Question
After the end of an advertising campaign, the daily sales of a product fell rapidly, with daily sales given by Sequals3300eSuperscript negative 0.14 x dollars, where x is the number of days from the end of the campaign.
a. What were daily sales when the campaign ended?
b. How many days passed after the campaign ended before daily sales were below half of what they were at the end of the campaign?
Question content area bottom
Part 1
a. The daily sales when the campaign ended were $
enter your response here.
(Simplify your answer. Round to the nearest cent as needed.)
a. What were daily sales when the campaign ended?
b. How many days passed after the campaign ended before daily sales were below half of what they were at the end of the campaign?
Question content area bottom
Part 1
a. The daily sales when the campaign ended were $
enter your response here.
(Simplify your answer. Round to the nearest cent as needed.)
Answers
GPT-4o mini
To find the daily sales at the end of the advertising campaign, we need to evaluate the given equation when \( x = 0 \).
The daily sales function is:
\[
S = 3300 e^{-0.14x}
\]
### Part a:
When the advertising campaign ended, \( x = 0 \):
\[
S = 3300 e^{-0.14(0)} = 3300 e^{0} = 3300 \times 1 = 3300
\]
Thus, the daily sales when the campaign ended were **$3300.00.**
---
### Part 2:
We’ll find out how many days passed after the campaign ended before daily sales fell below half of what they were at the end of the campaign. Half of the sales at the end is:
\[
\frac{3300}{2} = 1650
\]
Now, we set the sales function equal to 1650 and solve for \( x \):
\[
3300 e^{-0.14x} = 1650
\]
Dividing both sides by 3300:
\[
e^{-0.14x} = \frac{1650}{3300}
\]
This simplifies to:
\[
e^{-0.14x} = \frac{1}{2}
\]
Next, we take the natural logarithm of both sides:
\[
-0.14x = \ln\left(\frac{1}{2}\right)
\]
To isolate \( x \), we rearrange the equation:
\[
x = \frac{\ln\left(\frac{1}{2}\right)}{-0.14}
\]
Calculating \( \ln\left(\frac{1}{2}\right) \):
\[
\ln\left(\frac{1}{2}\right) \approx -0.693147
\]
Now plug this value into the equation for \( x \):
\[
x = \frac{-0.693147}{-0.14} \approx 4.95
\]
Thus, rounding to the nearest whole number, **5 days** passed after the campaign ended before daily sales were below half of what they were at the end of the campaign.
### Summary of Answers:
- a. Daily sales when the campaign ended: **$3300.00**
- b. Days until daily sales fell below half of that: **5 days**
The daily sales function is:
\[
S = 3300 e^{-0.14x}
\]
### Part a:
When the advertising campaign ended, \( x = 0 \):
\[
S = 3300 e^{-0.14(0)} = 3300 e^{0} = 3300 \times 1 = 3300
\]
Thus, the daily sales when the campaign ended were **$3300.00.**
---
### Part 2:
We’ll find out how many days passed after the campaign ended before daily sales fell below half of what they were at the end of the campaign. Half of the sales at the end is:
\[
\frac{3300}{2} = 1650
\]
Now, we set the sales function equal to 1650 and solve for \( x \):
\[
3300 e^{-0.14x} = 1650
\]
Dividing both sides by 3300:
\[
e^{-0.14x} = \frac{1650}{3300}
\]
This simplifies to:
\[
e^{-0.14x} = \frac{1}{2}
\]
Next, we take the natural logarithm of both sides:
\[
-0.14x = \ln\left(\frac{1}{2}\right)
\]
To isolate \( x \), we rearrange the equation:
\[
x = \frac{\ln\left(\frac{1}{2}\right)}{-0.14}
\]
Calculating \( \ln\left(\frac{1}{2}\right) \):
\[
\ln\left(\frac{1}{2}\right) \approx -0.693147
\]
Now plug this value into the equation for \( x \):
\[
x = \frac{-0.693147}{-0.14} \approx 4.95
\]
Thus, rounding to the nearest whole number, **5 days** passed after the campaign ended before daily sales were below half of what they were at the end of the campaign.
### Summary of Answers:
- a. Daily sales when the campaign ended: **$3300.00**
- b. Days until daily sales fell below half of that: **5 days**