# Thread: Correcting to 3 decimal places (3 d.p) and to 3 significant figures (3 s.f)

1. ## Correcting to 3 decimal places (3 d.p) and to 3 significant figures (3 s.f)

In correcting this number: "0.39" to 3 d.p and 3 s.f, would it be as follows?

3 d.p --> 0.390
3 s.f --> 0.390

Thanks.

2. Yes, you are correct.

For any further questions on significant digits, try this website:

Significant Figures

3. Originally Posted by SWEngineer

3 s.f --> 0.390

Thanks.
There are only two significant figures here as the number can be written as .39

4. 0.39 can also be expressed as $3.90*10^{-1}$, in which case there are three significant digits.

5. But, even that the last zero can be ommited after the decimal place, we are asked here for 3 s.f, which means there should be three numbers.

For example, what is the solution for this question:

Correct 0.0004298 to 3 s.f? When you solve it you will know what I mean.

Thanks.

6. 0.0004298 to 3 significant figures would be 0.000430

7. Originally Posted by rtblue
0.0004298 to 3 significant figures would be 0.000430
Yes, this is what I was trying to tell @pickslides

8. Originally Posted by SWEngineer
But, even that the last zero can be ommited after the decimal place, we are asked here for 3 s.f, which means there should be three numbers.
Yes, because the number of "significant digits" tells you the accuracy of the measurement. 0.39 means that, measured to the nearest hundreth, the number is 0.39- that is that the true value is somewhere between 0.385 and 0.395. 0.390 means that it was measured to thousandths place- the true value is between 0.3995 and 0.3905.

For example, what is the solution for this question:

Correct 0.0004298 to 3 s.f? When you solve it you will know what I mean.

Thanks.
0.0004298 to 3 significant figures is 0.000430 or, better, $4.30 \times 10^{-4}$. The fact that it has 3 significant figures means it is accurate to between 0.0004295 and 0.0004305.

9. Thanks a lot for your replies.