Answer:
\(\frac{42}{5}\text{ ft}^3\).
Step-by-step explanation:
We have been given that a carton has a length of fraction \(2\frac{1}{4}\) feet, width of fraction \(1\frac{3}{5}\) feet, and height of \(2\frac{1}{3}\) feet. We are asked to find the volume of the carton.
The volume of carton will be product of length, width and height.
Let us convert mixed fractions into improper fractions as:
\(2\frac{1}{4}=\frac{9}{4}\)
\(1\frac{3}{5}=\frac{8}{5}\)
\(2\frac{1}{3}=\frac{7}{3}\)
\(\text{Volume of the carton}=\frac{9}{4}\text{ ft}\times\frac{8}{5}\text{ ft}\times\frac{7}{3}\text{ ft}\)
\(\text{Volume of the carton}=\frac{3}{1}\times\frac{2}{5}\times\frac{7}{1}\text{ ft}^3\)
\(\text{Volume of the carton}=3\times\frac{2}{5}\times7\text{ ft}^3\)
\(\text{Volume of the carton}=\frac{42}{5}\text{ ft}^3\)
Therefore, the volume of the carton is \(\frac{42}{5}\text{ ft}^3\).
