Answer:
![V(t) = 425(0.952)^(t)](https://img.qammunity.org/2021/formulas/mathematics/college/thfcennmr1urkenon5bgqxfn867uk0fva8.png)
Explanation:
The amount of water in the bucket after t hours, in mL, can be modeled by an equation in the following format:
![V(t) = V(0)(1-r)^(t)](https://img.qammunity.org/2021/formulas/mathematics/college/jj65c2wy2mp7zttw6rb2v427q9k14q5kjx.png)
In which V(0) is the initial amount, and r is the constant decay rate, as a decimal.
Bucket contains 425 mL of water.
This means that
![V(0) = 425](https://img.qammunity.org/2021/formulas/mathematics/college/gk64oi1jo9a00umfc7bhepxkn69acpjelb.png)
The capacity of water in the bucket decreases 4.8% each hour.
This means that
![r = 0.048](https://img.qammunity.org/2021/formulas/mathematics/college/ijrf6zcr9078s9l5ghsc9kxjjozd0sa4l9.png)
So
![V(t) = V(0)(1-r)^(t)](https://img.qammunity.org/2021/formulas/mathematics/college/jj65c2wy2mp7zttw6rb2v427q9k14q5kjx.png)
![V(t) = 425(1-0.048)^(t)](https://img.qammunity.org/2021/formulas/mathematics/college/ezgqrxsl4a6u68kcyi7uzm7zwcrnvahv9n.png)
![V(t) = 425(0.952)^(t)](https://img.qammunity.org/2021/formulas/mathematics/college/thfcennmr1urkenon5bgqxfn867uk0fva8.png)