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Lorico [155]
3 years ago
11

a box was measured with a degree of accuracy to the nearest 2cm; 24cm x 24cm x 20cm. what is the largest possible volume of the

box to the nearest cm3
Mathematics
1 answer:
DochEvi [55]3 years ago
5 0
For any value, e.g. a, to the nearest x units, the upper and lower bounds are:
(a + x/2) and (a - x/2)
For your question, there are three dimensions so:
Dealing with the first one, we have 24 cm to the nearest 2 cm so the boundaries are:
Upper boundary: 24 + 2/2 = 25
Lower boundary: 24 - 2/2 = 23
The second dimension is the same as the first in value and is also given to the nearest 2 cm so the boundaries are the same as for the first.
The third dimension is 20 cm to the nearest 2 cm so the boundaries are:
Upper boundary: 20 + 2/2 = 21
Lower boundary: 20 - 2/2 = 19

To get the largest possible area, we take the upper bounds of all the dimensions and multiply them so:
25 * 25 * 21 = 13125 cm³
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