AC9M9M04Year 9MeasurementPractised on site

AC9M9M04 — Year 9 Measurement

Official Curriculum Content Description:

calculate and interpret absolute, relative and percentage errors in measurements, recognising that all measurements are estimates

Worked Examples (1)

Example 1Limits of Accuracy & Measurement Errors
A quantity is measured as 14.9 cm14.9\text{ cm} with reading precision to the nearest 0.1 cm. Find the lower limit of the true measurement in cm\text{cm}.
Common mistake: Subtracting the full precision interval instead of half the interval.
Correct Answer:14.85 cm14.85\text{ cm}
Accepted: A unit is optional (cm accepted).
Worked Solution:
1. Absolute error: 12×0.1 cm=0.05 cm\frac{1}{2} \times 0.1\text{ cm} = 0.05\text{ cm}. 2. Lower bound = 14.9−0.05=14.8514.9 - 0.05 = 14.85 (14.85 cm14.85\text{ cm}).

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