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Question Number 48090 by peter frank last updated on 19/Nov/18

Answered by tanmay.chaudhury50@gmail.com last updated on 19/Nov/18

1)true value of product of length×width =2.29×1.29=(2.30−0.01)(1.30−0.01) =2.30×1.30−0.01(2.30+1.30)+0.0001 =2.30×1.30−3.60×0.01+0.0001 product of estimated value =2.30×1.30 so absolute error =2.30×1.30−3.60×0.01+0.0001−2.30×1.30 =−0.0359 relatve error=((−0.0359)/(2.29×1.29))

$$\left.\mathrm{1}\right){true}\:{value}\:{of}\:{product}\:{of}\:{length}×{width} \\ $$$$=\mathrm{2}.\mathrm{29}×\mathrm{1}.\mathrm{29}=\left(\mathrm{2}.\mathrm{30}−\mathrm{0}.\mathrm{01}\right)\left(\mathrm{1}.\mathrm{30}−\mathrm{0}.\mathrm{01}\right) \\ $$$$=\mathrm{2}.\mathrm{30}×\mathrm{1}.\mathrm{30}−\mathrm{0}.\mathrm{01}\left(\mathrm{2}.\mathrm{30}+\mathrm{1}.\mathrm{30}\right)+\mathrm{0}.\mathrm{0001} \\ $$$$=\mathrm{2}.\mathrm{30}×\mathrm{1}.\mathrm{30}−\mathrm{3}.\mathrm{60}×\mathrm{0}.\mathrm{01}+\mathrm{0}.\mathrm{0001} \\ $$$${product}\:{of}\:{estimated}\:{value} \\ $$$$=\mathrm{2}.\mathrm{30}×\mathrm{1}.\mathrm{30} \\ $$$${so}\:{absolute}\:{error} \\ $$$$=\mathrm{2}.\mathrm{30}×\mathrm{1}.\mathrm{30}−\mathrm{3}.\mathrm{60}×\mathrm{0}.\mathrm{01}+\mathrm{0}.\mathrm{0001}−\mathrm{2}.\mathrm{30}×\mathrm{1}.\mathrm{30} \\ $$$$=−\mathrm{0}.\mathrm{0359} \\ $$$${relatve}\:{error}=\frac{−\mathrm{0}.\mathrm{0359}}{\mathrm{2}.\mathrm{29}×\mathrm{1}.\mathrm{29}} \\ $$

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