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Justify the number of significant figures for xby koat
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#1
Mar711, 01:15 PM

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hi we did an experiment at school and in my homework i am asked to justify the number of significant figures for x.
what do they mean with justify here? i dont understand the question. my x value has 3 s.f. thanks in advance 


#2
Mar711, 02:05 PM

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If your x value is a product or sum of different numbers, then it can only contain a certain amount of significant figures, and you can't justify any more than that. So if in your experiment you ended up multiplying 2.13 by 4.17, your calculator gives an answer of 8.8821, but per the sig fig rule, you can only justify the answer to 3 sig figs, so the answer is 8.88. Is that what you mean for your 'x' value?



#3
Mar711, 02:38 PM

P: 40

thanks but that's not actually what i mean with my x value.
the x value was an unknown in an equation and i had to work out this x value. it represents the mass of a mug. then after i have made the x value the subject of my equation they ask me that question which i have written in my first post. what do i have to do? and what does justify mean in this context? 


#4
Mar711, 02:52 PM

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Justify the number of significant figures for x



#5
Mar711, 03:11 PM

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Oh I think I understand .
But say the equation was y=0.6zx so x= y/0.6z x and y are given to 3 s.f But the 0.6 is just given to 1 s.f. will x still have 3 s.f. or does the 0.6 affect the number of s.f of x? 


#6
Mar711, 05:00 PM

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When multiplying or dividing, the answer is only as good as the number of the least significant digits of any numbers in the problem. Since 0.6 has only 1 sig fig, then the answer can only have one sig figure. Unless the '0.6' is an exact known number. As for example how much is 9 times 9....you can't say 80 , which is correct to one sig fig, or you'll flunk 2nd grade. But if you weighed something by experiment and it weighed 9 N, then 9 of those somethings would weigh 80 N. If you weighed it at 9.0 N, then 9 of them weigh 81 N. Sig fig rules get confusing.



#7
Mar811, 03:41 AM

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but in my equation the 0.6 is an exact known number and all the other numbers are measured to3 sf
how about now? 


#8
Mar811, 04:06 AM

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#9
Mar1011, 12:57 PM

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When I 've got y=0.6zx then you get x=y/0.6z At school we measured y and z to 3 sf. So does it mean that 0.6 is kind of a constant value and it doesn't affect the value of x which I am trying to work out? Because I didn't work out 0.6. This was the only number given in the equation but the other numbers which I have plugged into the equation are measured values. So does x also have 3 sf then? Can you explain that again please. I am slightly confused. Thanks in advance 


#10
Mar1011, 02:55 PM

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#11
Mar1011, 03:07 PM

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When you say 0.6 is an exact known number is that the same as saying it's a constant?
And are you trying to tell me that a constant doesn't affect the final result ? Am i right? If I would have this example: (0.333)(0.550)/0.2 where 0.2 is the value already given in the equation( so a constant?) and 0.333 and 0.550 are measured values. Is the answer going to have 3 sf???? 


#12
Mar1011, 03:20 PM

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Thus, if you were asked to calculate the circumference of a circle (C = pi(diameter)) having a measured diameter of 123.4 units (4 sig figures), and the problem said use pi= 3.14 (which has 3 sig figs), then the answer is C = 387 units (3 sig figs). But if it said calculate the circumference of the circle but did not put an approximation on pi, then C = 387.7 units. 


#13
Mar1011, 03:29 PM

P: 40

Ok I understand what you are saying but can you also answer the last bit of my question with the calculation please :)
And does 0.550 actually have 2 or 3 sf? And how do I know that 0.2 is an exact known number this value is just given in the equation ? thanks 


#14
Mar1011, 03:37 PM

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