rectangle Question (1 Viewer)

nsbrando

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The sides of a rectangle were recorded as 15m by 22m correct to the nearest metre.

i. Within what range of values do the actual length and breadth lie?

ii. Within what range of values does the exact area lie?
 

Vendetta VI

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Bit late for an answer but here:

i: Limits of accuracy, precision to nearest cm, therefore actual is + or - 0.5 cm:
L= 14.5<-->15.5 (cm) (minX), (maxX)
W= 21.5<-->22.5 (cm) (minY), (maxY)

ii: L*W=A, find actual, minX*minY and maxX*maxY
311.75 cm^2 <--> 348.75 cm^2

2 marks! :)
 

Girls

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Bit late for an answer but here:

i: Limits of accuracy, precision to nearest cm, therefore actual is + or - 0.5 cm:
L= 14.5<-->15.5 (cm) (minX), (maxX)
W= 21.5<-->22.5 (cm) (minY), (maxY)

ii: L*W=A, find actual, minX*minY and maxX*maxY
311.75 cm^2 <--> 348.75 cm^2

2 marks! :)
I don't do general so correct me if I'm wrong, but the question that OP posted is in metres, not centimetres.
Thus, wouldn't the range for (i) be 14.5-15.49 metres? I understand in general there is then some form of accuracy buffer you have to add in.
 

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