Count the significant figures in each number. Calculate the minimum of these numbers.
Do the multiplication
Round the product to the LEAST number of significant figures, determined above.
Wiki User
∙ 7y agoWhen multiplying, the number of significant numbers in the answer should be the same as the fewest significant figures in the problem. Both 13.5 and 3.00 have three significant figures, so the answer will have three significant figures. 13.5 x 3.00 = 40.5 exactly (no need to round).
There are 2 significant figures in 7.8x109^?
There are two significant figures which are the 5 and the 4. 0.054 = 5.4 x 10^-2
In both cases, there are 2 significant figures.
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.
4.884 has four significant figures and 2.25 has three significant figures. 4.884 x 2.25 = 10.989 = 11.0 rounded to three significant figures. When multiplying or dividing, the result must have the same number of significant figures as the number in the problem with the fewest significant figures.
When multiplying, the number of significant numbers in the answer should be the same as the fewest significant figures in the problem. Both 13.5 and 3.00 have three significant figures, so the answer will have three significant figures. 13.5 x 3.00 = 40.5 exactly (no need to round).
It has 5 significant figures - one trailing zero is significant.
The number 1.84 x 103 has three significant figures, 1.84. The 103 part of the number does not count when determining significant figures.
Four significant figures in 3.895.
6.5211 x 104 = 678.1944 678.1944 has 7 significant figures
5 significant figures Each figure that contributes to the accuracy of a value is considered significant. So 2.9979 has 5 significant figures. The 10^8 does not contribute to the accuracy as it simply indicates the number of trailing zeroes (i.e. 299,790,000) that are simply a result of rounding from the actual value (299,792,458)
Significant figures are very important when it comes to calculations. If the mass of an electron is 9.10939 x 10-31 then its significant figures are: 9 x 10^-31( correct 1 significant figure), 9.1 x 10^-31 kg ( correct to 2 significant figures), 9.11 x 10^-31 (correct to 3 significant figures), and 9.109 x 10^-31 (correct to 4 significant figures).
There are 2 significant figures in 7.8x109^?
There are two significant figures which are the 5 and the 4. 0.054 = 5.4 x 10^-2
(4.73*1000*0.568)+1.61 = 2688.25 meaning that it has 6 significant figures
Three significant figures of 1.702 x 10^5 are 1.70 x 10^5.