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

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: $$23÷6=3\text{ remainder }5$$ 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 2

it says its wrong

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: $$23÷6=3\text{(whole part)},\text{remainder }5$$ 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!