Answer:
(a) 3750 to 3 s.f
(b) 0.004710 to 4 s.f
(c) 5000 to 2 s.f
(d) 0.10000 to 2 s.f
0.10000 to 3 s.f
Step-by-step explanation:
<em>Things to note when rounding off a number to a number of significant figures:</em>
For a given number,
i. the most significant digit is the leftmost digit as long as it is not zero. For example, in 789, the most significant digit is 7.
ii. when rounding off to m significant figures, the least significant digit is the mth digit from the most significant digit. This can be zero. For example, to round off 789 to 2 significant figures, the least significant digit is 8 (since it's at position 2 starting from 7).
iii. every digit from the least significant digit is a non-significant digit. For example, to round off 789 to 2 significant digit, the digits after 8 (which is the least significant digit) are all non-significant. In this case, 9 is not significant. After rounding off, these non-significant digits are changed to zero.
iv. the first non-significant digit is the digit just after the least significant digit. For example, to round off 789 to 2 significant digit, the digit just after 8 (which is the least significant digit) is 9. Therefore, 9 is the first non-significant digit.
<em>Rounding off rules</em>
i. the least significant digit will be unchanged if the first non-significant digit is less than 5.
ii. the least significant digit will be increased by 1 if the first non-significant digit is greater than or equal to 5.
iii. all non-significant digits are either removed or changed to zero depending on whether the number is a decimal.
<em>(a) 37</em><u><em>4</em></u><em>8 (3 s.f)</em>
The least significant digit here is 4 since it is at position 3 starting from the most significant digit 3
The first non-significant digit is 8 which is greater than 5.
Therefore, the least significant digit 4 is increased by 1 to get 5 and the non-significant digits (8 in this case) are changed to 0.
The result is thus <em>3750.</em>
<em></em>
<em>(b) 0.00407</em><u><em>0</em></u><em>989 (4 s.f)</em>
The least significant digit here is 0 since it is at position 4 starting from the most significant digit 4
The first non-significant digit is 9 which is greater than 5.
Therefore, the least significant digit 0 is increased by 1 to get 1 and the non-significant digits (8 and 9 in this case) are changed to 0.
The result is thus <em>0.004071000. </em>
The ending zeros can be removed to give <em>0.004071</em>
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<em>(c) 4</em><u><em>9</em></u><em>71 (2 s.f)</em>
The least significant digit here is 9 since it is at position 2 starting from the most significant digit 4.
The first non-significant digit is 7 which is greater than 5.
Therefore, the least significant digit 9 is increased by 1 to get 10. This is a sort of overflow. This means that the least significant digit will be 0 and a 1 will be added to the significant digit before it (5 in this case which will become 6 due to the addition). The non-significant digits (7 and 1 in this case) are changed to 0.
The result is thus <em>5000</em>
<em>(d) 0.09</em><u><em>9</em></u><em>99 (2 s.f)</em>
The least significant digit here is 9 since it is at position 2 starting from the most significant digit 9.
The first non-significant digit is 9 which is greater than 5.
Therefore, the least significant digit 9 is increased by 1 to get 10. This is a sort of overflow.
This means that the least significant digit will be 0 and a 1 will be added to the significant digit before it (9 in this case which will become 10 due to the addition).
This addition also causes an overflow thereby causing 1 to be added to the digit before it (0 in this case which will become 1 due to the addition). The non-significant digits (9 and 9 in this case) are changed to 0.
The result is thus <em>0.10000</em>
The ending zeros can be removed to give <em>0.1</em>
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<em>d (ii) 0.099</em><u><em>9</em></u><em>9 (3 s.f)</em>
The least significant digit here is 9 since it is at position 3 starting from the most significant digit 9.
The first non-significant digit is 9 which is greater than 5.
Therefore, the least significant digit 9 is increased by 1 to get 10. This is a sort of overflow.
This means that the least significant digit will be 0 and a 1 will be added to the significant digit before it (9 in this case which will become 10 due to the addition).
This addition also causes an overflow thereby causing 1 to be added to the digit before it (9 in this case which will become 10 due to the addition).
This addition also causes an overflow thereby causing 1 to be added to the digit before it (0 in this case which will become 1 due to the addition). The non-significant digits (9 in this case) are changed to 0.
The result is thus <em>0.10000</em>
The ending zeros can be removed to give <em>0.1</em>
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