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The image shows a relationship between two sets of tables with the equation \( H/V = 75/975 = 1/13 \). The left side has a table of values corresponding to some function or relationship for values \(H = 25\) and \(H = 30\), at columns \(V = 0\) and \(V = 1\). Then, this table transforms into a second table that uses \(H/V = 1/13\).

We are asked to find the missing value for \(H = 25.8\) when \(H/V = 1/13\).

To summarize the tables:

**Initial Table:**
| H  | V=0  | V=1  |
|----|------|------|
| 25 | 1.35 | 0.54 |
| 30 | 1.59 | 0.59 |

**Transformed Table (with \( H/V = 1/13 \)):**
| H  | \( H/V = 1/13 \) |
|----|------------------|
| 25 | 1.29             |
| 30 | 1.51             |

**Third Table (Question for \( H = 25.8 \)):**
| H   | \( H/V = 1/13 \) |
|-----|------------------|
| 25.8 | ?               |

To find the missing value for \(H = 25.8\), we can use linear interpolation between the values at \(H = 25\) and \(H = 30\).

Let me calculate that for you.The interpolated value for \( H = 25.8 \) when \( H/V = 1/13 \) is approximately \( 1.325 \).

Would you like more details or have any other questions?

Here are 5 related questions for further exploration:
1. How does linear interpolation work in practice?
2. Can we apply quadratic interpolation for more accuracy with additional points?
3. What other methods can we use for estimating values between known data points?
4. How would the result change if the data were non-linear?
5. Could we model this data with a function, such as a polynomial fit?

**Tip:** Linear interpolation assumes the data changes at a constant rate between points, which is useful for simple estimates between small ranges.

Find the missing value for H = 25.8 when H/V = 1/13 using the given table of values and linear interpolation.

Learn how to find the missing value for H = 25.8 using the table and linear interpolation with the equation H/V = 1/13. Detailed step-by-step solution included, ideal for Grades 9-12 students.

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