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What is the approximate value of `(1.02)^(8)` ?A. `1.171`B. `1.175`C. `1.177`D. `1.179` |
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Answer» Correct Answer - A `(1.02)^(8)=(1+0.02)^(8)` `(1+x)^(n)=1+nx+(n(n-1))/(2!)x^(2)+(n(n-1)(n-2))/(3!)x^(3)+....` `n=8, x=0.02` `(1+0.02)^(8)` `=1+8xx0.02+(8xx7)/(2!).(0.02)^(2)+(8.7.6)/(3!)(.02)^(3)` Neglecting higher terms `=1+0.16+28xx0.0004+56xx0.000008` `cong 1+0.16+0.0112=1.171` |
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