Do I base my sig figs on the initial data given or calculations throughout?

Question

I am solving a problem for volume, using V=m/d. Based on my initial data, I should end up with two sig figs after the decimal in my answer. But before performing this operation I had to calculate V, which only gave me 1 sig fig after the decimal place. So should I have one or two sig figs in my answer?

Answer 1

sig fig rules...



> in multiplication and division, your answer has the same number of sig figs as the factor with the least sig figs.



> in addition and subtraction, your answer can only be as precise as the number with the lowest precision.



> in intermediate steps, you should carry 1 extra sig fig....



> in the event of both mult/div AND add/sub, do what's in parenthesis first.



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examples....



#1...

3.45 / 2.1 = 1.6

why?

3.45 has 3 sig figs

2.1 has 2 sig figs.

so 1.6 can only have 2



#2...

3.045 x 2.10 = 6.39.... why? 3.045 has 4, 2.10 has 3.. therefore 6.39 can only have 3 sigfigs



#3

3.045 + 2.10 = 5.14...

because...

3.045 is precise to the 0.001's column

2.10 is precise to the 0.01's column

so 5.14 can only be precise to the 0.01's column

and when we round...

if the # to be round is even and followed by 5 we drop the 5

if the # to be round is odd and followed by 5 we round up....

so 5.145 rounds to 5.14



#4...

3.055 + 2.10 = 5.16...

same as #3 except we round 5.155 up to 5.16



#5...add extra zeros if you need extra sig figs...

2.160 / 1.08 = 2.00... 3 sig figs... not 2 with 1 sig fig



#6...

if this was an intermediate step...

3.45 / 2.1 = 1.65... we'd carry 1 extra sig fig

same for this...

3.055 + 2.10 = 5.165.... with 1 extra sig fig.



#7...

10.25 x (1.1 + 10.2) = 10.25 x (11.3) = 116

because...

10.25 has 4 sig figs

11.3 has 3 sig figs

so the final answer can only have 3



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questions?

Answer 2

You usually don't round until you are completely done with the calculation. In other words, you carry your sig figs to the very end and then go with the least common denominator (the original data that has the least significant figures). You look at your calculations and come up with a delta after the fact (it's 1294 + or - 7, for example)



There's a whole study of what your + and - delta is depending on the operations you apply and the original significant figures of the data.



So, with initial data...



1000 has one significant figure

1001 has four

0.1 has one

0.011 has two

0.0110 has three

1000.0 has five



1000 + 1 in engineering is technically 1000 after the calculations are done. There are caveats to this, but in most simple calculations it's a simple rule to follow. 1000 * 101 = 101000, but if that 1 was supposed to be a zero, it's only off by a 1000/101000, or a factor of about 10^-2, so that second 1 is not overly critical in most cases.



As most software engineers know too well, though, is that multiplicative errors are a boon in any nonlinear system, which is why analyzing error with respect to the actual calculation is important.



Answer: don't round your in between steps during the calculation, but round to the lowest sig fig in the final answer. Your computer/calculator will round in between to some extent anyway, but that's just something you have to live with.

Answer 3

Okay for example if you were given this information...

lets say you had to convert the following to cm



1.32km + 13.254cm notice how you have 3 sig figs, then 5 sig figs, a common rule states that your answer should have the least amount of sig figs, so your answer should have 3 significant figures. so

1.32km = 132000

now + 13.254 cm so your answers is 132013.254cm but this answer has 9 sig figs. so you would have to change your answer to have 3 significant figures. Your final answer would be 132000cm



Sorry for giving a weird example to get to the point your answer should have 2 sig figs.

Answer 4

nicely, at the start, you opt to renowned the least volume of substantial numbers. that's 2 (0.28 (the 1st extensive form would not count extensive form)) Now purely calculate it: a million.68616^23 may be the respond, yet if you consider which you purely want 2 substantial numbers it is going to grow to be a million.7x10^23