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?

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.