A. When you are rounding to the nth significant figure, we look at the (n +1) figure and make a decision as to whether we round up or down based on the value of that.  In our particular case we want to round to one significant figure so we look at the second nonzero number which is 9. 9 is clearly greater than or equal to 5 therefore we would round up our first figure, so in this case 69,212 to one significant figure would be 70,000.


A. We are working to 3 significant figures so we start from the left and move across to the fourth one, the fourth figure is one, which will round down therefore our number 16,412 to 3 significant figures become 16,400.


A. Remember that the first significant figure is always the first nonzero number, so in this case it will be (working from left to right) the number 9. In this case the 4th significant figure (remember n +1) is 5, which has to be rounded up. Therefore our number 0.00097151 rounded to 3 significant figures becomes 0.000972


A. Remember once again that our first significant figure is our first nonzero digit starting from the left and moving to the right, in this case our first nonzero digit is 8 so this is our first significant figure. As we are rounding to one significant figure we look at the second one which is 9 and so we realise we have to round up. Therefore the density of hydrogen gas in kilograms per cubic metre to one significant figure is 0.09


A. Looking at the number 238,000 and moving to the second significant figure we can see that this is a 3. The third significant figure is an 8 which will round up so the distance between the Earth and the moon in kilometres to 2 significant figures is 240,000.


Sometimes we need to just estimate numbers, just to give a guideline or an "rule of thumb" approximation where a simple guess will do pending (presumably) a further, more accurate calculation. For example if you were buying turf to re-lay your back garden you probably wouldn't measure it to the inch, probably not even to the nearest 6 inches or foot (that is perhaps the nearest centimetre or half meter/meter if we are working in SI units). For example my back garden is approximately 10 m wide and about 8 m from bottom to top, so I would need approximately 80 m² of turf. This will be sufficient for me to go to the nearest garden centre and obtain an approximate cost. When it came to an actual purchase I would probably get out a tape measure and be a little bit more accurate in my measurements for my final order.


So there is an example of why we might approximate. Using our knowledge of decimal places and significant figures we can now start to make approximations, let us take a look at a few examples:


A. 

(a) Our first non-zero number in 102.2 is one, therefore this number to one significant figure would be one hundred. In the case of 4.2 our first significant figure is 4 and because the second significant figure in this case is lower than 5 we round downwards. The answer therefore to 102.2×4.2 approximates to 100×4 which equals four hundred 400.


(b) 3.9 to 1 significant figure is 4.0, 5.1 to one significant figure is 5 so the answer to our problem 3.9÷5.1 approximates to 4/5 or 0.8.


(c) In the number 142.75 our first significant figure is 1, and because the second one is below five we round downwards so 142.75 to 1 significant figure is one hundred 100. The second number 9.56 has a first significant figure of 9 so we look to the second significant figure which is five, in this case we round upwards so that 9.5621 significant figure becomes ten 10. The answer to our problem therefore approximates to 100×10 which is one thousand 1000.


(d) 542.04 has a first significant figure of 5. The next number is 4 which is less than 5, so in this case we round down. 542.04 to 1 significant figure is therefore five hundred 500. Our second number is 1.88 having a first significant figure of 1. As the second significant figure is above five (being 8) we round our first significant figure upwards to make it 2. Therefore the approximated solution to the problem 542.04 multiplied by 1.88 becomes five hundred multiplied by two which is a thousand 1000.


(e) 494.275 has a first significant figure of four, the next significant figure is nine which will therefore be rounded up, so 494.27521 significant figure becomes five hundred 500. 5.05 has a first significant figure of five, and the next figure which is significant would be the other five. But if we round it up in this case we turn a 0 into 1 and as 1 is less than 5 it makes no difference. The number 5.05 to 1 significant figure is therefore 5. Approximation 494.275÷5.05 then becomes 500÷5 which is 100.


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