Answered

In the first week of its release, the latest blockbuster movie sold \[tex]$16.3 million in tickets. The movie's producers use the formula \( P_t = P_0 e^{-0.4t} \) to predict the number of ticket sales \( t \) weeks after a movie's release, where \( P_0 \) is the first week's ticket sales.

What are the predicted ticket sales to the nearest \$[/tex]0.1 million for the sixth week of this movie's release? (Note: [tex]\( t=0 \)[/tex] for the first week.)

A. \[tex]$13.3 million
B. \$[/tex]0.2 million
C. \[tex]$2.2 million
D. \$[/tex]1.5 million



Answer :

GinnyAnswer
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]

Other Questions