III. Calculate the absolute errors and the percent relative errors for each of the following using propagation of uncertainty rules, then round the results to the correct number of significant figures. 1. 100.60 (±0.08) - 99.82 (±0.06) + 0.600 (±0.004) = 2. -1.15 (±0.02) ×107 +3.94 (±0.01)×108 = 3. 1.63 (±0.02) ×105 ÷ 2.74 (±0.03) ×104 = 4. [102.34 (±0.06) + 21.444 (±0.004)] ÷ 5.33 (±0.02) =
III. Calculate the absolute errors and the percent relative errors for each of the following using propagation of uncertainty rules, then round the results to the correct number of significant figures. 1. 100.60 (±0.08) - 99.82 (±0.06) + 0.600 (±0.004) = 2. -1.15 (±0.02) ×107 +3.94 (±0.01)×108 = 3. 1.63 (±0.02) ×105 ÷ 2.74 (±0.03) ×104 = 4. [102.34 (±0.06) + 21.444 (±0.004)] ÷ 5.33 (±0.02) =
Biochemistry
9th Edition
ISBN:9781319114671
Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Chapter1: Biochemistry: An Evolving Science
Section: Chapter Questions
Problem 1P
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![III. Calculate the absolute errors and the percent relative errors for each of the following using
propagation of uncertainty rules, then round the results to the correct number of significant figures.
1. 100.60 (±0.08) - 99.82 (±0.06) + 0.600 (±0.004) =
2. -1.15 (±0.02) ×107 +3.94 (±0.01)×108 =
3. 1.63 (±0.02) ×105 ÷ 2.74 (±0.03) ×104 =
4. [102.34 (±0.06) + 21.444 (±0.004)] ÷ 5.33 (±0.02) =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa5822045-42a2-44b8-8b3c-cf2f483f17e7%2F312b1a75-949f-4e71-911b-e40d534c80ff%2F1etmm5q_processed.jpeg&w=3840&q=75)
Transcribed Image Text:III. Calculate the absolute errors and the percent relative errors for each of the following using
propagation of uncertainty rules, then round the results to the correct number of significant figures.
1. 100.60 (±0.08) - 99.82 (±0.06) + 0.600 (±0.004) =
2. -1.15 (±0.02) ×107 +3.94 (±0.01)×108 =
3. 1.63 (±0.02) ×105 ÷ 2.74 (±0.03) ×104 =
4. [102.34 (±0.06) + 21.444 (±0.004)] ÷ 5.33 (±0.02) =
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