Answer:
[tex]\frac{42}{5}\text{ ft}^3[/tex].
Step-by-step explanation:
We have been given that a carton has a length of fraction [tex]2\frac{1}{4}[/tex] feet, width of fraction [tex]1\frac{3}{5}[/tex] feet, and height of [tex]2\frac{1}{3}[/tex] 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:
[tex]2\frac{1}{4}=\frac{9}{4}[/tex]
[tex]1\frac{3}{5}=\frac{8}{5}[/tex]
[tex]2\frac{1}{3}=\frac{7}{3}[/tex]
[tex]\text{Volume of the carton}=\frac{9}{4}\text{ ft}\times\frac{8}{5}\text{ ft}\times\frac{7}{3}\text{ ft}[/tex]
[tex]\text{Volume of the carton}=\frac{3}{1}\times\frac{2}{5}\times\frac{7}{1}\text{ ft}^3[/tex]
[tex]\text{Volume of the carton}=3\times\frac{2}{5}\times7\text{ ft}^3[/tex]
[tex]\text{Volume of the carton}=\frac{42}{5}\text{ ft}^3[/tex]
Therefore, the volume of the carton is [tex]\frac{42}{5}\text{ ft}^3[/tex].
