Answer :
Sure! Let's solve the problem step-by-step.
Given that the number [tex]\( x \)[/tex] has been truncated to a whole number and the result is 16, the truncation implies removing the decimal part without rounding. This means the number [tex]\( x \)[/tex] is between 16 and just under 17.
### Step-by-Step Solution:
1. Understanding Truncation:
- When we truncate a number to the whole number 16, this implies that [tex]\( x \)[/tex] is at least 16 because truncating any number between 16 (inclusive) and 17 (exclusive) results in 16.
2. Defining the Interval:
- Therefore, the lower bound of [tex]\( x \)[/tex] is 16.
- The upper bound of [tex]\( x \)[/tex] is just less than 17, since if [tex]\( x \)[/tex] were 17 or more, its truncated whole number would be 17, not 16.
3. Expressing the Error Interval:
- Combining these bounds together, the interval can be written as:
[tex]\[ 16 \leq x < 17 \][/tex]
Thus, the error interval for [tex]\( x \)[/tex] is:
[tex]\[ 16 \leq x < 17 \][/tex]
This properly captures the set of all numbers that would truncate to 16 when the decimal portion is removed.
Given that the number [tex]\( x \)[/tex] has been truncated to a whole number and the result is 16, the truncation implies removing the decimal part without rounding. This means the number [tex]\( x \)[/tex] is between 16 and just under 17.
### Step-by-Step Solution:
1. Understanding Truncation:
- When we truncate a number to the whole number 16, this implies that [tex]\( x \)[/tex] is at least 16 because truncating any number between 16 (inclusive) and 17 (exclusive) results in 16.
2. Defining the Interval:
- Therefore, the lower bound of [tex]\( x \)[/tex] is 16.
- The upper bound of [tex]\( x \)[/tex] is just less than 17, since if [tex]\( x \)[/tex] were 17 or more, its truncated whole number would be 17, not 16.
3. Expressing the Error Interval:
- Combining these bounds together, the interval can be written as:
[tex]\[ 16 \leq x < 17 \][/tex]
Thus, the error interval for [tex]\( x \)[/tex] is:
[tex]\[ 16 \leq x < 17 \][/tex]
This properly captures the set of all numbers that would truncate to 16 when the decimal portion is removed.
