Answer:
84 cubes
Step-by-step explanation:
Given,
The dimension of the rectangular prism are,
[tex]1\frac{1}{3}\text{ ft }\times 1\text{ ft }\times 2\frac{1}{3}\text{ ft }[/tex]
Hence, the volume of the prism,
[tex]V=1\frac{1}{3}\times 1\times 2\frac{1}{3}[/tex]
[tex]=\frac{4}{3}\times \frac{7}{3}[/tex]
[tex]=\frac{28}{9}\text{ cube ft}[/tex]
Now, the volume of a cube = side³,
If side = [tex]\frac{1}{3}[/tex] ft,
Then the volume of each cube,
[tex]V'=(\frac{1}{3})^3=\frac{1}{27}\text{ cube ft}[/tex]
Hence, the number of cubes that can be packet in the prism
[tex]=\frac{V}{V'}[/tex]
[tex]=\frac{28/9}{1/27}[/tex]
[tex]=\frac{27\times 28}{9}[/tex]
[tex]=3\times 28[/tex]
[tex]=84[/tex]
