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(C) $1$. The given readings are $5.50\, mm, 5.55\, mm, 5.45\, mm,$ and $5.65\, mm$. All these readings have $3$ significant figures.$2$. The average v

Vedclass · Accepted Answer

$(5.54 \pm 0.07)\, mm$

Vedclass · Answer

(C) $1$. The given readings are $5.50\, mm, 5.55\, mm, 5.45\, mm,$ and $5.65\, mm$. All these readings have $3$ significant figures.$2$. The average value is $5.5375\, mm$. According to the rules of significant figures,the result should be rounded off to the same number of significant figures as the measured values,which is $3$. Thus,$5.5375$ is rounded to $5.54\, mm$.$3$. The standard deviation is $0.07395\, mm$. The uncertainty (error) should generally be expressed to one or two significant figures. Rounding $0.07395$ to one significant figure gives $0.07\, mm$.$4$. Therefore,the diameter should be recorded as $(5.54 \pm 0.07)\, mm$.

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