Question

what errors did monica make? check all that apply. she should have counted only 2 unit cubes instead of 4. she should have counted 8 one - third cubes instead of only 4. she should have found the total value of the one - third cubes by multiplying by $\frac{1}{3}$ instead of by $\frac{1}{2}$. each fractional cube is worth $\frac{2}{3}$ instead of the $\frac{1}{2}$ that she used.

Explanation:

Step1: Analyze the side - length of the fractional cubes

The height of the large prism is \(1\frac{2}{3}=\frac{5}{3}\text{cm}\). If we consider the unit of the small cubes along the height, since the height is divided into 5 equal parts (because \(\frac{5}{3}\div\frac{1}{3} = 5\)), but the base of the prism has side - lengths of \(2\text{cm}\). The volume of a cube with side - length \(s\) is \(V = s^{3}\). The side - length of the small cubes (along the height direction) is \(\frac{1}{3}\text{cm}\), and the side - lengths in the other two directions are \(1\text{cm}\) (since \(2\div2 = 1\)).

Step2: Calculate the number of one - third cubes

The number of one - third cubes along the height: \(\frac{5}{3}\div\frac{1}{3}=5\), along the length: \(2\div1 = 2\), along the width: \(2\div1=2\). The total number of one - third cubes is \(5\times2\times2 = 20\). But if we just consider the non - height direction (length and width), for a \(2\times2\) base (in terms of \(1\text{cm}\) units), and height divided into \(\frac{1}{3}\text{cm}\) units. The volume of the prism \(V=l\times w\times h=2\times2\times\frac{5}{3}=\frac{20}{3}\). The volume of a one - third cube \(v = 1\times1\times\frac{1}{3}=\frac{1}{3}\). The number of one - third cubes \(n=\frac{V}{v}=\frac{\frac{20}{3}}{\frac{1}{3}} = 20\).

The volume of a cube with side - length \(s\): If \(s=\frac{1}{3}\), \(V_{cube}=(\frac{1}{3})^{3}=\frac{1}{27}\) (incorrect approach for this problem, we use the formula \(V = l\times w\times h\) for the prism and \(V_{small}=s_{1}\times s_{2}\times s_{3}\) where \(s_{1}=1,s_{2}=1,s_{3}=\frac{1}{3}\)).

If we assume Monica was using a wrong unit - cube value. The volume of a unit cube with side - length \(a\):
Let's check each option:

• Option 1: The base has \(2\times2 = 4\) unit squares (in \(1\text{cm}\times1\text{cm}\) units), so this is wrong.
• Option 2: Wrong, as we calculated above.
• Option 3: The volume of a small cube (with side - lengths \(1\text{cm},1\text{cm},\frac{1}{3}\text{cm}\)) is \(V=1\times1\times\frac{1}{3}\), so if we count the number of such cubes and multiply by \(\frac{1}{3}\) (the height factor, since \(l = w=1\)), this is correct.
• Option 4: The volume of the small cube (relevant part) is \(\frac{1}{3}\) (since \(V=l\times w\times h\) with \(l = w = 1\) and \(h=\frac{1}{3}\)), not \(\frac{2}{3}\)

Answer:

She should have found the total value of the one - third cubes by multiplying by \(\frac{1}{3}\) instead of by \(\frac{1}{2}\)