Rounding to required level of accuracy

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The discussion centers on calculating results to the required level of accuracy based on significant figures. For part (a), the calculated answer of 94.5 is considered correct since 1.95 limits the significant figures to three, while the textbook suggests 94.54. In part (b), the participant's answer of 0.480 is deemed correct due to the three significant figures of the factors involved, but the textbook states 0.4806, indicating a discrepancy in rounding rules. The conversation highlights confusion over the application of significant figure rules, particularly in relation to addition and multiplication. Participants express concern about the textbook's unusual rules and the potential for unexpected answers in engineering math problems.
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Homework Statement



Calculate each of the following to the required level of accuracy given that each number, other than those indicated in brackets, has been obtained by measurement.

a) (3.142*1.95)/6*(3*5.44^2+1.95^2)
(power 2, divisor 6, multiplier 3)

b) (3.142*1.234)/12*( 0.424^2+0.424*0.951+0.951^2)
(power 2 and divisor 12)

Homework Equations





The Attempt at a Solution



a)My calculator gives me 94.5414368

So I think 94.5 should be the answer since 1.95 has the least number of significant figures- three. My book, though, says it is 94.54

b) The calculator gives 0.4805827337

So my answer is 0.480 because 0.424 and 0.951 each have only 3 sig figs- least number of sig figures. But my book says the answer is 0.4806


Please, show me where I am wrong.

Thanks.
 
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The rules for sig. fig.s are different for (addition&subtraction) than they are for (multiplication&division).

Your answer (a) looks right to me. 1.95 is a factor of all the rest, so it should limit the result to 3 sig. fig.

For (b) however, each factor in ( 0.4242+0.424*0.951+0.9512) is accurate to the 0.001 place, so the result of adding these three factors is also accurate to the 0.001 place. That give 4 sig. fig. for that result. Each of the other two factors, 3.142 & 1.234 also has 4 sig. fig.s.
 
SammyS said:
The rules for sig. fig.s are different for (addition&subtraction) than they are for (multiplication&division).

Your answer (a) looks right to me. 1.95 is a factor of all the rest, so it should limit the result to 3 sig. fig.

For (b) however, each factor in ( 0.4242+0.424*0.951+0.9512) is accurate to the 0.001 place, so the result of adding these three factors is also accurate to the 0.001 place. That give 4 sig. fig. for that result. Each of the other two factors, 3.142 & 1.234 also has 4 sig. fig.s.

Thank you for the answer. I tried to apply the rule that used for (b) to another problem and got another wrong answer.

[4.62^2-(7.16-2.35)]/ [2.63+1.89* √(73.24)]

The answer is 0.8793. I thought the answer had to be rounded to 3 sig fig.
 
roger12 said:
Thank you for the answer. I tried to apply the rule that used for (b) to another problem and got another wrong answer.

[4.62^2-(7.16-2.35)]/ [2.63+1.89* √(73.24)]

The answer is 0.8793. I thought the answer had to be rounded to 3 sig fig.
Beats me!

I agree with you on this one.

Does your textbook have some unusual rules for these?
 
Maybe the author treats "required level of accuracy" different than "significant figures".
?
 
SammyS said:
Beats me!

I agree with you on this one.

Does your textbook have some unusual rules for these?

It's Engineering Math by Stroud. It's a great book. Sometimes unexpected problems with strange answers pop up. These are some of the few.
 
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