Evaluate (0.08)raised to power 3
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There are many methods to solve this, but I suppose it is to be solved using Binomial Theorem. (I would be expressing it in the form of nCr•a^(n-r)•b^r)
(0.08)^3 = (1 - 0.92)^3
=> 3C0•(1)³•(-0.92)^(0) + 3C1•(1)²•(-0.92)¹ + 3C2•(1)¹•(-0.92)² + 3C3•(1)^(0)•(-0.92)³
=> 1 + 3×(-0.92) + 3×(-0.92)² + (-0.92)³
=> 1 - 2.76 + 2.5392 - 0.778688
=> 0.000512
(0.08)^3 = (1 - 0.92)^3
=> 3C0•(1)³•(-0.92)^(0) + 3C1•(1)²•(-0.92)¹ + 3C2•(1)¹•(-0.92)² + 3C3•(1)^(0)•(-0.92)³
=> 1 + 3×(-0.92) + 3×(-0.92)² + (-0.92)³
=> 1 - 2.76 + 2.5392 - 0.778688
=> 0.000512
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