Question

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)


Skip to navigation

Answer 1

hello!

Answer 2

Part C is wrong.

Answer 3

Let's go through the problem step by step.

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

The rate of decay is given as 6%. Therefore:

**Part A: The rate of decay is 6%**

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

According to the problem, a typical cup of coffee contains about 100 mg of caffeine.

**Part B: The initial amount of caffeine consumed in this scenario is 100 mg.**

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

To model the caffeine decay, we use the formula for exponential decay:

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

where:
- \( C_0 \) is the initial amount of caffeine (100 mg),
- \( r \) is the decay rate (0.06),
- \( t \) is the time in hours.

Putting these values into the formula:

\[ C(t) = 100 \cdot (1 - 0.06)^t \]
\[ C(t) = 100 \cdot (0.94)^t \]

**Part C: C(t) = 100 * (0.94)^t**

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

Using the equation we derived, we will now calculate \( C(t) \) for \( t = 0, 1, 3, 5 \):

1. For \( t = 0 \):
\[
C(0) = 100 \cdot (0.94)^0 = 100 \cdot 1 = 100.00
\]

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

3. For \( t = 3 \):
\[
C(3) = 100 \cdot (0.94)^3 \approx 100 \cdot 0.830584 = 83.06
\]

4. For \( t = 5 \):
\[
C(5) = 100 \cdot (0.94)^5 \approx 100 \cdot 0.735091 = 73.51
\]

Now, we can fill in the table:

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

So, the completed table is:

**t: 0, 1, 3, 5**
**C(t): 100.00, 94.00, 83.06, 73.51**