
Concept explainers
(a)
Copy and complete the table
(a)

Explanation of Solution
Given:
Concept Used:
(b)
Find the absolute error and range of absolute error.
(b)

Answer to Problem 9QC
Absolute error and range are: 0.05 cm ; 12.75 cm to 12.85 cm
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: ± 12 · 0.1 = 0.05
Range: 12.80−0.05 = 12.75 to 12.80 + 0.05=12.85
Calculation:
Thus, absolute error and range are: 0.05 cm ; 12.75 cm to 12.85 cm
(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 9QC
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 9QC
The relative errors: 0.0077 + 0.0069 + 0.00049 = 0.01509 ≈ 0.015 cm
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 | absolute errorexpected measurement = 0.05 cm6.5 cm ≈ 0.0077 |
7.2 cm | 0.05 cm | absolute errorexpected measurement = 0.05 cm7.2 cm ≈ 0.0069 |
10.25cm | 0.005 cm | absolute errorexpected measurement = 0.05 cm10.25 cm ≈ 0.00049 |
Add the relative errors: 0.0077 + 0.0069 + 0.00049 = 0.01509 ≈ 0.015 cm
Thus, the relative errors: 0.0077 + 0.0069 + 0.00049 = 0.01509 ≈ 0.015 cm
Chapter 6 Solutions
Algebra 1
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