Question

Keeping in mind the rules for rounding, significant digits, and scientific notation, what is 975.0321/0.0003

Answer 1

3x10^6 Let's take this problem one step at a time. First, just simply do the division, the numbers are small enough to handle, so x = 975.0321/0.0003 = 3250107 But you have to pay attention to significant digits. So how many significant digits do we have? Looking at 975.0321, I see a total of 7 significant digits. Looks like we have a rather precise value. Now to look at the next number. But how many significant digits does 0.0003 have? Only 1. You can't count the leading zeros as significant even if they come after the decimal point. So the only significant digit is that 3. So now, you have a division problem where 1 number has 7 significant digits and the other has only 1. Just as a chain is only as strong as its weakest link, the number of significant digits you have is only as large as the smallest number. So the number of significant digits you have is only 1. So we need to round 3250107 to 1 significant digit. The first digit is a 3 and second digit is less than 5, so the 3 stays a 3 and all the other digits turn into 0, giving 3000000 So let's convert 3000000 into scientific notation. It's easy, here it is 3000000 x 10^0 But we have a problem. The number has to be at least 1 and less than 10, and of course 3000000 is way too large. So let's get rid of one of those zeros by dividing by 10, and to keep track of that division, add 1 to the exponent of the power of 10, giving 300000 x 10^1 Still not there. Let's divide again by 100. And to keep track of that division, we'll add 2 to the exponent. 3000 x 10^3 Closer, but not quite done yet. Let's divide again by 1000 and add 3 to the exponent. 3 x 10^6 Got it! We have 1 significant digit and the correct power of 10. Of course we could have done that division in one step by counting the number of decimal places we needed to move (6) and then doing it all at once, but I wanted to illustrate that you can also kinda sneak up on the answer and it's OK for intermediate results to be not properly normalized.