標題:
Numerical Estimation
發問:
Please see if below my tentative answer is correct or not.Question is: (a) The plastic box storing cookies has dimensions of 96mm x 72mm x 25 mm, correct to the nearest miilimetre. (i) Estimate the volume of the box correct to the nearest cm3 (cubic)my tentative answer => 210 cm3 (ii) Find the upper limit... 顯示更多 Please see if below my tentative answer is correct or not. Question is: (a) The plastic box storing cookies has dimensions of 96mm x 72mm x 25 mm, correct to the nearest miilimetre. (i) Estimate the volume of the box correct to the nearest cm3 (cubic) my tentative answer => 210 cm3 (ii) Find the upper limit of the volume of the box, correct to the nearest cm3 (cubic) my tentative answer => 240 cm3 (b) By using a kitchen scale, the total weight of ten of those plastic boxes in above (a) is found to be 200g, correct to the nearest 10g. (i) Estimate the weight of each box. my tentative answer => 20 g (?????) (ii) State the degree of accuracy of the estimate in (b)(i) my tentative answer => under estimation (iii) Describe how you could reduce the error in measuring the weight of a box using the kitchen scale. my tentative answer => use an electronic scale instead (???)
ai) incorrect 96 mm x 72 mm x 25 mm = 9.6 cm x 7.2 cm x 2.5 cm i.e. The volume of the box : ( 9.6 )( 7.2 )( 2.5 ) = 173 cm^3 (Corr. to the nearest cm^3) aii) incorrect The absolute error is 0.5 mm, i.e. The max. dimensions of the box : 96.5 mm x 72.5 mm x 25.5 mm = 9.65 cm x 7.25 cm x 2.55 cm Therefore, the upper limit of the volume of the box : ( 9.65 )( 7.25 )( 2.55 ) = 178 cm^3 (Corr. to the nearest cm^3) bi) correct 200 / 10 = 20 g bii) incorrect The degree of accuracy tells you to round to hundreds / tens / tenths, etc. e.g. 175472 (Corr. to the nearest hundreds) = 175500 So, the degree of accuracy of the estimate in bi), should be correct to the nearest g . biii) incorrect The question said that how you could reduce the error in measuring the weight of a box using the kitchen scale. So, we can't use other materials. One of the method is: Do not measure the ten plastic boxes together to counting the total weight. We can measure the ten boxes one by one, then, it become more accurate and reduce the error.
其他解答:
Thank you very much
Numerical Estimation
發問:
Please see if below my tentative answer is correct or not.Question is: (a) The plastic box storing cookies has dimensions of 96mm x 72mm x 25 mm, correct to the nearest miilimetre. (i) Estimate the volume of the box correct to the nearest cm3 (cubic)my tentative answer => 210 cm3 (ii) Find the upper limit... 顯示更多 Please see if below my tentative answer is correct or not. Question is: (a) The plastic box storing cookies has dimensions of 96mm x 72mm x 25 mm, correct to the nearest miilimetre. (i) Estimate the volume of the box correct to the nearest cm3 (cubic) my tentative answer => 210 cm3 (ii) Find the upper limit of the volume of the box, correct to the nearest cm3 (cubic) my tentative answer => 240 cm3 (b) By using a kitchen scale, the total weight of ten of those plastic boxes in above (a) is found to be 200g, correct to the nearest 10g. (i) Estimate the weight of each box. my tentative answer => 20 g (?????) (ii) State the degree of accuracy of the estimate in (b)(i) my tentative answer => under estimation (iii) Describe how you could reduce the error in measuring the weight of a box using the kitchen scale. my tentative answer => use an electronic scale instead (???)
此文章來自奇摩知識+如有不便請留言告知
最佳解答:ai) incorrect 96 mm x 72 mm x 25 mm = 9.6 cm x 7.2 cm x 2.5 cm i.e. The volume of the box : ( 9.6 )( 7.2 )( 2.5 ) = 173 cm^3 (Corr. to the nearest cm^3) aii) incorrect The absolute error is 0.5 mm, i.e. The max. dimensions of the box : 96.5 mm x 72.5 mm x 25.5 mm = 9.65 cm x 7.25 cm x 2.55 cm Therefore, the upper limit of the volume of the box : ( 9.65 )( 7.25 )( 2.55 ) = 178 cm^3 (Corr. to the nearest cm^3) bi) correct 200 / 10 = 20 g bii) incorrect The degree of accuracy tells you to round to hundreds / tens / tenths, etc. e.g. 175472 (Corr. to the nearest hundreds) = 175500 So, the degree of accuracy of the estimate in bi), should be correct to the nearest g . biii) incorrect The question said that how you could reduce the error in measuring the weight of a box using the kitchen scale. So, we can't use other materials. One of the method is: Do not measure the ten plastic boxes together to counting the total weight. We can measure the ten boxes one by one, then, it become more accurate and reduce the error.
其他解答:
Thank you very much
全站熱搜
留言列表
