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To find the volume (\(V\)) corresponding to a sounding of 4.13 meters, we can use **linear interpolation** since the table gives us the values for soundings at 4.10 meters and 4.15 meters.

### Given dataTo find the volume (\(V\)) corresponding to a sounding of 4.13 meters, we can use **linear interpolation** since the table gives us the values for soundings at 4.10 meters and 4.15 meters.

### Given 
- Sounding at 4.10 meters corresponds to a volume of 120.91 cubic meters.
- Sounding at 4.15 meters corresponds to a volume of 124.27 cubic meters.

The formula for linear interpolation is:

\[
V = V_1 + \frac{(V_2 - V_1)}{(S_2 - S_1)} \times (S - S_1)
\]

Where:
- \(S_1 = 4.10 \, \text{m}\), \(V_1 = 120.91 \, \text{m}^3\)
- \(S_2 = 4.15 \, \text{m}\), \(V_2 = 124.27 \, \text{m}^3\)
- \(S = 4.13 \, \text{m}\) (the value for which we want to find the volume)

Let me calculate this for you.The volume corresponding to a sounding of 4.13 meters is approximately **122.93 cubic meters**.

Would you like more details on how the interpolation works, or do you have any further questions?

Here are 5 related questions that can expand on this topic:
1. What is the importance of interpolation in engineering calculations?
2. How can interpolation be extended to multiple dimensions (e.g., bilinear interpolation)?
3. Can interpolation be used for non-linear data? If so, how?
4. What is the difference between interpolation and extrapolation?
5. How would this method change if we had a larger data set?

**Tip:** Interpolation assumes the data between two known points follows a linear trend. Always check the appropriateness of this assumption for your data.

Find the volume (V) corresponding to a sounding of 4.13 m using the values from the fuel oil sounding table.

Learn how to use linear interpolation to find the volume (V) corresponding to a sounding of 4.13 meters. This solution includes the formula for interpolation using values from the fuel oil sounding table and explains each step in detail.

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