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

3 answers

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

hello!
Part C is wrong.