Concept explainers
(a)
Copy and complete the table
(a)
Answer to Problem 38PPS
Explanation of Solution
Given:
Concept Used:
(b)
Find the absolute error and range of absolute error.
(b)
Answer to Problem 38PPS
Absolute error and range are:
Explanation of Solution
Given:
You measured a length of 12.8 am. Compute the absolute error and then write the range of possible measures.
Concept Used:
The absolute error of a measurement is equal to one half of the unit of measure
Measurement of length is 12.8.
Absolute error:
Range:
Calculation:
Thus, absolute error and range are:
(c)
Find what precision you would have to measure a length in cm to have an absolute error of less than 0.05 cm.
(c)
Answer to Problem 38PPS
To the nearest hundredth place
Explanation of Solution
Given:
To what precision would you have to measure a length in cm to have an absolute error of less than 0.05 cm?
Concept Used:
The absolute error of a measurement is equal to one half of the unit of measure
If the absolute error of loss is 0.05 cm, then the error is to the nearest hundredth place.
Calculation:
Thus, to the nearest hundredth place
(d)
Find the relative error of the volume of the box.
(d)
Answer to Problem 38PPS
Therelative errors:
Explanation of Solution
Given:
To find the relative error of an area or volume calculation, add the relative errors of each linear measure. If the measures of the sides of a rectangular box are 6.5 cm, 7.2 cm and 10.25 cm, what is the relative error of the volume of the box?
Concept Used:
The absolute error of a measurement is equal to one half of the unit of measure and the relative error is of a measure is the ratio of the absolute error to the expected measure.
Calculation:
Measure | Absolute error | Relative error |
6.5 cm | 0.05 cm | |
7.2 cm | 0.05 cm | |
10.25cm | 0.005 cm |
Add the relative errors:
Thus, the relative errors:
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