Weight of the piece of aluminium is 5.3 lbs
Explanation:We consider the aluminium as a rectangular prism with a cylindrical hole
The formula for volume of a rectangular prism:
[tex]Volume\text{ = width }\times length\text{ }\times\text{ height}[/tex]When we look at the aluminium, we see that has inner opening aside the cylindrical hole. The volume of the aluminium without the cylindrical hole:
[tex]\begin{gathered} \text{Volume of the opened prism= volume of outer rectangular prism - Volume of the inner rectangular prism} \\ \end{gathered}[/tex][tex]\begin{gathered} \text{outer prism: 6 in by 6 in by 71/2 in} \\ \text{length = 6 in, width = 6 in, height = 7 1/2 in} \\ \text{Volume of the outer prism = }6\times6\times\frac{15}{2} \\ \text{Volume of the outer prism = 270 }in^3 \\ \\ In\text{ner prism: } \\ width\text{ = 5 in, length = 6 in, } \\ \text{height = 7 in} \\ \text{Volume of inner prism = 5}\times6\times7 \\ \text{Volume of inner prism =}210\text{ }in^3 \end{gathered}[/tex][tex]\begin{gathered} \text{Volume of the opened prism without the hole = 270 - 210} \\ \text{Volume of the opened prism without the hole = }60in^3 \end{gathered}[/tex]The opened aluminuim piece with the cylindrical hole:
Volume of aluminium piece = volume of opened prism - volume of cylindrical hole
[tex]\begin{gathered} \text{volume of cylinder = }\pi r^{2}h \\ \text{diameter of hole = 3 in} \\ \text{radius = diameter/2 = }\frac{3}{2\text{ }} \\ \text{radius = 1.5 in} \\ \text{height = 1/2 in = 0.5 in} \\ \text{let }\pi\text{ = 3.14} \\ \text{Volume of cylinder = }3.14\times1.5^2\times0.5 \\ \text{Volume of cylinder = 3.5325 in}^3 \end{gathered}[/tex][tex]\begin{gathered} \text{Volume of aluminium piece = 60 - 3.5325} \\ \text{Volume of aluminium piece = }54.4675\text{ in}^3 \end{gathered}[/tex]Density of the aluminium piece = 0.0974 lbs/cubic inch
Volume of the aluminium piece = 54.4675 in³
weight = ?
[tex]\begin{gathered} \text{Density = }\frac{mass}{\text{volume}} \\ let\text{ mass = weight} \\ 0.0974lbs/in^3\text{ = }\frac{\text{weight}}{54.4675in^3} \\ \text{weight = }0.0974\frac{lbs}{in^3\text{ }}\text{ }\times\text{ }54.4675in^3 \\ \text{weight = 5.3051 lbs} \\ \\ To\text{ the nearest tenth, weight of the aluminium piece = 5.3 lbs} \end{gathered}[/tex]