小学数学 有效数字和四舍五入 Significant Figures 英语双语练习题Word文档格式.docx

1、Significant Figures Examples with Answers重要数字题目示例（附答案）Check different problems impose on Significant Figures and get a grip on every concept available in the Significant Figures.检查对有效数字所问的不同问题，掌握有效数字中可用的每个概念。1. Find the Number of a Significant Figure in Each of the Following （a） 8.4 （b） 173.6 m （c）

2、407 g （d） 4.68 m （e） 8.1165 kg （f） 0.054 km （g） 0.00343 l （h） 93.040 mg （i） 30.030300 g （j） 40.00 km （k） 3.600 ml （l） 70.0021.在下列各项中找出有效数字的数字：（a）8.4（b）173.6 m（c）407 g（d）4.68 m（e）8.1165 kg（f）0.054 km（g）0.00343 l（h）93.040 mg（i）30.030300 g（j）40.00 km（k）3.600 ml（l）70.002Solution: （a） There are 2 signifi

3、cant figures available in 8.4. They are 8, 4. （b） There are 4 significant figures available in 173.6 m. They are 1, 7, 3, and 4. （c） There are 3 significant figures available in 407 g. They are 4, 0, and 7. （d） There are 3 significant figures available in 4.68 m. They are 4, 6, and 8. （e） There are

4、5 significant figures available in 8.1165 kg. They are 8, 1, 1, 6, and 5. （f） There are 2 significant figures available in 0.054 km. They are 5, and 4. （g） There are 3 significant figures available in 0.00343 l. They are 3, 4, and 3. （h） There are 5 significant figures available in 93.040 mg. They a

5、re 9, 3, 0, 4, and 0. （i） There are 8 significant figures available in 30.030300 g. They are 3, 0, 0, 3, 0, 3, 0, and 0. （j） There are 4 significant figures available in 40.00 km. They are 4, 0, 0, and 0. （k） There are 4 significant figures available in 3.600 ml. They are 3, 6, 0, and 0. （l） There a

6、re 5 significant figures available in 70.002. They are 7, 0, 0, 0, and 2.解决方案：（a）8.4中有两个有效数字。它们是8，4。（b）173.6 m中有4个有效数字。它们是1、7、3和4。（c）407g中有3个有效数字。它们是4、0和7。（d）4.68 m中有3个有效数字。它们是4、6和8。（e）8.1165 kg中有5个有效数字。它们是8、1、1、6和5。（f）在0.054 km范围内有2个有效数字。它们是5和4。（g）在0.00343 l中有3个有效数字。它们是3、4和3。（h）在93.040 mg中有5个有效数字可用

7、。它们是9、3、0、4和0。（i）30.030300 g中有8个有效数字。它们是3、0、0、3、0、3、0和0。（j）40.00 km内有4个有效数字。它们是4、0、0和0。（k）3.600 ml中有4个有效数字。它们是3、6、0和0。（l）70.002中有5个有效数字。它们是7、0、0、0和2。2. Round off each of the following correct up to 3 significant figures （a） 57.3628 g （b） 6.31874 kg （c） 44.422 km （d） 60.001 cm （e） 0.0023596 m （f） 0.0

8、024030 l2.将下列各有效数字四舍五入至3位：（a）57.3628 g（b）6.31874 kg（c）44.422 km（d）60.001 cm（e）0.0023596 m（f）0.002430 l （a） Given that 57.3628 g. It has 6 significant figures. To round off the given number into 3 significant digits, we need to round it off to 1 place after the decimal. The digit 6 is greater than 5.

9、 So, the digit 3 becomes 4, and the digits 6, 2, and 8 disappear. Therefore, 57.3628 g = 57.4 g rounded off to 3 significant figures. （b） Given that 6.31874 kg. It has 6 significant figures. To round off the given number into 3 significant digits, we need to round it off to 2 places after the decima

10、l. The digit 8 is greater than 5. So, the digit 1 becomes 2, and the digits 8, 7, and 4 disappear. Therefore, 6.31874 kg = 6.32 kg rounded off to 3 significant figures. （c） Given that 44.422 km. It has 5 significant figures. To round off the given number into 3 significant digits, we need to round i

11、t off to 1 place after the decimal. The digit 2 is less than 5. So, the digit 4 remains 4, and the digits 2, and 2 disappear. Therefore, 44.422 km = 44.4 km rounded off to 3 significant figures. （d） Given that 60.001 cm. It has 5 significant figures. To round off the given number into 3 significant

12、digits, we need to round it off to 1 place after the decimal. The digit 0 is less than 5. So, the digit 0 remains 0, and the digits 0, and 1 disappear. Therefore, 60.001 cm = 60.0 cm rounded off to 3 significant figures. （e） Given that 0.0023596 m. It has 5 significant figures. To round off the give

13、n number to 3 significant digits, we need to round it off to 5 places after the decimal. The digit 9 is greater than 5. So, the digit 5 becomes 6, and the digits 9, and 6 disappear. Therefore, 0.0023596 m = 0.00236 m rounded off to 3 significant figures. （f） Given that 0.0024030 l. It has 5 signific

14、ant figures. To round off the given number to 3 significant digits, we need to round it off to 5 places after the decimal. The digit 3 is less than 5. So, the digit 0 remains 0, and the digits 3, and 0 disappear. Therefore, 0.0024030 l = 0.00240 l rounded off to 3 significant figures.（a）假设57.3628克。它

15、有6个重要数字。要将给定的数字四舍五入为3位有效数字，我们需要将它四舍五入到小数点后的1位。数字6大于5。因此，数字3变为4，数字6、2和8消失。因此，57.3628 g=57.4 g四舍五入为3位有效数字（b） 考虑到6.31874千克。要将给定的数字四舍五入为3位有效数字，我们需要将它四舍五入到小数点后的2位。数字8大于5。因此，数字1变为2，数字8、7和4消失。因此，6.31874 kg=6.32 kg四舍五入为3位有效数字（c） 考虑到44.422公里。它有5个重要数字。数字2小于5。因此，数字4保持为4，数字2和2消失。因此，44.422 km=44.4 km四舍五入为3位有效数字（

16、d） 考虑到60.001厘米。数字0小于5。因此，数字0保持为0，数字0和1消失。因此，60.001 cm=60.0 cm四舍五入为3个有效数字（e） 考虑到0.0023596米。要把给定的数字四舍五入到3位有效数字，我们需要把它四舍五入到小数点后的5位。数字9大于5。因此，数字5变为6，数字9和6消失。因此，0.0023596 m=0.00236 m四舍五入为3位有效数字（f） 考虑到0.002430升。数字3小于5。因此，数字0保持为0，数字3和0消失。因此，0.0024030 l=0.00240 l四舍五入为3位有效数字。3. Round off （a） 16.367 g correct

17、 to the three significant figures. （b） 0.00949 mg correct to the two significant figures. （c） 0.005618 g correct to the one significant figures. （d） 28.303 mm correct to the four significant figures. （e） 33.422 km correct to the four significant figures. （f） 4.0832 kg correct to the two significant

18、figures. （g） 0.004628 mm correct to the one significant figures.3.四舍五入（a）16.367 g，更正为三位有效数字（b） 0.00949 mg，校正为两个有效数字（c） 0.005618 g，精确到一位有效数字（d） 28.303 mm，四位有效数字（e） 33.422公里，四位有效数字正确（f） 4.0832 kg，校正为两个有效数字（g） 0.004628 mm，精确到一位有效数字。 （a） Given that 16.367 g. It has 5 significant figures. To round off th

19、e given number into 3 significant digits, we need to round it off to 1 place after the decimal. The digit 6 is greater than 5. So, digit 3 becomes 4, and the digits 6, and 7 disappear. Therefore, 16.367 g = 16.4 g rounded off to 3 significant figures. （b） Given that 0.00949 mg. It has 3 significant

20、figures. To round off the given number into 2 significant digits, we need to round it off to 4 places after the decimal. The digit 9 is greater than 5. So, the digit 4 becomes 5, and the digit 9 disappear. Therefore, 0.00949 mg = 0.0095 mg rounded off to 2 significant figures. （c） Given that 0.00561

21、8 g. It has 4 significant figures. To round off the given number into 1 significant digit, we need to round it off to 3 places after the decimal. The digit 6 is greater than 5. So, the digit 5 becomes 6, and the digits 6, 1, and 8 disappear. Therefore, 0.005618 g = 0.006 g rounded off to 3 significa

22、nt figures. （d） Given that 28.303 mm. It has 5 significant figures. To round off the given number into 4 significant digits, we need to round it off to 2 places after the decimal. The digit 3 is less than 5. So, the digit 0 remains 0, and the digits 3 disappear. Therefore, 28.303 mm = 28.30 mm round

23、ed off to 4 significant figures. （e） Given that 33.422 km. It has 5 significant figures. To round off the given number to 4 significant digits, we need to round it off to 2 places after the decimal. The digit 2 is less than 5. So, digit 2 remains 2, and the digits 2 disappear. Therefore, 33.422 km =

24、 33.42 km rounded off to 4 significant figures. （f） Given that 4.0832 kg. It has 5 significant figures. To round off the given number to 2 significant digits, we need to round it off to 1 place after the decimal. The digit 8 is greater than 5. So, the digit 0 becomes 1, and the digits 8, 3, and 2 di

25、sappear. Therefore, 4.0832 kg = 4.1 kg rounded off to 2 significant figures. （g） Given that 0.004628 mm. It has 4 significant figures. To round off the given number to 1 significant digit, we need to round it off to 3 places after the decimal. The digit 6 is greater than 5. So, the digit 4 becomes 5

26、, and the digits 6, 2, and 8 disappear. Therefore, 0.004628 mm = 0.005 mm rounded off to 1 significant figure.（a）假设为16.367克。因此，数字3变为4，数字6和7消失。因此，16.367 g=16.4 g四舍五入为3位有效数字（b） 考虑到0.00949毫克。它有三个重要的数字。要将给定的数字四舍五入为2位有效数字，我们需要将它四舍五入到小数点后的4位。因此，数字4变为5，数字9消失。因此，0.00949 mg=0.0095 mg四舍五入为2位有效数字（c） 考虑到0.00561

27、8克。它有4个有效数字。要将给定的数字四舍五入为1位有效数字，我们需要将它四舍五入到小数点后的3位。因此，数字5变为6，数字6、1和8消失。因此，0.005618 g=0.006 g四舍五入为3位有效数字（d） 考虑到28.303毫米。要将给定的数字四舍五入为4位有效数字，我们需要将它四舍五入到小数点后的2位。因此，数字0保持为0，数字3消失。因此，28.303 mm=28.30 mm四舍五入为4位有效数字（e） 考虑到33.422公里。要将给定的数字四舍五入到4位有效数字，我们需要将它四舍五入到小数点后的2位。因此，数字2保持为2，数字2消失。因此，33.422 km=33.42 km四舍五

28、入为4位有效数字（f） 考虑到4.0832千克。要将给定的数字四舍五入到2位有效数字，我们需要将它四舍五入到小数点后的1位。因此，数字0变为1，数字8、3和2消失。因此，4.0832 kg=4.1 kg四舍五入为2个有效数字（g） 假设是0.004628毫米。要将给定的数字四舍五入到1位有效数字，我们需要将它四舍五入到小数点后的3位。因此，数字4变为5，数字6、2和8消失。因此，0.004628 mm=0.005 mm四舍五入为1个有效数字。4. Round off （a） $ 3067.665 to the nearest cents. （b） 0.00588 m to the neares

29、t cm. （c） 18.0333 kg to the nearest g. （d） $ 49.63 to the nearest dollar.4.四舍五入（a）$3067.665精确到美分（b） 0.00588 m精确到cm（c） 18.0333千克，精确到克（d） 49.63美元精确到1美元。 （a） Given that $ 3067.665. It has 7 significant figures. To round off the $ 3067.665 to the nearest cents, we need to round it off to 3 places after the decimal. The digit 5 is equal to 5. So, digit 6 becomes 7, and the digit 5 disappear. Therefore, $ 3067.665 = $ 3067.67 （b） Given that 0.00588 m. It has 3 significant figures. To round off the 0.00