How many sig figs should this quotient have?
((2.89 X 10^4)(10^3)(10^6)) / 45.6
My answer is 633 x 10^8.
I was told I was wrong.
Wouldn’t it depend on which figures were measured? Assuming that 2.89 and 45.6 are the measured figures, I don’t see why that’d be wrong unless you rounded incorrectly or made some other math error.
Well, this excersize as you presented it is highly questionable. What’s the purpose of finding the correct number of sigfigs in an expression with no context? None. The only thing to assume is that it is pure mathematical expression… in which case the answer is roughly 6.337719e+11.
Was there any context to this problem?
I’m starting to think now that the answer to the question is 1 significant figure.
It’s (311)/3
and 1 is the smallest, so the answer has to have 1 sig fig
Well, it depends… how many sig figs are in 2[symbol]p[/symbol]? Answer: an infinite number. Is it 1.*10[sup]3[/sup], a quantity with uncertainty (in which case there’s 1 sig fig) or is 10[sup]3[/sup], an exact number (like 2), in which case there’s 3 sig figs (coming from 2.89 and 45.6, unless, as JS pointed out, 2.89 and 45.6 are also exact numbers, in which case there are an infinite number of sig figs).
Really, I’ve always thought the entire thing was a little arbitrary and just try to take a reasonable number of digits without worrying too much about significant figures. But of course, you can’t do that in a class.
Normally in multiplication and division you are limited to the number of digits contained in the number with the fewest significant figures. Has your instructor given you a different idea?
I see three sig figs as well. Check with your teacher.
Also see this thread for a discussion of why sig figs are dumb.
I think it should be 6.34 x 10^11. Here’s my raw number
633771929824.561403508771929824561
When I rounded it I rounded the 3 to a 4 since the numbers after it make more than 1/2. Also I maybe haven’t counted the number of “tens” incorrectly but it seems to me I have to shift the decimal point over 11 spots to get it to the right place.(Normally I’d expect only one digit in front of the decimal place.)
When you were told you were wrong, was it specified that you were wrong because of the sigfigs, or because you didn’t put it into scientific notation and truncated rather than rounding?
For the record, I’m now entering my third year of studying physics at University. With that said, we’re taught to take the lowest number of sig figs (3, in this case - I’m assuming the factors of ten come from unit conversions, and thus are exact) and then reduce it by one to get the appropriate number of sig figs for the answer. Thus, 6.3 x 10^11.
One thing to be careful of - where did the 3-figure numbers come from? If you calculated them separately, you should use the numbers to as much precision as you calculated. Number of sig figs is determined by any numbers supplied to you in the problem.
If we write the equation out:
28900(3sigfig)*1000(1sigfig)*1000000(1sigfig) / 45.6(3sigfig)
we see that 1000 and 1000000 have 1 sig fig each. 1 is the smallest number of sig figs, ergo, I’d assume answer needs 1 sig fig?
That is not correct. 10[sup]3[/sup] = 1000 exactly. There is no rounding off involved, so it has infinitely many sig figs.
This thread is an excellent illustration for how significant figures have been made a needless waste of time in certain educational settings.
Muldoon III, my advice to you is to simply move on with your life. There is actually no particular value in figuring out the arbitrary way your teacher wishes to approach significant figures.
They are a ROUGH guide, nothing more. They actually have no practical meaning. They are meant so that you do not blindly copy down a ridiculous string of digits in an answer akin to the “raw numbers” presented by Dave_D. I would say it is time to move on from this project.
Am I right in assuming there to be no context to this question other than the mathematical expression? If that’s the case, if your science teacher is assigning context-less arithmetic problems do be done to establish “significant figures”, it is completely pointless work. If there is no context for a calculation there is no reason to be persnickety about precision. It’s as simple as that.
Okay
I have to disagree with this statement. Sig figs do have a practical meaning, and aren’t just a guide. They give you the maximum precision that you can expect to discern from a value. If I measure a voltage as 1.035 V using my little volt-meter, and unless the specs for the volt-meter tell me otherwise, I can assume that my reading is accurate to within +/- 0.001V, which in this case is about 0.01% of the value. If I then plug that value into a bunch of equations to calculate, say, the resistence over this voltage drop occurs, I need to preserve that notion of precision. If I choose to select a number of sig figs for my answer other than 4, I’m misrepresenting the accuracy of my calculation. If I only use 2 sig figs, I’m saying that my value is only accurate to within roughly 1%. Now certainly this isn’t precise, as it only tells you the order of magnitude of the error in measurement, but it’s far from “having no practical meaning”.
If you want to get really anal about it, you can explicitly carry the measurement error throughout all your calculations. So you have (1.035 +/-0.001)*(5) = (5.165 +/-0.005), but in most applications this is excessive.
One other thing (and this is just how I was taught, so it shouldn’t be taken as gospel), but if Muldoon III’s prof is being particularly formal about this problem, then even in the absence of context it has an answer. 2.89, absent context, has 3 sig figs, unless explicitly stated other wise. If it was meant to convey more than that, it would be written as 2.890 (4 sig figs), 2.8900000 (8 sig figs), and so on. So given this, the answer to this question in a vacuum would have 1 sig fig.
Jeff
Jeff, but if there is no measurement made… i.e. you’re given a bunch of numbers with no uncertainties or measurement units… it’s anybody’s guess as to what is expected. This is NOTHING like your example because that’s actually a measurement with uncertainty. This is simply an expression without context.
It’s akin to asking, “How many significant figures in the reduced decimal form of 1/3?” Okay, there’s no context so can I simply say “One significant figure, the answer is 0.3”? But that’s also the answer to 3/10. Are we saying 1/3=3/10? Are we saying that 9/10=1? Well, sometimes that’s a good assumption to make, it all DEPENDS ON THE CONTEXT. As there is no context to the problem there is no way to determine what the answer is (or, one can assume whatever threshhold one wishes… it’s completely and utterly arbitrary).
Also, Jeff, you should check out the link provided by ultrafilter. Crafter_Man is a little brash, and I don’t agree with his complete dismissal of the utility of teacing about significant figures, but he basically knows what he’s talking about. Sig figs are only a pedagogical technique for getting students to think about uncertainty… NOTHING MORE. Just to warn you if Crafter_Man comes by and reads your post he’ll go nuts. The number of significant figures is irrelevent once you establish the uncertainty. Ironically, the real danger in such reporting is NOT REPORTING ENOUGH significant figures.
See what really getting anal means:
You confuse precision with accuracy. Accuracy is how close A reading is close to the actual. And Precision is, if you measure the thing again and again - how close are your values distributed (the standard deviation).
Also since you made such a huge cry, instruments don’t usually measure to x% (like 0.01% in your above case) accuracy of the READING but x% of the full scale. So when your Voltmeter says its accuracy is 0.01% - it means 0.01 % of the full scale (If your full scale is 0-5V and accuracy is 10%, if you measure 2 volts, it means the accuracy is 0.5V NOT 0.2V).
I can go on and on this. So sig figs are better left to the particular measurements and generalizing the concept creates problems.
Argh!!! Nah, just kidding. 
I’ve calmed down a little since that previous thread.
“Calculating the number of significant digits” offers a fun, intellectual exercise, but has no value in real life. If a co-op came to me with this problem, and assuming the quantities in question represented output from IM&TE[sup]a[/sup], I would ask the following questions:
Do the “10^x” quantities represent measured quantities, or are they scale factors? (I would assume the latter, but you never know.)
What is the readability of the IM&TE that gave you “2.89”? Is it X.XX?
What is the readability of the IM&TE that gave you “45.6”? Is it XX.X?
What is the measurement uncertainty of the IM&TE that gave you “2.89”?
What is the measurement uncertainty of the IM&TE that gave you “45.6”?
What is the confidence interval?
Based on the above, I would calculate the answer, including measurement uncertainty. Once I know the uncertainty, I can then determine how many “significant digits” to report. This is how you do it when you sell data for a living.
And I completely agree there’s little harm in displaying too many significant digits. In fact, if you properly calculate and report the final uncertainty value, there’s no such thing as reporting too many significant digits.
[sup]a[/sup][sub]Instrument, Measurement, and Test Equipment[/sub]
Readability is not such an easy thing to figure for a co-op and you conviniently skipped the calculations. Oh and BTW - uncertainty is again a “very grey” area.
Calculus tells you, how to calculate the sig figs (warning : its still a futile exercise). this is how calculus tells you :
A = x / z ( x = 2.89 X 10^4)(10^3)(10^6)) and y = 45.6 , assuming the 10^ s are just scale factors)
Take logs of both sides
ln(A) = ln(x) - ln(z)
Take the total differential
delta(A)/A = delta(x)/x - delta(z)/z
Now, consider the worst case scenario i.e. the errors add up.
delta(A)/[A] =delta(x)/ + delta(z)/[z] ( is for the absolute value)
put in the delta(x) and delta(z), i.e. the uncertainty in your measurement and get delta(A).
Hope that helps