Question

The water level of a river is below normal levels. Its water level is at 1 1/2 meters below sea level. During the summer, the river drops another 2 1/3 meters below sea level. What is the water level at the end of summer? Remember, the elevation is below sea level. Account for this in your answer.

Answer 1

it says its wrong

Answer 2

To find the water level of the river at the end of summer, we need to add the two negative elevations:

1. **Initial Water Level**: \( 1 \frac{1}{2} \) meters below sea level
- Converting this to an improper fraction:
\[
1 \frac{1}{2} = \frac{3}{2} \text{ meters below sea level}
\]

2. **Summer Drop**: \( 2 \frac{1}{3} \) meters below sea level
- Converting this to an improper fraction:
\[
2 \frac{1}{3} = \frac{7}{3} \text{ meters below sea level}
\]

Next, we need to add these two values together. Since both are below sea level, we will combine them, keeping in mind that they are both negative values.

We need a common denominator to perform the addition. The least common multiple of 2 and 3 is 6.

3. Convert both fractions to have a denominator of 6:
- For \(\frac{3}{2}\):
\[
\frac{3}{2} = \frac{3 \times 3}{2 \times 3} = \frac{9}{6}
\]
- For \(\frac{7}{3}\):
\[
\frac{7}{3} = \frac{7 \times 2}{3 \times 2} = \frac{14}{6}
\]

4. Now, we can add the fractions:
\[
\frac{9}{6} + \frac{14}{6} = \frac{9 + 14}{6} = \frac{23}{6}
\]

This result, \(\frac{23}{6}\) meters, means that at the end of summer, the river is \( \frac{23}{6} \) meters below sea level.

5. Converting \(\frac{23}{6}\) to a mixed number:
\[
23 \div 6 = 3 \text{ remainder } 5
\]
Therefore,
\[
\frac{23}{6} = 3 \frac{5}{6} \text{ meters below sea level}
\]

Thus, the water level at the end of summer is \( 3 \frac{5}{6} \) meters below sea level.

Answer 3

Let’s carefully re-evaluate the calculations to make sure we arrive at the correct answer step-by-step.

1. **Initial Water Level**: The initial water level is given as \( 1 \frac{1}{2} \) meters below sea level, which can be expressed as:
\[
1 \frac{1}{2} = \frac{3}{2} \text{ meters below sea level}
\]

2. **Summer Drop**: The river drops another \( 2 \frac{1}{3} \) meters below sea level, which can be converted as:
\[
2 \frac{1}{3} = \frac{7}{3} \text{ meters below sea level}
\]

3. **Combining the Values**: To find the total drop from sea level, we need to add these two quantities:
\[
Total = \frac{3}{2} + \frac{7}{3}
\]

4. **Finding a Common Denominator**: The least common multiple of 2 and 3 is 6.

- Convert \(\frac{3}{2}\) to a denominator of 6:
\[
\frac{3}{2} = \frac{3 \times 3}{2 \times 3} = \frac{9}{6}
\]

- Convert \(\frac{7}{3}\) to a denominator of 6:
\[
\frac{7}{3} = \frac{7 \times 2}{3 \times 2} = \frac{14}{6}
\]

5. **Adding the Converted Fractions**:
\[
\frac{9}{6} + \frac{14}{6} = \frac{9 + 14}{6} = \frac{23}{6}
\]

6. **Final Calculation**:

The water level at the end of summer is at \(\frac{23}{6}\) meters below sea level.

To express \(\frac{23}{6}\) as a mixed number:
- Divide 23 by 6, which gives 3 with a remainder of 5:
\[
23 \div 6 = 3 \quad \text{(whole part)}, \quad \text{remainder } 5
\]
Thus, \(\frac{23}{6} = 3 \frac{5}{6}\).

Therefore, the water level at the end of summer is **3 \(\frac{5}{6}\) meters below sea level**.

If you believe there is still a misunderstanding or if the problem's requirements change, please provide more detail, and I'll be glad to assist further!