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If x and y are related as 4x + 3y = 11 and mean deviation of x is 5.40. What is the mean deviation of y ?

From \(4x + 3y = 11\), we have \(y = \frac{11}{3} - \frac{4}{3}x\). Therefore, \(\text{MD}(y) = \left|-\frac{4}{3}\right| \text{MD}(x) = \frac{4}{3} \times 5.40 = 7.2\).

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If x and y are related as 4x + 3y = 11 and mean deviation of x is 5.40. What is the mean deviation of y ?

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Correct answer A

From 4x+3y=114x + 3y = 11, we have y=113−43xy = \frac{11}{3} - \frac{4}{3}x. Therefore, MD(y)=∣−43∣MD(x)=43×5.40=7.2\text{MD}(y) = \left|-\frac{4}{3}\right| \text{MD}(x) = \frac{4}{3} \times 5.40 = 7.2.

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  • Rajasthan Clerk Grade 2

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