true
But according to the rules of significant figures, the least number of significant figures in any number of the problem determines the number of significant figures in the answer which, in this case, would be 11.
according to my physics teacher, there are four significant figures in 0.01500
273.821 contains 6 significant figures. Rounding that to Hundredths (5 sig. figures) it becomes 273.82 Rounding it to Tenths (4 sig. figures) it becomes 273.8 And Rounding it to the nearest whole number (3 sig. figures) it becomes 274 etc., etc.
There are actually 6 significant figures in 1.02300 because the trailing zeros after a non-zero number are significant.
1. All non-zero digits are significant. For example, 295 has three significant figures. 2. Leading zeroes in front of a decimal are not significant. For example 0.295 has three significant figures. 3. Zeroes between other significant figures are significant. For example 2095 has four significant figures. 4. Trailing zeroes after a decimal are significant. For example 295.0 has four significant figures. And 2950 has three significant figures because the trailing zero does not occur after a decimal.
Five, everything up to the zero in the hundredths place is a significant figure.
No, because it has 3 significant figures
There are some rules for finding significant figures. here there is a problem how many significant figures in 8.00. here in 8.00 have three significant figures. Because after decimal point they may have zeros. but we have to take this as significant figures. There are some rules for finding significant figures. here there is a problem how many significant figures in 8.00. here in 8.00 have three significant figures. Because after decimal point they may have zeros. but we have to take this as significant figures. there are three significant figures because three decimals points these question answering from anjaneyulu
There are five significant figures in the given value. It is according to the rule of significant figures which say that zeros right to the decimal point are significant and all non zero digits are significant So , all the digits in the given value are significant figures i.e 5 significant figures.
But according to the rules of significant figures, the least number of significant figures in any number of the problem determines the number of significant figures in the answer which, in this case, would be 11.
There are four significant figures in the number 0.006709.
There are three significant figures in 0.00701 because the initial zeros are non-significant and the 0 is significant because it is sandwiched between two significant numbers.
Yes, because the zero is between two other significant figures, it is significant.
9843.6 has five significant figures because all non-zero numbers are significant.
0.007001 has four significant figures. The leading zeroes are not significant. The zeroes between the 7 and 1 are significant because they are between two significant figures (7 and 1).
Just leave them alone because they are already in 4 significant figures.
None of them because they have 3 and 2 significant figures respectively