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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.(1 point) The water level is meters below sea level.

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

  1. The initial water level is \( 1 \frac{1}{2} \) meters below sea level. This can be converted to an improper fraction: \[ 1 \frac{1}{2} = \frac{3}{2} \text{ meters} \]

  2. During the summer, the river drops another \( 2 \frac{1}{3} \) meters below sea level. This can also be converted to an improper fraction: \[ 2 \frac{1}{3} = \frac{7}{3} \text{ meters} \]

Next, we need to add these two fractions together to find the total drop in water level:

\[ \frac{3}{2} + \frac{7}{3} \]

To add these fractions, we need to find a common denominator. The least common multiple of 2 and 3 is 6.

Now let's convert each fraction: \[ \frac{3}{2} = \frac{3 \times 3}{2 \times 3} = \frac{9}{6} \] \[ \frac{7}{3} = \frac{7 \times 2}{3 \times 2} = \frac{14}{6} \]

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

So, the total drop in water level is \( \frac{23}{6} \) meters below sea level.

To express this as a mixed number: \[ \frac{23}{6} = 3 \frac{5}{6} \]

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

So the final answer is: \[ \text{The water level is } 3 \frac{5}{6} \text{ meters below sea level.} \]

rong

I apologize for any confusion in my previous response. Let's carefully recalculate the water level step-by-step.

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

  2. Summer Drop: The additional drop is \(2 \frac{1}{3}\) meters below sea level, which can be expressed as: \[ 2 \frac{1}{3} = \frac{7}{3} \text{ meters} \]

  3. Adding the Drops: We need to add \(\frac{3}{2}\) and \(\frac{7}{3}\). To do this, we need a common denominator. The least common multiple of 2 and 3 is 6.

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

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

  4. Adding these fractions: \[ \frac{9}{6} + \frac{14}{6} = \frac{9 + 14}{6} = \frac{23}{6} \]

  5. Final Water Level: Now, since the water level is below sea level, we express the total drop as: \[ \frac{23}{6} \text{ meters below sea level} \]

    Converting \(\frac{23}{6}\) into a mixed number gives: \[ 23 \div 6 = 3 \quad \text{(with a remainder of 5)} \] Thus: \[ \frac{23}{6} = 3 \frac{5}{6} \]

Therefore, at the end of summer, the final water level of the river is: \[ \text{3} \frac{5}{6} \text{ meters below sea level.} \]

Thank you for your patience, and I hope this clarified the calculations!

thats still the rong answer

I apologize for the oversight. Let's go through the calculation step by step again to find the correct water level below sea level at the end of summer.

  1. Initial Water Level: The river is initially at \(1 \frac{1}{2}\) meters below sea level, which is: \[ 1 \frac{1}{2} = 1.5 \text{ meters} \text{ below sea level}. \]

  2. Additional Summer Drop: During the summer, the river drops another \(2 \frac{1}{3}\) meters below sea level, which is: \[ 2 \frac{1}{3} = 2.3333 \text{ meters (or } \frac{7}{3} \text{ meters)} \text{ below sea level}. \]

  3. Adding Both Drops: To find the final water level, we add the initial level \(1.5\) meters and the summer drop \(2.3333\) meters together:

    • Convert \(1.5\) to a fraction: \[ 1.5 = \frac{3}{2} \]

    • Convert \(2.\overline{3}\) to a fraction: \[ 2 \frac{1}{3} = \frac{7}{3} \]

  4. Finding a common denominator: The least common multiple of 2 and 3 is 6.

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

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

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

  6. Final Water Level: The total drop is \(\frac{23}{6}\) meters below sea level. We can convert this back into a decimal for clarity: \[ \frac{23}{6} = 3.8333\ldots \text{ or } 3 \frac{5}{6} \text{ meters} \]

In conclusion, the water level of the river at the end of summer is: \[ \text{The water level is } 3 \frac{5}{6} \text{ meters below sea level.} \]

If you still find discrepancies, please specify any particular points of confusion or clarify any additional parameters needed for the solution. Thank you for your patience!