Track accumulated rounding error and model storage with powers of ten
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Track accumulated rounding error and model storage with powers of ten
Track accumulated rounding error and model storage with powers of ten
Example 1
Each of 3900 particles has mass 5.32×10−5 grams. An instrument rounds the mass of each particle to 2 significant figures before adding the masses. Compared with the exact total, what is the difference in nanograms? Use 1 gram =109 nanograms. Enter a positive number for an overestimate and a negative number for an underestimate.
Show worked answer
Answer: −780000
The rounded coefficient is 5.3.
Subtract the exact coefficient, multiply by 3900
convert grams to nanograms: (−0.02)×10−5×3900×109=−780000.
Example 2
A recorder receives 5×104 events per second. Each event uses 2×102 bytes, but only 60% of events are retained. A drive holds 14×108 bytes, with 1×107 bytes reserved. How many complete seconds of recording can it store?
Show worked answer
Answer: 231
Retained data uses 60×105 bytes per second.
Available space is 14×108−1×107 bytes.
Divide space by rate and round down to 231 complete seconds.
Example 3
Each of 3800 particles has mass 5.31×10−5 grams. An instrument rounds the mass of each particle to 2 significant figures before adding the masses. Compared with the exact total, what is the difference in nanograms? Use 1 gram =109 nanograms. Enter a positive number for an overestimate and a negative number for an underestimate.
Show worked answer
Answer: −380000
The rounded coefficient is 5.3.
Subtract the exact coefficient, multiply by 3800
convert grams to nanograms: (−0.01)×10−5×3800×109=−380000.
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