Chapter 6, Problem 9QC

(a)

To determine

Copy and complete the table

Explanation of Solution

Given:

  

Concept Used:

  

(b)

To determine

Find the absolute error and range of absolute error.

Answer to Problem 9QC

Absolute error and range are: $\text{0}\text{.05}\text{cm}\text{;}\text{12}\text{.75}\text{cm}\text{to}\text{12}\text{.85}\text{cm}$

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: $\pm \frac{1}{2}·0.1=0.05$

Range: $12.80-0.05=12.75\text{to}12.80+0.05=12.85$

Calculation:

Thus, absolute error and range are: $\text{0}\text{.05}\text{cm}\text{;}\text{12}\text{.75}\text{cm}\text{to}\text{12}\text{.85}\text{cm}$

(c)

To determine

Find what precision you would have to measure a length in cm to have an absolute error of less than 0.05 cm.

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)

To determine

Find the relative error of the volume of the box.

Answer to Problem 9QC

The relative errors: $0.0077+0.0069+0.00049=0.01509\approx 0.015\text{cm}$

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	$\frac{\text{absolute}\text{error}}{\text{expected}\text{measurement}}=\frac{0.05\text{cm}}{6.5\text{cm}}\approx 0.0077$
7.2 cm	0.05 cm	$\frac{\text{absolute}\text{error}}{\text{expected}\text{measurement}}=\frac{0.05\text{cm}}{7.2\text{cm}}\approx 0.0069$
10.25cm	0.005 cm	$\frac{\text{absolute}\text{error}}{\text{expected}\text{measurement}}=\frac{0.05\text{cm}}{10.25\text{cm}}\approx 0.00049$

Add the relative errors: $0.0077+0.0069+0.00049=0.01509\approx 0.015\text{cm}$

Thus, the relative errors: $0.0077+0.0069+0.00049=0.01509\approx 0.015\text{cm}$