00:02
For these problems, we need to perform the calculation, round to the correct number of significant digits, and include the proper units.
00:10
For problem a, we're dividing.
00:13
When we divide for significant digits, we're going to count the number of significant digits.
00:18
For this first one, we have three significant digits and the denominator has four significant digits.
00:29
So in our answer, we're going to round to the fewest, which is three significant digits.
00:34
So if we put it on the calculator, you get 2 .05865...
00:41
It continues on.
00:43
We want to round to the correct number of significant digits, which is only three.
00:47
So we're going to round 2 .06.
00:50
So the eight after the five rounds it up to six.
00:54
Our units, we're just going to follow the units that were in the problem.
00:58
So grams per milliliter.
01:00
For the second one, when we add, we look at decimal places.
01:05
So 4 .02 has two numbers after the decimal place.
01:09
0 .001 has three numbers after the decimal place.
01:14
We're going to take the shortest.
01:15
So when you put it on the calculator, you would get 4 .021, but we're going to turn that into having only two places after the decimal point, which is 4 .02.
01:32
Our unit milliliters plus milliliters is just milliliters.
01:36
It's like combining like terms in math.
01:39
All right.
01:40
For c, we have a numerator, which we are going to subtract.
01:45
And both of these have only one decimal place.
01:48
So we can round our answer to only one decimal place, which is 14 .1 grams on the top for the numerator.
01:56
For the denominator, both of them have three decimal places.
02:01
So we are going to keep three decimal places in our answer there.
02:06
So 1 .140 milliliters.
02:09
Include that zero to show that it is significant.
02:12
Now, when we're dividing, we count all our significant digits.
02:16
So this has three significant digits.
02:18
This has four significant digits...