4. Exponential Functions in Context(1 point)

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)

Answer 1

Part A:

The rate of decay is 6%.

Part B:

The initial amount of caffeine consumed in this scenario is 100 mg.

Part C:

The amount of caffeine remaining in the body can be modeled using the formula for exponential decay:

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

Where:

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

Given \(C_0 = 100\) mg and a decay rate of \(r = 0.06\) (since 6% metabolized means 94% remains), we can plug in these values:

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

Part D:

Now we'll calculate \(C(t)\) for the given time values \(t = 0, 1, 3, 5\).

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

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

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

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

Final Table

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