Sig Figs, and Scientific Notation
How to handle numbers without doing more work than you need to, or exaggerating how
precisely you know something. This topic is deceptively simple, so heads up!
1. How many sig figs (significant figures) are in each of the following quantities? In other words, how well do
we know these numbers? Count the number of digits you are sure about.
A. 1 mile
B. 5281 feet
C. 0.0025 mm
D. 200,000,000 m/s
Notice that the last # (D) is a little ambiguous. Is the speed really known to the nearest m/s or are all those zeros
placeholders? Since you can’t tell by looking we use a more definitive notation: Scientific Notation. For
example, let’s say we think that something is moving at about two hundred million m/s. Then we would write
that as 2 x 108 m/s. (The exponent in scientific notation is a count of how many decimal places you have to
move: the exponent is positive if you move to the left and negative if you move to the right.) Check it out:
Since we don’t know this number very well (it is only “about two hundred million m/s”)
we only have one digit to write down: this “2” and we say this number has 1 sig fig!
200,000,000 m/s = 2 x 108 m/s
We say this number has 1 sig fig!
Move decimal 8 places to the left so write the exponent “8” here.
OK, but what if you actually clocked the speed of something at two hundred million m/s to the nearest 1 m/s?
Then in scientific notation, you would write 2.00000000 x 108 m/s since we know there are 9 sig figs. Get it?
2. Write the following in scientific notation, and pay attention to the number of sig figs.
A. 2001 =
B. 0.010 =
C. 93,000,000 (assume 2 sig figs) =
D. 0.0000000250 =
How should you handle sig figs when you have a product or quotient? Check your PJ…
Product/Quotient Rule: Go with the least number of sig figs in a product or quotient.
Example: 3 x 2.3 = 7 . The answer has 1 sig fig because the “3” is only one sig fig.
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Answer the following problems in scientific notation:
3. 41 x 391 =
4. 3.14159 ÷ 2 =
5. 2/3 =
6. 25 x 2 =
7. A defective elevator cable will snap if there are more than 4,500 lbs (assume 4 sig figs) of cargo in it. Can
you safely load twenty eight crates that are 160 lbs (assume 3 sig figs) each in the elevator without the cable
snapping?
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