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To solve this problem, we need to determine the error interval for \( x \), given that \( x \), when rounded to 2 decimal places, is 15.31.

### Step-by-step solution:

1. **Understanding rounding**: 
   - A number rounded to 2 decimal places means it is accurate to the nearest hundredth.
   - The actual value of \( x \) could be slightly less or slightly more than 15.31 but would still round to 15.31.

2. **Determine the range**:
   - The maximum error when rounding to 2 decimal places is \( \pm 0.005 \) (half of 0.01).
   - This means \( x \) could be as small as 15.305 or as large as 15.315.

3. **Error interval**:
   - The lower bound of \( x \) is 15.305.
   - The upper bound of \( x \) is 15.315 (but not inclusive because at exactly 15.315, the number would round to 15.32).

Thus, the **error interval for \( x \)** is:
\[
15.305 \leq x < 15.315
\]

Would you like further details or have any questions?

Here are 5 follow-up questions to consider:
1. How is the error interval affected if \( x \) were rounded to 3 decimal places?
2. What would the error interval be if \( x \) were rounded to the nearest whole number?
3. Can you apply this concept to other forms of rounding, like significant figures?
4. What is the effect of rounding in practical applications, such as measurements?
5. How can you ensure rounding errors don’t affect important calculations?

**Tip**: Always keep in mind the impact of rounding when dealing with precision, especially in scientific or financial calculations.

A number, x, rounded to 2 decimal places is 15.31. Write down the error interval for x.

Learn how to calculate the error interval for a number rounded to two decimal places, using the example of 15.31. This problem covers key concepts in rounding and error intervals, suitable for students in Grades 7-10.

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