The deviations from the mean of a set of numbers are \((k+3)^{2}, (k+7), -2, \text{k and (} k+2)^{2}\), where k is a constant. Find the value of k.
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Correct Answer: Option D
Explanation:
The sum of deviations from the mean of a set of numbers equals 0.
\((k+3)^{2} + (k+7) + (-2) + k + (k+2)^{2} = 0\)
\((k^2 + 6k + 9) + (k+7) - 2 + k + (k^2 + 4k + 4) = 0\)
\(2k^{2} + 12k + 18 = 0\)
\(2k^{2} + 6k + 6k + 18 = 2k(k + 3) + 6(k + 3) = 0\)
\(k = -3 (twice)\)
The sum of deviations from the mean of a set of numbers equals 0.
\((k+3)^{2} + (k+7) + (-2) + k + (k+2)^{2} = 0\)
\((k^2 + 6k + 9) + (k+7) - 2 + k + (k^2 + 4k + 4) = 0\)
\(2k^{2} + 12k + 18 = 0\)
\(2k^{2} + 6k + 6k + 18 = 2k(k + 3) + 6(k + 3) = 0\)
\(k = -3 (twice)\)