To determine the water level at the end of summer, we first need to convert the fractions to improper fractions so we can easily add them together.
-
The initial water level is \(1 \frac{1}{2}\) meters below sea level. Converting this to an improper fraction: \[ 1 \frac{1}{2} = \frac{2 \times 1 + 1}{2} = \frac{3}{2} \] So, the initial water level is \(-\frac{3}{2}\) meters (since it is below sea level).
-
The additional drop in water level during summer is \(2 \frac{1}{3}\) meters. Converting this to an improper fraction: \[ 2 \frac{1}{3} = \frac{3 \times 2 + 1}{3} = \frac{7}{3} \] Therefore, the summer drop is \(-\frac{7}{3}\) meters.
-
Now we can add these two values together to find the total water level at the end of summer: \[ -\frac{3}{2} - \frac{7}{3} \] To add these fractions, we need a common denominator. The least common multiple of 2 and 3 is 6.
Converting \(-\frac{3}{2}\) to have a denominator of 6: \[ -\frac{3}{2} = -\frac{3 \times 3}{2 \times 3} = -\frac{9}{6} \]
Converting \(-\frac{7}{3}\) to have a denominator of 6: \[ -\frac{7}{3} = -\frac{7 \times 2}{3 \times 2} = -\frac{14}{6} \]
-
Now we add the two fractions: \[ -\frac{9}{6} - \frac{14}{6} = -\frac{9 + 14}{6} = -\frac{23}{6} \]
-
The final water level at the end of summer is \(-\frac{23}{6}\) meters from sea level.
To express \(-\frac{23}{6}\) as a mixed number: \[ -\frac{23}{6} = -3 \frac{5}{6} \]
Thus, the water level is 3 5/6 meters below sea level.
So the answer is: The water level is 3.833... meters (or 3 5/6 meters) from sea level.