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.
Three significant figures of 1.702 x 10^5 are 1.70 x 10^5.
The LEFT zeros are place holders, which are not contibuting to significance.0.0360 = 3.60 X 10-2So, actually, there are 3 significant digits. The '3' and '6' and the '0'; this one -AFTER the 6- is not a place holderbut an indicator of the precision of the number - a significant digit.0.036 has 2 significants, and 0.0360 has 3 significant numbers.
The appropriate number of significant figures to use in expressing the result of 51.6 x 3.1416 is three. This is because the factors each have three significant figures, so the result should also have three significant figures. The answer would be 162.
Assuming 5 is not an exact value, the 5 has one significant figure and the 5.364 has four significant figures. 5.364 x 5 = 26.82 However, since 5 only has one significant figure, you would round the 26.82 to be 30 or 3 x 101 30 has one significant figure.
It has 5 significant figures - one trailing zero is significant.
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).
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)
There are two significant figures which are the 5 and the 4. 0.054 = 5.4 x 10^-2
There are 2 significant figures in 7.8x109^?
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).
(4.73*1000*0.568)+1.61 = 2688.25 meaning that it has 6 significant figures
In both cases, there are 2 significant figures.
Three significant figures of 1.702 x 10^5 are 1.70 x 10^5.