To determine how far Joey is from home, we need to find the total distance he walked east and west and then calculate the net distance from home.
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Convert the mixed numbers to improper fractions:
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For 15 2/3: \[ 15 \times 3 + 2 = 45 + 2 = 47 \quad \text{So, } 15 2/3 = \frac{47}{3} \]
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For 21 3/4: \[ 21 \times 4 + 3 = 84 + 3 = 87 \quad \text{So, } 21 3/4 = \frac{87}{4} \]
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Now, calculate the distance Joey walked in one direction (east) and subtract the distance he walked in the opposite direction (west):
- Joey walked east: \(\frac{47}{3}\)
- Joey walked west: \(\frac{87}{4}\)
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We need a common denominator to subtract these fractions. The least common multiple of 3 and 4 is 12.
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Convert \(\frac{47}{3}\) to a denominator of 12: \[ \frac{47}{3} = \frac{47 \times 4}{3 \times 4} = \frac{188}{12} \]
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Convert \(\frac{87}{4}\) to a denominator of 12: \[ \frac{87}{4} = \frac{87 \times 3}{4 \times 3} = \frac{261}{12} \]
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Now we can subtract the two fractions: \[ \frac{188}{12} - \frac{261}{12} = \frac{188 - 261}{12} = \frac{-73}{12} \]
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Since we are interested in the distance, we take the absolute value: \[ \frac{73}{12} \]
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Convert \(\frac{73}{12}\) to a mixed number:
- Divide 73 by 12: \[ 73 \div 12 = 6 \quad \text{(which is the whole number part)} \] \[ 73 - (12 \times 6) = 73 - 72 = 1 \quad \text{(which is the remainder)} \]
- So, \(\frac{73}{12} = 6 \frac{1}{12}\).
Thus, Joey is \( \boxed{6 \frac{1}{12}} \) meters from home.