Jean had \( 8 \frac{4}{9} \) lbs of apples. She used \( 2 \frac{1}{9} \) lbs to make pies. How many lbs of apples does Jean still have? a \( \quad 105 / 9 \) pounds b 6 pounds c \( 61 / 3 \) pounds d 6 apples

To find out how many pounds of apples Jean still has after using some to make pies, we need to subtract the amount she used from the total amount she had. 1. **Convert the mixed numbers to improper fractions:** - For \( 8 \frac{4}{9} \): \[ 8 \frac{4}{9} = \frac{8 \times 9 + 4}{9} = \frac{72 + 4}{9} = \frac{76}{9} \] - For \( 2 \frac{1}{9} \): \[ 2 \frac{1}{9} = \frac{2 \times 9 + 1}{9} = \frac{18 + 1}{9} = \frac{19}{9} \] 2. **Subtract the two fractions:** \[ \frac{76}{9} - \frac{19}{9} = \frac{76 - 19}{9} = \frac{57}{9} \] 3. **Simplify the fraction if possible:** \[ \frac{57}{9} = 6 \frac{3}{9} = 6 \frac{1}{3} \] Now, let's check the answer choices: - a \( \quad \frac{105}{9} \) pounds - b 6 pounds - c \( \frac{61}{3} \) pounds - d 6 apples To convert \( 6 \frac{1}{3} \) to an improper fraction: \[ 6 \frac{1}{3} = \frac{6 \times 3 + 1}{3} = \frac{18 + 1}{3} = \frac{19}{3} \] Now, let's convert \( \frac{57}{9} \) to a fraction with a denominator of 3: \[ \frac{57}{9} = \frac{57 \div 3}{9 \div 3} = \frac{19}{3} \] Thus, Jean still has \( 6 \frac{1}{3} \) lbs of apples, which is equivalent to \( \frac{19}{3} \) lbs. The correct answer is: c \( \frac{61}{3} \) pounds.