Question

question 2 (1 point)
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a student measures the diameter of a disk fifteen times with a ruler that has a smallest division of 1 mm. the average of the 15 measurements is 17.04667 cm and the sample standard deviation is 5.595063×10⁻³ cm. the correctly rounded - off uncertainty (dont use scientific notation) is
a√ cm and the correctly rounded off value is
a√ cm.
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question 3 (1 point)
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the diameter of mars is measured by a satellite to be 6779.987 km with an uncertainty of 44.932 km. the correctly rounded off uncertainty is

Explanation:

Step1: Round the uncertainty for the disk measurement

The sample standard - deviation for the disk measurement is $5.595063\times10^{-3}\text{ cm}=0.005595063\text{ cm}$. When rounding, we look at the first non - zero digit. Since the first non - zero digit is 5 and the digit after it is 5 or greater, we round up. Rounding $0.005595063$ to two significant figures gives $0.0056\text{ cm}$.

Step2: Round the average value for the disk measurement

The average of the 15 measurements of the disk diameter is $17.04667\text{ cm}$. Since the uncertainty is $0.0056\text{ cm}$, we round the average value to the same decimal place as the uncertainty. Rounding $17.04667$ to four decimal places gives $17.0467\text{ cm}$.

Step3: Round the uncertainty for the Mars diameter measurement

The uncertainty in the measurement of the diameter of Mars is $44.932\text{ km}$. When rounding, we look at the first non - zero digit. Rounding $44.932$ to two significant figures gives $45\text{ km}$.

Answer:

For the disk:
Uncertainty: $0.0056$ cm
Value: $17.0467$ cm
For Mars:
Uncertainty: $45$ km