A num ber y , when trunc ated to 3 d ecima I place s, is eq ual to 0.004 Find th e erro r inte rval fo r u.

## Step 1<br />The question mentions that the number, \(y \), is truncated to 3 decimal places and the result is 0.004. Truncation involves discarding the digits beyond a certain point without rounding to the nearest possible value.<br /><br />## Step 2<br />Truncation understandably introduces an error in measurement. Knowing the position till which we've truncated allows us to estimate this error. <br /><br />## Step 3<br />Given that \( y \) was truncated to 3 decimal places, this means our error will come from digits in the fourth decimal place and beyond.<br /><br />## Step 4<br />Therefore, the actual value of \( y \) could be as much as 0.00005 more than the recorded value of 0.004, or 0.00005 less. The reasoning here is straightforward: if we had a situation where the number that was truncated was in fact, say, 0.00405, then the actual number was 0.00005 more than we recorded it as (0.004); likewise, if we had a number like say 0.00395, our report would have shown it as 0.004 and the actual number would have been 0.00005 less than what we recorded.<br /><br />### **Formulas:**<br />Based on these discoveries, the limits of the true value \( y \) likely represents due to possible error, therefore gives us our error interval: <br />To calculate the upper limit, the formula used would be: \( y_{upper} = y_{recorded} + 0.00005 \)<br />To calculate the lower limit, the formula to use would be: \( y_{lower} = y_{recorded} - 0.00005 \)<br /><br />## Step 5<br />By substituting \( y_{recorded} = 0.004 \) into the above formulas,<br />we'll have the upper bound as \( y_{upper} = 0.004 + 0.00005 = 0.00405 \),<br />and the lower limit as \( y_{lower} = 0.004 - 0.00005 = 0.00395 \).