2300.m has four significant figures.
The least number of significant figures in any number of the problem determines the number of significant figures in the answer.
There are 2 significant figures in this measurement.
Two. When multiplying or dividing the answer is rounded to the fewest significant figures in the given measurements. 0.55 has only two significant figures, so the answer can have only two significant figures.
Three significant figures: two before the decimal point and one after.
Just one.
2300.m has four significant figures.
125 m in two significant figures is: 120 m
2 significant figures.
There are three significant figures in 4.00
Six significant figures.
The least number of significant figures in any number of the problem determines the number of significant figures in the answer.
When rounding 1286 to 2 significant figures, you must consider the first two non-zero digits from the left. In this case, those digits are 1 and 2. The digit to the right of 2 (8) is greater than 5, so you round up. Therefore, 1286 rounded to 2 significant figures is 1300.
There are 4 significant figures in this measurement.
There are 3 significant figures in this measurement.
There are 2 significant figures in this measurement.
Two. When multiplying or dividing the answer is rounded to the fewest significant figures in the given measurements. 0.55 has only two significant figures, so the answer can have only two significant figures.