1. The problem asks for the total amount of glitter used to make the paper chains, given a line plot showing the number of containers used for each chain. 2. The line plot shows the following data points with frequencies: - 1 container: 2 chains - 1 \frac{1}{8} containers: 2 chains - 1 \frac{1}{4} containers: 2 chains - 1 \frac{3}{8} containers: 1 chain - 1 \frac{1}{2} containers: 0 chains - 1 \frac{5}{8} containers: 0 chains - 1 \frac{3}{4} containers: 2 chains - 1 \frac{7}{8} containers: 2 chains - 2 containers: 0 chains 3. To find the total amount of glitter used, multiply each amount by its frequency and sum all results: $$\text{Total} = 2 \times 1 + 2 \times \frac{9}{8} + 2 \times \frac{5}{4} + 1 \times \frac{11}{8} + 0 + 0 + 2 \times \frac{7}{4} + 2 \times \frac{15}{8} + 0$$ 4. Convert mixed numbers to improper fractions: - 1 \frac{1}{8} = \frac{9}{8} - 1 \frac{1}{4} = \frac{5}{4} - 1 \frac{3}{8} = \frac{11}{8} - 1 \frac{3}{4} = \frac{7}{4} - 1 \frac{7}{8} = \frac{15}{8} 5. Calculate each term: - $2 \times 1 = 2$ - $2 \times \frac{9}{8} = \frac{18}{8} = \frac{9}{4}$ - $2 \times \frac{5}{4} = \frac{10}{4} = \frac{5}{2}$ - $1 \times \frac{11}{8} = \frac{11}{8}$ - $2 \times \frac{7}{4} = \frac{14}{4} = \frac{7}{2}$ - $2 \times \frac{15}{8} = \frac{30}{8} = \frac{15}{4}$ 6. Sum all terms: $$2 + \frac{9}{4} + \frac{5}{2} + \frac{11}{8} + \frac{7}{2} + \frac{15}{4}$$ 7. Find a common denominator, which is 8: $$2 = \frac{16}{8}, \quad \frac{9}{4} = \frac{18}{8}, \quad \frac{5}{2} = \frac{20}{8}, \quad \frac{7}{2} = \frac{28}{8}, \quad \frac{15}{4} = \frac{30}{8}$$ 8. Rewrite the sum: $$\frac{16}{8} + \frac{18}{8} + \frac{20}{8} + \frac{11}{8} + \frac{28}{8} + \frac{30}{8} = \frac{16 + 18 + 20 + 11 + 28 + 30}{8} = \frac{123}{8}$$ 9. Convert $\frac{123}{8}$ to a mixed number: $$123 \div 8 = 15 \text{ remainder } 3 \Rightarrow 15 \frac{3}{8}$$ 10. Therefore, the total amount of glitter used is $15 \frac{3}{8}$ containers.