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 6
C(t)


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Answer 1

To solve the problem, let’s go through each part step by step.

### Part A: The rate of decay is ___%.

The problem states that approximately 6% of the amount of caffeine is metabolized and eliminated every hour. Thus, the rate of decay is **6%**.

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

The problem specifies that a typical cup of coffee contains about **100 mg** of caffeine. Thus, the initial amount of caffeine is **100 mg**.

### Part C: Write C as a function of t.

The amount of caffeine in the body decreases exponentially. We can use 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 in decimal form,
- \( t \) is the time in hours.

Given:
- \( C_0 = 100 \) mg (initial amount)
- \( r = 0.06 \) (6% decay rate per hour)

We have:

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

Thus:

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

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

We will calculate \( C(t) \) for \( t = 0 \), \( t = 1 \), \( t = 3 \), and \( t = 6 \).

1. For \( t = 0 \):
\[ C(0) = 100 \cdot (0.94)^0 = 100 \text{ mg} \]

2. For \( t = 1 \):
\[ C(1) = 100 \cdot (0.94)^1 = 100 \cdot 0.94 = 94.00 \text{ mg} \]

3. For \( t = 3 \):
\[ C(3) = 100 \cdot (0.94)^3 \]
\[ C(3) = 100 \cdot 0.830584 = 83.06 \text{ mg} \] (rounded to 83.06)

4. For \( t = 6 \):
\[ C(6) = 100 \cdot (0.94)^6 \]
\[ C(6) = 100 \cdot 0.598736 = 59.87 \text{ mg} \] (rounded to 59.87)

Now, putting it all together:

| t | C(t) |
|---|--------|
| 0 | 100.00 |
| 1 | 94.00 |
| 3 | 83.06 |
| 6 | 59.87 |

### Summary

- **Part A:** 6%
- **Part B:** 100 mg
- **Part C:** \( C(t) = 100 \cdot (0.94)^t \)
- **Part D:**

| t | C(t) |
|---|--------|
| 0 | 100.00 |
| 1 | 94.00 |
| 3 | 83.06 |
| 6 | 59.87 |