## Solution.

Let's create a table.

| $t$ | 0 | 5 | 10 | 15 | 20 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $Q(t)$ | 330 | 165 | 82.5 | 41.25 | 20.625 |

The function $Q(t)$ is an exponential function of the form $Q(t)=Q_{0} b^{\frac{t}{5}}$.

From the given data $Q(0)=330$, we easily get $Q_{0}=330$, and from $Q(5)=165$ we get the base $b$

$$
165=330 b
$$

so $b=\frac{1}{2}$. The amount of caffeine, in mg, in the bloodstream after $t$ hours is $Q(t)=330 \cdot 2^{\frac{-t}{5}}$. From the coffee a person drinks at 8 AM, the amount remaining in the bloodstream at 20 hours, or more precisely at 8 PM, is $Q(12)=330 \cdot 2^{\frac{-12}{5}} \mathrm{mg}$.

From the coffee a person drinks at 10 AM, the amount remaining in the bloodstream at 10 hours, or more precisely at 8 PM, is $Q(10)=330 \cdot 2^{\frac{-10}{5}} \mathrm{mg}$.

From the coffee a person drinks at 12 PM, the amount remaining in the bloodstream at 8 hours, or more precisely at 8 PM, is $Q(8)=330 \cdot 2^{\frac{-8}{5}} \mathrm{mg}$.

At $20 \mathrm{~h}$, the total amount of caffeine remaining in the bloodstream will be $330 \cdot 2^{\frac{-12}{5}}+330 \cdot 2^{-2}+330 \cdot 2^{\frac{-8}{5}}=330 \cdot 2^{\frac{-8}{5}}\left(2^{\frac{-4}{5}}+2^{\frac{-2}{5}}+1\right) \mathrm{mg}$.
Task B-2.3. The amount of caffeine in the bloodstream decreases by $50 \%$ every 5 hours. A double espresso contains $330 \mathrm{mg}$ of caffeine. Assume that the entire double espresso is consumed at once. If a person drinks a double espresso at $8 \mathrm{~h}, 10 \mathrm{~h}$ and $12 \mathrm{~h}$ in one day, determine how many mg of caffeine that person will have in their bloodstream at 20 hours.