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 1The question mentions that the number, , 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.## Step 2Truncation understandably introduces an error in measurement. Knowing the position till which we've truncated allows us to estimate this error. ## Step 3Given that was truncated to 3 decimal places, this means our error will come from digits in the fourth decimal place and beyond.## Step 4Therefore, the actual value of 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.### **Formulas:**Based on these discoveries, the limits of the true value likely represents due to possible error, therefore gives us our error interval: To calculate the upper limit, the formula used would be: To calculate the lower limit, the formula to use would be: ## Step 5By substituting into the above formulas,we'll have the upper bound as ,and the lower limit as .