To find the water level of the river at the end of summer, we need to add both negative values together since both measurements are below sea level.
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The river's initial water level is \( 1 \frac{1}{2} \) meters below sea level, which can be converted to an improper fraction: \[ 1 \frac{1}{2} = \frac{3}{2} \text{ meters} \]
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The additional drop during summer is \( 2 \frac{1}{3} \) meters below sea level, which can also be converted to an improper fraction: \[ 2 \frac{1}{3} = \frac{7}{3} \text{ meters} \]
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Now, we will calculate the total drop in water level: \[ \text{Total drop} = \frac{3}{2} + \frac{7}{3} \]
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To add these fractions, we need a common denominator. The least common multiple of 2 and 3 is 6. We can convert each fraction: \[ \frac{3}{2} = \frac{3 \times 3}{2 \times 3} = \frac{9}{6} \] \[ \frac{7}{3} = \frac{7 \times 2}{3 \times 2} = \frac{14}{6} \]
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Now we can add the two fractions: \[ \frac{9}{6} + \frac{14}{6} = \frac{9 + 14}{6} = \frac{23}{6} \]
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Since the water level is below sea level, we denote it as: \[ \frac{23}{6} \text{ meters below sea level} \]
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To convert \(\frac{23}{6}\) to a mixed number:
- \(23 \div 6 = 3\) remainder 5, which means: \[ \frac{23}{6} = 3 \frac{5}{6} \]
Thus, at the end of summer, the water level of the river is \( \frac{23}{6} \) or \( 3 \frac{5}{6} \) meters below sea level.
So the final answer is: The water level is \( 3 \frac{5}{6} \) meters below sea level.