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 \( t = 6 \).
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For \( t = 0 \): \[ C(0) = 100 \cdot (0.94)^0 = 100 \text{ mg} \]
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For \( t = 1 \): \[ C(1) = 100 \cdot (0.94)^1 = 100 \cdot 0.94 = 94.00 \text{ mg} \]
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For \( t = 3 \): \[ C(3) = 100 \cdot (0.94)^3 \] \[ C(3) = 100 \cdot 0.830584 = 83.06 \text{ mg} \] (rounded to 83.06)
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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 |