Free math worksheets

- Identify the measured value and the accepted value before choosing a fraction. - The numerator should represent the size of the measurement error. - Percent error compares that error with the accepted value rather than the measured value.

1. Percent error compares the absolute error with the accepted value. 2. The absolute error is \(\lvert52-50\rvert\), and the accepted value is \(50\). 3. Therefore, choice a) correctly represents the percent error.

a) \(\frac{\lvert52-50\rvert}{50}\cdot100\%\)

- First find the difference between the measured and accepted lengths. - Percent error compares that difference with the accepted value. - The accepted value here makes the percent calculation especially direct.

1. The absolute error is \(\lvert102-100\rvert=2\,\text{cm}\). 2. Compare the error with the accepted value: \(\frac{2}{100}\cdot100\%=2\%\). 3. The percent error is \(2\%\).

\(2\%\)

- First compare the recorded measurement with the accepted distance. - The percent should describe the size of the error relative to the accepted value. - Check that your result is reasonable for a difference of only one tenth of a mile.

1. The absolute error is \(\lvert2.4-2.5\rvert=0.1\,\text{mile}\). 2. Compare the error with the accepted distance: \(\frac{0.1}{2.5}\cdot100\%=4\%\).

The percent error is \(4\%\).

- Read the values of markers A and M from the number line. - Find the distance between the measured value and the accepted value. - Compare that error with the accepted value, not with the measured value.

1. The number line shows accepted value \(80\) at A and measured value \(84\) at M. 2. The absolute error is \(\lvert84-80\rvert=4\). 3. The percent error is \(\frac{4}{80}\cdot100\%=5\%\).

\(5\%\)

- Decide how the scale reading compares with the accepted weight when it reads high. - Represent the reading as a percent of the accepted weight. - Work backward from the displayed weight.

1. A reading that is \(4\%\) high is \(104\%\) of the accepted weight. 2. Let the accepted weight be \(w\). Then \(1.04w=78\). 3. Solve: \(w=78 \div 1.04=75\).

The accepted weight is \(75\,\text{pounds}\).

- Absolute errors from different-sized quantities are not enough by themselves for a fair comparison. - Compare each error with the accepted quantity from its own situation. - After finding both relative errors, compare the percentages.

1. Jordan's absolute error is \(3\,\text{seconds}\), so the percent error is \(\frac{3}{60}\cdot100\%=5\%\). 2. Casey's absolute error is \(0.5\,\text{foot}\), so the percent error is \(\frac{0.5}{8}\cdot100\%=6.25\%\). 3. Since \(5\%<6.25\%\), Jordan's estimate has the smaller percent error.

Jordan's estimate is relatively more accurate: \(5\%\) error compared with Casey's \(6.25\%\) error.

- Identify which number represents the accepted value and which number is the measurement. - Ask what quantity the error should be relative to. - Check whether the denominator in the student's fraction matches that reference quantity.

1. The absolute error is \(\lvert240-250\rvert=10\,\text{grams}\). 2. Percent error compares the error with the accepted value, not the measured value. 3. The correct calculation is \(\frac{10}{250}\cdot100\%=4\%\).

The student used the measured value in the denominator. The accepted value should be the reference, so the correct percent error is \(4\%\).

- Compare each measured length with the target length. - The inspection rule is based on relative error, not just whether a part is long or short. - Check each result against the stated maximum percentage.

1. For \(7.82\,\text{cm}\), the error is \(0.18\,\text{cm}\) and the percent error is \(\frac{0.18}{8.00}\cdot100\%=2.25\%\). 2. For \(8.18\,\text{cm}\), the error is also \(0.18\,\text{cm}\), so the percent error is \(2.25\%\). 3. For \(8.24\,\text{cm}\), the error is \(0.24\,\text{cm}\) and the percent error is \(\frac{0.24}{8.00}\cdot100\%=3\%\). 4. Only the first two measurements are at or below the \(2.5\%\) limit.

The \(7.82\,\text{cm}\) and \(8.18\,\text{cm}\) parts pass. The \(8.24\,\text{cm}\) part does not.

- Use the accepted mass as the reference value for percent error. - Compare the calculated percentage directly with the lab's maximum allowed percentage. - In part c, think about what happens when the same absolute error is compared with a larger accepted value.

1. The absolute error is \(\lvert196-200\rvert=4\,\text{g}\). 2. The percent error is \(\frac{4}{200}\cdot100\%=2\%\). 3. Since \(2\%>1.5\%\), the reading does not meet the lab requirement. 4. With the same \(4\,\text{g}\) error and an accepted mass of \(400\,\text{g}\), the percent error is \(\frac{4}{400}\cdot100\%=1\%\), which is smaller.

a) \(2\%\) b) No. \(2\%\) is greater than the \(1.5\%\) limit. c) Smaller; the percent error would be \(1\%\).

- First determine the size of the allowed error in the same units as the length. - Percent error gives the magnitude of the difference, not its direction. - Consider both sides of the accepted value.

1. An error of \(5\%\) of \(120\,\text{millimeters}\) is \(0.05\cdot120=6\,\text{millimeters}\). 2. A measurement \(6\,\text{millimeters}\) below the accepted value is \(120-6=114\,\text{millimeters}\). 3. A measurement \(6\,\text{millimeters}\) above the accepted value is \(120+6=126\,\text{millimeters}\).

The two possible measured lengths are \(114\,\text{mm}\) and \(126\,\text{mm}\).