Let's break down the problem step by step.
Firstly, we need to convert the mixed numbers into improper fractions or decimals for easier calculation.
1. The initial amount of water in the bathtub is [tex]\(38 \frac{1}{3}\)[/tex] gallons.
Converting [tex]\(38 \frac{1}{3}\)[/tex] to an improper fraction:
[tex]\[
38 \frac{1}{3} = 38 + \frac{1}{3}
\][/tex]
[tex]\[
38 \frac{1}{3} = 38 + 0.3333333333333333 \approx 38.333333333333336
\][/tex]
2. The amount of water that splashes out is [tex]\(2 \frac{3}{5}\)[/tex] gallons.
Converting [tex]\(2 \frac{3}{5}\)[/tex] to an improper fraction:
[tex]\[
2 \frac{3}{5} = 2 + \frac{3}{5}
\][/tex]
[tex]\[
2 \frac{3}{5} = 2 + 0.6 = 2.6
\][/tex]
3. Finally, we calculate the remaining amount of water in the tub by subtracting the splashed out water from the initial water.
[tex]\[
\text{Remaining water} = 38.333333333333336 - 2.6
\][/tex]
[tex]\[
\text{Remaining water} = 35.733333333333334
\][/tex]
So, the amount of water left in the tub is approximately [tex]\(35.733333333333334\)[/tex] gallons.
