
So this is driving me crazy. In my Chemistry class, we are learning about retaining significant figures and digits of precision in problems with Mixed Operations. We are told not to round until the end, but to use superscripts to retain significant figures/digits of precision. I have submitted this answer and it was marked incorrect. PLEASE HALP!! Thank you!!!
0.5230 73.152 - 2.35
-------- - ---------------- * 5.72
1.2941 0.001
First I take care of my fractions:
0.5230/1.2941 = 0.4041
41 (4 SFs retained)
73.152-2.35 = 70.80
2 (2 digits of precision retained)
70.80
2/0.001 = 70802 (w/b 70000 for 1 SF)
Problem is now:
0.4041
41 - 70000
(70802) * 5.72
Multiplication first:
70802 * 5.72 = 404987
.44 (0 digits of precision)
Then Subtraction:
0.4041
41 - 404987
.44 = -404987.0359
-404987.0359 = -404987.04 (2 digits of precision)
I've tried submitting -404987.04 AND -404987 and both are incorrect. I can't figure it out! Ripping my hair out! Please help
Why would you submit -404987.04? That has 8 SF. You started with a maximum of 5 SF.
Why would you submit -404987? That has 6 SF. You started with maximum of 5 SF.
You seem to be confusing FIGURES with DIGITS to the right of the decimal point.
You may have mistaken one of your tiny '2's for a real '2'.
Please identify the number of significant figures in each value:
0.5230 -- (4 SF)
73.152 -- (5 SF)
2.35 -- (3 SF)
5.72 -- (3 SF)
1.2941 -- (5 SF)
0.001 <== Personally, I would consider this a Scaling Factor, and not anything else, unless I know something else about it. Example: 2.73 and 2.73E4 both have 3 Significant Figures.
4, 5, 3, 3, 5 -- You should end up with 3 Significant Figures
0.5230/1.2941 = 0.404141875 ==> 0.4041 (4)
73.152 - 2.35 = 70.802 ==> 70.8 (3)
70.8 / 0.001 = 70.8E-3 ==> 70,800 (3)
70,800 * 5.72 = 404,976 ==> 405,000 (3)
0.4041 - 405,000 = -404,999.959 ==> -405,000 (3)
Keep this in mind when adding values of substantially different magnitudes:
Mass of the Earth + One Grain of Sand = Mass of the Earth
Try another one and show us your clean and careful work.