Using the binomial theorem,the value of (0.99 | vedclass

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(C) We can write $(0.999)^3$ as $(1 - 0.001)^3$. Using the binomial theorem,$(a - b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} (-b)^k$. For $(1 - 0.001)

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$0.997$

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(C) We can write $(0.999)^3$ as $(1 - 0.001)^3$. Using the binomial theorem,$(a - b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} (-b)^k$. For $(1 - 0.001)^3$,we have: $(1 - 0.001)^3 = \binom{3}{0}(1)^3 - \binom{3}{1}(1)^2(0.001) + \binom{3}{2}(1)(0.001)^2 - \binom{3}{3}(0.001)^3$ $= 1 - 3(0.001) + 3(0.000001) - 0.000000001$ $= 1 - 0.003 + 0.000003 - 0.000000001$ $= 0.997 + 0.000002999$ $= 0.997002999$ Rounding to $3$ decimal places,we get $0.997$.

Using the binomial theorem,the value of $(0.999)^3$ correct to $3$ decimal places is:

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