Answer:
![C = \frac{C_(0) }{2^{(t)/(3.5) } }](https://img.qammunity.org/2020/formulas/mathematics/high-school/du81a11ezmjmshts3yp3va5ch7fdafje3r.png)
Explanation:
As the initial amount decreases by half every 3.5 hours, considering
as the initial amount you have that:
at t=0 the amount would be
![C_(0)](https://img.qammunity.org/2020/formulas/mathematics/college/kalakdarxc8gist69eaho8x908pqiniul0.png)
at t=3.5 the amount would be
![C_(0)/2](https://img.qammunity.org/2020/formulas/mathematics/high-school/qa8ltw02nx069gv2k5krrvnl7c9b2y67pd.png)
at t=7 the amount would be
![C_(0)/4](https://img.qammunity.org/2020/formulas/mathematics/high-school/rzqev1bavwja013804kin6etcid4l2auar.png)
at t=10.5 the amount would be
![C_(0)/8](https://img.qammunity.org/2020/formulas/mathematics/high-school/q5xtr4xrnlzf0mr246289e53dihnk695zs.png)
and so on...
this behavior can the modeled as:
![C = \frac{C_(0) }{2^{(t)/(3.5) } }](https://img.qammunity.org/2020/formulas/mathematics/high-school/du81a11ezmjmshts3yp3va5ch7fdafje3r.png)
This matches all the points described above.