Question

4. Exponential Functions in Context(1 point)
A typical cup of coffee contains about 100 milligrams (mg) of caffeine.

Every hour, approximately 6%
of the amount of caffeine in the body is metabolized and eliminated.



Part A: The rate of decay is
%

Part B: The initial amount of caffeine consumed in this scenario is
mg.

Part C: Let C
represent the amount of caffeine in the body, in mg, and t
represents the number of hours since a cup of coffee was consumed. Write C
as a function of t
.

C(t) =

Part D: Fill out the table below. Round your answer to the nearest hundredth.

t
0 1 3 5
C(t)

Answer 1

Let's break down the problem step by step:

### Part A: The rate of decay
The problem states that approximately 6% of the caffeine in the body is metabolized and eliminated every hour. Consequently, the rate of decay of caffeine is \( 6\% \).

### Part B: The initial amount of caffeine consumed
The initial amount of caffeine in a typical cup of coffee is given as \( 100 \, \text{mg} \).

### Part C: Write C(t) as a function of t
The amount of caffeine in the body can be described by the exponential decay formula:

\[
C(t) = C_0 \cdot (1 - r)^t
\]

Where:
- \( C_0 \) is the initial amount of caffeine,
- \( r \) is the decay rate (as a decimal),
- \( t \) is the time in hours.

In this scenario:
- \( C_0 = 100 \) mg,
- \( r = 0.06 \).

Thus, the function becomes:

\[
C(t) = 100 \cdot (1 - 0.06)^t
\]

\[
C(t) = 100 \cdot (0.94)^t
\]

### Part D: Fill out the table
Now, we need to calculate \( C(t) \) for the given values of \( t \).

- For \( t = 0 \):

\[
C(0) = 100 \cdot (0.94)^0 = 100 \cdot 1 = 100 \, \text{mg}
\]

- For \( t = 1 \):

\[
C(1) = 100 \cdot (0.94)^1 = 100 \cdot 0.94 = 94.00 \, \text{mg}
\]

- For \( t = 3 \):

\[
C(3) = 100 \cdot (0.94)^3 \approx 100 \cdot 0.830584 = 83.06 \, \text{mg} \quad (\text{rounded to the nearest hundredth})
\]

- For \( t = 5 \):

\[
C(5) = 100 \cdot (0.94)^5 \approx 100 \cdot 0.735091 = 73.51 \, \text{mg} \quad (\text{rounded to the nearest hundredth})
\]

Now let's summarize the results in the table:

| t | C(t) |
|-----|--------------|
| 0 | 100.00 mg |
| 1 | 94.00 mg |
| 3 | 83.06 mg |
| 5 | 73.51 mg |

### Summary of Answers
- **Part A**: 6%
- **Part B**: 100 mg
- **Part C**: \( C(t) = 100 \cdot (0.94)^t \)
- **Part D**:
- \( C(0) = 100.00 \, \text{mg} \)
- \( C(1) = 94.00 \, \text{mg} \)
- \( C(3) = 83.06 \, \text{mg} \)
- \( C(5) = 73.51 \, \text{mg} \)