
[tex]\[ P_{6} = 16.3 \cdot e^{-0.4 \cdot 6} \][/tex]
This simplifies to:
[tex]\[ P_{6} = 16.3 \cdot e^{-2.4} \][/tex]
Now calculate the value of [tex]\( e^{-2.4} \)[/tex], and then multiply by 16.3.
The predicted ticket sales for the sixth week is:
[tex]\[ P_{6} \approx 1.5 \][/tex]
Thus, to the nearest \[tex]$0.1 million, the predicted ticket sales for the sixth week is \$[/tex]1.5 million.
Therefore, the correct answer is:
[tex]\[ \$ 1.5 \text{ million} \][/tex]
[tex]\[ P_t = P_0 e^{-kt} \][/tex]
where:
- [tex]\( P_t \)[/tex] is the ticket sales at week [tex]\( t \)[/tex].
- [tex]\( P_0 \)[/tex] is the initial ticket sales (first week).
- [tex]\( e \)[/tex] is the base of the natural logarithm (approximately 2.71828).
- [tex]\( k \)[/tex] is the decay rate.
- [tex]\( t \)[/tex] is the number of weeks after the release.
Given values:
- [tex]\( P_0 = 16.3 \)[/tex] million dollars (initial ticket sales).
- [tex]\( k = 0.4 \)[/tex] (decay rate).
- [tex]\( t = 6 \)[/tex] (number of weeks after release).
We need to find [tex]\( P_t \)[/tex] for [tex]\( t = 6 \)[/tex].
Substitute the given values into the formula:
[tex]\[ P_{6} = 16.3 \cdot e^{-0.4 \cdot 6} \][/tex]
This simplifies to:
[tex]\[ P_{6} = 16.3 \cdot e^{-2.4} \][/tex]
Now calculate the value of [tex]\( e^{-2.4} \)[/tex], and then multiply by 16.3.
The predicted ticket sales for the sixth week is:
[tex]\[ P_{6} \approx 1.5 \][/tex]
Thus, to the nearest \[tex]$0.1 million, the predicted ticket sales for the sixth week is \$[/tex]1.5 million.
Therefore, the correct answer is:
[tex]\[ \$ 1.5 \text{ million} \][/tex]
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