1. **State the problem:** We need to find the range of the values $\frac{7}{3}$, $2 \frac{5}{12}$, $-1 \frac{7}{12}$, and $-\frac{5}{3}$. The range is the difference between the maximum and minimum values. 2. **Convert all mixed numbers to improper fractions:** - $2 \frac{5}{12} = \frac{2 \times 12 + 5}{12} = \frac{24 + 5}{12} = \frac{29}{12}$ - $-1 \frac{7}{12} = -\frac{1 \times 12 + 7}{12} = -\frac{12 + 7}{12} = -\frac{19}{12}$ 3. **List all values as improper fractions:** - $\frac{7}{3}$ - $\frac{29}{12}$ - $-\frac{19}{12}$ - $-\frac{5}{3}$ 4. **Find the maximum and minimum values:** - Convert $\frac{7}{3}$ and $-\frac{5}{3}$ to twelfths to compare: - $\frac{7}{3} = \frac{7 \times 4}{3 \times 4} = \frac{28}{12}$ - $-\frac{5}{3} = -\frac{5 \times 4}{3 \times 4} = -\frac{20}{12}$ - Values in twelfths: - $\frac{28}{12}$ - $\frac{29}{12}$ - $-\frac{19}{12}$ - $-\frac{20}{12}$ - Maximum is $\frac{29}{12}$ - Minimum is $-\frac{20}{12}$ 5. **Calculate the range:** $$\text{Range} = \text{Maximum} - \text{Minimum} = \frac{29}{12} - \left(-\frac{20}{12}\right) = \frac{29}{12} + \frac{20}{12} = \frac{49}{12}$$ 6. **Convert the range to a mixed number:** $$\frac{49}{12} = 4 \frac{1}{12}$$ **Final answer:** The range of the given values is $4 \frac{1}{12}$.