4300.728 can be round off to 3 figures by
firstly 4300.73
then 4300.7
then 4301
then 430 now i have round off that into 3 figures
thanx
3
It has 3 significant figures.
it is 3 if you're talking about sig figs. It is 6 if you're talking about an int object in cse.
Ah, a lovely question about significant figures. In the number 0.800, there are three significant figures. Remember, zeros between non-zero digits are always significant. Keep painting with those numbers, my friend!
It depends to how many significant figures: 1 sig fig => 10 (technically 1 x 101 or 1e1) 2 sig fig => 13 3 sig fig => 12.7 4 sig fig => 12.74 5 sig fig => 12.744
45.0
2.0
there are 3 sig figs. 4, 0, and 5 are the sig figs
When adding or subtracting, follow these steps to find a sig figs answer: 1) Add/Subtract numbers regularly. 2) Determine which measurement has the least decimal places. 3) Round final answer to the same number of decimal places. When multiplying or dividing, follow these: 1) Count the number of sig figs in the numbers you are multiplying/dividing. 2) Multiply/Divide regularly. 3) Round final answer to the same number of sig figs as the measurement with the fewest sig figs.
There are 3 sig figs becasue when you multiply two numbers you take the same number of sig figs as the multiple with the fewest sig figs. 0.280 and 3.70 both have 3 sig figs.
12.0 if you want 3 sig-figs, or 12 if you only want 2.
Depending on the number of significant figures, it could be the following:4.251x10^3 (4 sig figs)4.25x10^3 (3 sig figs)4.3x10^3 (2 sig figs)4x10^3 (1 sig fig)
Depending on how many significant figures you want, it can be2.421200x10^6 (7 sig figs)2.42120x10^6 (6 sig figs)2.4212x10^6 (5 sig figs)2.421x10^6 (4 sig figs)2.42x10^6 (3 sig figs)2.4x10^6 (2 sig figs)2x10^6 (1 sig fig)
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