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The problem asks for the **error interval** of the length \(L\), given that \(L = 7.36\), correct to **2 decimal places**. Let's walk through the solution step-by-step.

### Key Concept: 
When a value is rounded to **2 decimal places**, the error can be up to **half of the next decimal place**. In this case:
- The next decimal place is the **thousandths place** (0.001).
- Half of 0.001 is **0.0005**.

### Error Interval:
This means the **actual length \(L\)** lies within **±0.005** of the given value 7.36. 

So:
- **Lower bound** = \( 7.36 - 0.005 = 7.355 \)  
- **Upper bound** = \( 7.36 + 0.005 = 7.365 \)  

Since the measurement is **rounded to 2 decimal places**, we express the bounds as:
- **Lower bound**: \( 7.355 \leq L \)
- **Upper bound**: \( L < 7.365 \)

### Final Answer:
\[
7.355 \leq L < 7.365
\]

Do you have any questions or want further details?

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**Here are five related questions:**

1. What is the error interval for 5.42, correct to 2 decimal places?
2. How would the interval change if the number were correct to 1 decimal place?
3. Why do we use strict inequality for the upper bound in error intervals?
4. What would the interval be if the number were 7.360, correct to 3 decimal places?
5. How do rounding and truncation affect error intervals?

**Tip:**  
When working with error intervals, always remember to halve the value of the next decimal place for accurate precision.

The length, L, of a steel rod is 7.36, correct to 2 decimal places. Complete the error interval for length L.

Learn how to calculate the error interval for a steel rod with a length of 7.36, correct to 2 decimal places. Understand the concept of rounding and error intervals, key for high school math students.

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