What is the weight of the piece of aluminum shown below at 0.0974 lbs/in.^3 (hint:subtract the volume of the cylindrical hole.) Round to the nearest tenth as needed.

We will have the following:
First, we will determine the volume of the side plate:
[tex]V_1=0.5\cdot4\cdot5\Rightarrow V_1=10[/tex]Now, we find the volume of the front plate:
[tex]V_2=0.5\cdot4\cdot4-\pi(2)^2(0.5)\Rightarrow V_2=8-2\pi[/tex]Now, we calculate the total volume:
[tex]V_t=2V_1+V_2\Rightarrow V_t=2(10)+(8-2\pi)[/tex][tex]\Rightarrow V_t=28-2\pi[/tex]Now, we calculate its weight:
[tex]w=0.0974\cdot\frac{b}{in^3}\cdot(28-2\pi)\cdot in^3\Rightarrow w=2.115217751\ldots[/tex][tex]\Rightarrow w\approx2.1[/tex]So, it's weight is approximately 2.1 Lb.