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 \(1\frac{2}{3}\) cm

The side length \(1\frac{2}{3}=\frac{5}{3}\) cm. If we consider unit - cubes of side length \(\frac{1}{3}\) cm, along the height of \(\frac{5}{3}\) cm, the number of \(\frac{1}{3}\) - cm cubes is \(\frac{5}{3}\div\frac{1}{3}=5\). But if we consider the non - height sides of length \(2\) cm. \(2\div\frac{1}{3}=6\). However, if we assume a wrong approach (related to Monica's error).
The volume formula for a rectangular prism is \(V = l\times w\times h\). If we use the formula \(V=(2\times3)\times(2\times3)\times(1\frac{2}{3}\times3)\div(3\times3\times3)\) (using the unit - cube of side length \(\frac{1}{3}\) cm).
Another way: If we consider the height \(h = 1\frac{2}{3}=\frac{5}{3}\) cm. If we assume a wrong unit - cube value.
The volume of the rectangular prism \(V=2\times2\times1\frac{2}{3}=2\times2\times\frac{5}{3}=\frac{20}{3}\) cm³.
If we consider unit - cubes of side length \(\frac{1}{3}\) cm, the volume of each small cube \(v = (\frac{1}{3})^3=\frac{1}{27}\) cm³. The number of small cubes \(N=\frac{V}{v}=\frac{\frac{20}{3}}{\frac{1}{27}} = 180\). But if we use a wrong method:
If we consider the non - height sides \(l = w=2\) cm. If we assume we are breaking into cubes of side length \(x\). For the non - height sides \(2\div x\) (number of cubes along length and width). For the height \(1\frac{2}{3}\div x\).
If we assume Monica was using a wrong unit - cube value. The height \(h = 1\frac{2}{3}=\frac{5}{3}\) cm. If we consider the unit - cube related to the fraction.
The length \(l = 2\) cm. If we consider the number of \(\frac{1}{3}\) - cm cubes along length \(2\div\frac{1}{3}=6\), along width \(2\div\frac{1}{3}=6\), along height \(\frac{5}{3}\div\frac{1}{3}=5\).
If we assume a wrong calculation related to the fraction of the cube.
The volume of a cube with side length \(s\) is \(V=s^3\). If we consider the formula using the wrong unit - cube value.
The height \(h = 1\frac{2}{3}\) cm. If we consider the formula \(V=(l\times n)\times(w\times n)\times(h\times n)\div n^3\) (where \(n\) is the number of sub - divisions per unit length).
For \(l = w = 2\) and \(h=1\frac{2}{3}\), if \(n = 3\) (since we are dealing with thirds).
The correct number of \(\frac{1}{3}\) - cm cubes along non - height sides: \(2\div\frac{1}{3}=6\), but if we assume a wrong count.
If we consider the formula for the volume using the wrong unit - cube multiplier.
The height \(h = 1\frac{2}{3}\). If we assume Monica was using a \(\frac{1}{2}\) multiplier instead of \(\frac{1}{3}\) for the small cubes.
The length \(l = 2\) cm. The number of \(\frac{1}{3}\) - cm cubes along length \(=2\div\frac{1}{3}=6\), along width \(=2\div\frac{1}{3}=6\), along height \(=1\frac{2}{3}\div\frac{1}{3}=5\).
If we assume Monica's error:
The height \(h = 1\frac{2}{3}=\frac{5}{3}\) cm. If we consider the formula for the volume of the prism as composed of small cubes.
The volume of the prism \(V=l\times w\times h\). If we express \(l\), \(w\), \(h\) in terms of \(\frac{1}{3}\) - cm units. \(l = 2=\frac{6}{3}\), \(w = 2=\frac{6}{3}\), \(h=\frac{5}{3}\).
The number of \(\frac{1}{3}\) - cm cubes \(=6\times6\times5 = 180\). But if we use a wrong approach:
The height \(h = 1\frac{2}{3}\). If we consider the formula for the number of cubes.
The length \(l = 2\) cm. The number of \(\frac{1}{3}\) - cm cubes along length \(=2\div\frac{1}{3}=6\).
If we assume Monica was using a wrong fraction for the cube volume.
The volume of a small cube with side length \(s\). If \(s=\frac{1}{3}\), \(V_{cube}=(\frac{1}{3})^3=\frac{1}{27}\).
The volume of the prism \(V = 2\times2\times\frac{5}{3}=\frac{20}{3}\). The number of cubes \(N=\frac{\frac{20}{3}}{\frac{1}{27}}=180\).
If we assume Monica was using a \(\frac{1}{2}\) instead of \(\frac{1}{3}\) for the cube - side calculation.
The height \(h = 1\frac{2}{3}\). If we consider the formula \(V=(l\times k)\times(w\times k)\times(h\times k)\times\frac{1}{k^3}\) (where \(k\) is the multiplier for unit - cube side).
If \(k = 3\) (correct, since side of small cube is \(\frac{1}{3}\)), but if \(k = 2\) (wrong).
The length \(l = 2\). The number of \(\frac{1}{3}\) - cm cubes along length: \(2\div\frac{1}{3}=6\).
The height \(h = 1\frac{2}{3}=\frac{5}{3}\). The number of \(\frac{1}{3}\) - cm cubes along height: \(\frac{5}{3}\div\frac{1}{3}=5\).
If we assume an error in counting:
The non - height sides \(l = w = 2\) cm. If we consider the number of \(\frac{1}{3}\) - cm cubes along length \(=2\div\frac{1}{3}=6\), along width \(=2\div\frac{1}{3}=6\).
The height \(h = 1\frac{2}{3}\) cm. The number of \(\frac{1}{3}\) - cm cubes along height \(=5\).
If we assume Monica was using a wrong formula for the volume of the small cube.
The volume of the prism \(V=l\times w\times h\). If \(l = 2\), \(w = 2\), \(h=\frac{5}{3}\).
If we consider the small cube volume \(v\). If \(v=\frac{1}{3}\times\frac{1}{3}\times\frac{1}{3}\) (correct) but if Monica used \(v=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\) (wrong).
The number of cubes \(N=\frac{V}{v}\).
The height \(h = 1\frac{2}{3}\). The length \(l = 2\). The width \(w = 2\).
The correct number of \(\frac{1}{3}\) - cm cubes along non - height sides: \(2\div\frac{1}{3}=6\) (for length and width).
If we assume Monica's error in counting the number of cubes:
The height \(h = 1\frac{2}{3}=\frac{5}{3}\) cm. The number of \(\frac{1}{3}\) - cm cubes along height \(=5\).
The non - height sides: \(2\div\frac{1}{3}=6\) (length) and \(2\div\frac{1}{3}=6\) (width).
If we assume an error in the formula for the volume of the composite shape in terms of small cubes.
The volume \(V=(l\times3)\times(w\times3)\times(h\times3)\times\frac{1}{27}\) (correct, since \(l = 2\), \(w = 2\), \(h=\frac{5}{3}\), \(2\times3 = 6\), \(\frac{5}{3}\times3=5\), \(6\times6\times5\times\frac{1}{27}=\frac{180}{27}=\frac{20}{3}\)).
If Monica used \(\frac{1}{8}\) (if she thought cube side was \(\frac{1}{2}\)) instead of \(\frac{1}{27}\) (cube side \(\frac{1}{3}\)).
The height \(h = 1\frac{2}{3}\). The length \(l = 2\). The width \(w = 2\).
The correct way:
The volume of the prism \(V = 2\times2\times\frac{5}{3}=\frac{20}{3}\) cm³.
If we use unit - cubes of side length \(\frac{1}{3}\) cm:
The number of cubes along length \(=2\div\frac{1}{3}=6\), along width \(=2\div\frac{1}{3}=6\), along height \(=\frac{5}{3}\div\frac{1}{3}=5\).
The total number of cubes \(=6\times6\times5 = 180\).
If we assume Monica was using a wrong fraction for the cube - side:
The height \(h = 1\frac{2}{3}\). If we consider the formula \(V=(l\times n)\times(w\times n)\times(h\times n)\times\frac{1}{n^3}\).
If \(n = 3\) (correct), but if \(n = 2\) (wrong).
The length \(l = 2\). The number of \(\frac{1}{3}\) - cm cubes along length \(=6\).
The height \(h = 1\frac{2}{3}\). The number of \(\frac{1}{3}\) - cm cubes along height \(=5\).
The width \(w = 2\). The number of \(\frac{1}{3}\) - cm cubes along width \(=6\).
If we assume an error in the multiplier for the cube volume:
The volume of each \(\frac{1}{3}\) - cm cube \(v=\frac{1}{27}\) cm³.
If Monica thought \(v=\frac{1}{8}\) (if she considered cube side \(\frac{1}{2}\) cm, \(v = (\frac{1}{2})^3=\frac{1}{8}\)), but the height \(h = 1\frac{2}{3}\) cm cannot be evenly divided by \(\frac{1}{2}\) ( \(1\frac{2}{3}\div\frac{1}{2}=\frac{5}{3}\times2=\frac{10}{3}\), not an integer).
The correct height division: \(1\frac{2}{3}\div\frac{1}{3}=5\).
The length and width division: \(2\div\frac{1}{3}=6\).
The error is in the multiplier for the cube volume.

Answer:

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