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answer it with solution using interpolation

To solve this problem, we can apply **linear interpolation**. This method helps to estimate the volume corresponding to a sounding of 4.13 meters, using the two given data points from the table.

The given points are:
- Sounding 1: \( x_1 = 4.10 \) meters, \( V_1 = 120.91 \) cubic meters
- Sounding 2: \( x_2 = 4.15 \) meters, \( V_2 = 124.27 \) cubic meters
- Desired sounding: \( x = 4.13 \) meters

The formula for linear interpolation is:

\[
V = V_1 + \frac{(V_2 - V_1)(x - x_1)}{(x_2 - x_1)}
\]

Now, let's plug in the values:

\[
V = 120.91 + \frac{(124.27 - 120.91)(4.13 - 4.10)}{(4.15 - 4.10)}
\]

Let me calculate that for you.The volume \( V \) corresponding to a sounding of 4.13 meters is approximately **122.93 cubic meters**.

Would you like further details or explanations on this? 

Here are some related questions to explore:
1. How does linear interpolation work in different contexts?
2. What other methods could estimate volumes between data points?
3. How would this approach differ if the data were non-linear?
4. Can you apply interpolation for more than two data points?
5. How could errors affect interpolation results in real-world scenarios?

**Tip:** Linear interpolation assumes a straight-line relationship between two points. Always verify if this assumption is valid for your data!

Find the volume (V) corresponding to a sounding of 4.13 m, given the table: Sounding: 4.10 meters -> Volume: 120.91 m³, Sounding: 4.15 meters -> Volume: 124.27 m³.

Learn how to calculate the volume corresponding to a sounding of 4.13 meters using linear interpolation. Given a table with soundings of 4.10 meters (120.91 m³) and 4.15 meters (124.27 m³), this problem involves key algebraic concepts. Suitable for students in Grades 10-12.

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