To find the total water level of the river at the end of summer, we will add the two distances below sea level.
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The initial water level is \(1 \frac{1}{2}\) meters below sea level. In improper fraction form, this is: \[ 1 \frac{1}{2} = \frac{3}{2} \text{ meters} \]
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The additional drop during the summer is \(2 \frac{1}{3}\) meters below sea level. In improper fraction form, this is: \[ 2 \frac{1}{3} = \frac{7}{3} \text{ meters} \]
Now, we need to add these two values: \[ \frac{3}{2} + \frac{7}{3} \]
To perform the addition, we need a common denominator. The least common multiple of 2 and 3 is 6. We will convert both fractions:
\[ \frac{3}{2} = \frac{3 \times 3}{2 \times 3} = \frac{9}{6} \]
\[ \frac{7}{3} = \frac{7 \times 2}{3 \times 2} = \frac{14}{6} \]
Now we can add the two fractions: \[ \frac{9}{6} + \frac{14}{6} = \frac{9 + 14}{6} = \frac{23}{6} \]
The resulting fraction \(\frac{23}{6}\) represents the total water level below sea level. Converting it to a mixed number, we divide 23 by 6: \[ 23 \div 6 = 3 \text{ R } 5 \quad \Rightarrow \quad 3 \frac{5}{6} \]
Therefore, the water level at the end of summer is: \[ 3 \frac{5}{6} \text{ meters below sea level} \]
Answering the question directly: The water level is 3 5/6 meters from sea level (below sea level).