To add the mixed numbers \(-3 \frac{1}{6}\) and \(5 \frac{3}{4}\), we first convert them to improper fractions.
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Convert \(-3 \frac{1}{6}\) to an improper fraction: \[ -3 \frac{1}{6} = -\left(3 \times 6 + 1\right)/6 = -\left(18 + 1\right)/6 = -\frac{19}{6} \]
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Convert \(5 \frac{3}{4}\) to an improper fraction: \[ 5 \frac{3}{4} = 5 \times 4 + 3 / 4 = \frac{20 + 3}{4} = \frac{23}{4} \]
Now we need to add \(-\frac{19}{6}\) and \(\frac{23}{4}\). To do this, we need a common denominator. The least common multiple of 6 and 4 is 12.
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Convert \(-\frac{19}{6}\) to a denominator of 12: \[ -\frac{19}{6} = -\frac{19 \times 2}{6 \times 2} = -\frac{38}{12} \]
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Convert \(\frac{23}{4}\) to a denominator of 12: \[ \frac{23}{4} = \frac{23 \times 3}{4 \times 3} = \frac{69}{12} \]
Now we can add the two fractions: \[ -\frac{38}{12} + \frac{69}{12} = \frac{69 - 38}{12} = \frac{31}{12} \]
Now, we convert \(\frac{31}{12}\) back to a mixed number: \[ 31 ÷ 12 = 2 \quad \text{(whole part)} \] \[ 31 - (12 \times 2) = 31 - 24 = 7 \quad \text{(remainder)} \]
Thus, \[ \frac{31}{12} = 2 \frac{7}{12} \]
The answer, expressed as a reduced mixed number, is: \[ \boxed{2 \frac{7}{12}} \]