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Published byTyler Walsh Modified over 6 years ago

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When we count we use exact numbers If we count people we have exactly 4 people There is no uncertainty about this number of people. Measurements using an interval scale (like a ruler) has some uncertainty Is it 29.68cm or is it 29.67cm? This involves some estimation on the last digit

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Invented in order to compare quantities The metre was once defined as the distance between two scratches on a bar of platinum- iridium alloy at 0 o C All other measurements of distance are estimated Not all measurements, however, are known to the same accuracy Accuracy is how close the measurement is to the real value. (Sometimes the real value is not known) A micrometer, for example, will yield a more accurate measurement of a hair's diameter than will a metre stick

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Significant digits tell us something about how the measurement was made A better instrument allows us to make better measurements and we record a greater number of significant digits in reporting this value A metre stick would only allow us to record the thickness of a hair as 0.1mm A micrometer allows us to record the thickness as 0.137 mm This measurement is closer to the true value and is a more accurate measurement

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Is it possible to be too accurate? When is it not suitable to use a more precise measuring tool? For each measurement would you use a micrometer or a metre stick? Length of table Width of fruit fly Thickness of a penny Width of hand Distance to the sun Having the length of a table recorded at 1.2345678909876m is not needed for most cases A recording of 1.23m is sufficient most of the time

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2m 20cm406080 3m Suppose the length of a table is measured with a ruler calibrated to 10 centimetres The table is definitely less than 2.8 metres but greater than 2.7 metres

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Is the extra length 0.04 or is it 0.05? We make the best possible estimate A proper measurement would be recorded as 2.74 metres. This has 3 Sig-Figs Where did the 0.04 come from? I thought the ruler could only do 10 centimeters at a time? This indicates the table has a length of 2 metres plus 70 centimetres plus a little bit more A measurement of the same table with a ruler calibrated to centimetres could yield 2.742 metres. This has 4 Sig-Figs 2m 20cm406080 3m

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1. All non-zero digits are significant a. Eg.374 (3 sig-figs) b. 8.1 (2 sig-figs) c. 8.365 X 10 4 (4 sig-figs) 2. All zeroes between non-zero digits are significant (Captured Zeros) a. Eg.50407 (5) b. 8.001 (4) c. 9.05 X 10 4 (3)

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3. Leading zeroes in a decimal are not significant a. Eg.0.54 (2) b. 0.0098 (2) c. 0.05 X 10 -7 (1) ( Not proper Scientific notation) 4. Trailing zeroes are significant only if they are to the right of a decimal point a. Eg.2370 (3) b. 16000 (2) c. 16.000 (5) 5. In numbers greater than 1, trailing zeroes are not significant unless stated so a. Eg. 37000 (2)

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The last three zeroes may or may not be part of the measurement. To show that they are, we use scientific notation. All the zeroes written in the number in scientific notation are significant. 37000 with 3 sig. digits would be 3.70 x 10 4 37000 with 4 sig. digits would be 3.700 x 10 4 37000 with 5 sig. digits would be 3.7000 x 10 4 37000.0 has 6 sig. digits Exact Numbers do not affect Sig-Figs If I count 4 people, I have exactly 4, not 4.01 or 3.99 If I take the average of 7 tests, that 7 is an exact number, will not affect sig-figs in any way.

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Round each value to 3 sig figs 1.234 1.23 9.865 9.87 0.07888 0.0789 0.5399 0.540 12990 1.30 x 10 4

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General Rules: Add or subtract as normal. Count the number of digits to the right of the decimal. The answer must be rounded to contain the same number of decimal places as the value with the LEAST number of decimal places

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Example 12.0 + 131.59 + 0.2798 = ? Add as normal: 12.0 + 131.59 + 0.2798 = 143.8698 The least number of decimal places is 12.0, with one decimal place Round to the least number of decimal places Final Answer = 143.9 0.0998 – 1.0 = -0.9002 = -0.9 5.4 x 10 2 + 2.8 x 10 1 = 568 = 570

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General Rules: Multiply or divide as normal. Count the number of significant figures in each number. The answer must be rounded to contain the same number of significant figures as the number with the LEAST number of significant figures

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Example 51.3 × 13.75 = ? Multiply as normal: 51.3 × 13.75 = 705.375 The least number of significant figures is 3 in 51.3 Round to the least number of significant figures Final Answer= 705

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Example 2.00 x 7.00 = 14 = 14.0 (3.0×10 12 ) / (6.02×10 23 )= ? Put the numbers into your calculator as usual: 3.0×10 12 ÷ 6.02×10 23 = 4.98338×10 −12 The least number of significant figures belongs to 3.0×10 12 with 2 sig figs Final Answer = 5.0×10 −12 250 x 4.0 = 1000 = 1.0 x 10 3

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Using proper sig-figs, find how many seconds are in 1 year (365.25 days) = 31557600 = 31558000

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