A student measures a distance several times. The readings lie
between 49.8 cm and 50.2 cm. This measurement is best recorded as
A. (49.8 +0.2) cm. B. (49.8 +0.4) cm.
C. (50.0 +0.2) cm. D.( 50.0 + 0.4 )cm.
Answers
Answer:
c is your answer
hope it helps
Correct Question: A student measures a distance several times. The readings lie between 49.8 cm and 50.2 cm. This measurement is best recorded as
(A.) (49.8 ± 0.2) cm (B.) (49.8 ± 0.4) cm
(C.) (50.0 ± 0.2) cm (D.) (50.0 ± 0.4) cm
→ The correct answer is (C.) (50.0 ± 0.2) cm.
→ (50.0 ± 0.2) cm is the best way to record the distance because it contains all the readings taken and it includes both the maxima and minima of the readings as its boundary value. It includes both 49.8 cm as well as 50.2 cm as its boundary values.
→ (50.0 ± 0.2) cm will contain all of the measurements recorded and will not contain any value other than the recorded values.
→ During scientific experiments, the value of a quantity (Q) is recorded in the following format: (Q ± ΔQ), where 'Q' is the mean of all the observations and 'ΔQ' is called the error of limit. The magnitude of the quantity lies between (Q - ΔQ) and (Q + ΔQ).
→ Hence (Q ± ΔQ) contains all the readings taken and it includes both the maxima and minima of the readings as its boundary. It does not contains any value other than the recorded values.
Therefore (50.0 ± 0.2) cm is the best way to record the distance.
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