The mean of numbers a, b, 8, 5, 10 is 6 and t | vedclass

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(D) Given the mean of $a, b, 8, 5, 10$ is $6$: $\frac{a+b+8+5+10}{5} = 6 \implies a+b+23 = 30 \implies a+b = 7$. The variance is $6.80$: $\frac{a^2+b^

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$\operatorname{Tan}^{-1} \frac{7}{11}$

Vedclass · Answer

(D) Given the mean of $a, b, 8, 5, 10$ is $6$: $\frac{a+b+8+5+10}{5} = 6 \implies a+b+23 = 30 \implies a+b = 7$. The variance is $6.80$: $\frac{a^2+b^2+8^2+5^2+10^2}{5} - (6)^2 = 6.80$. $\frac{a^2+b^2+64+25+100}{5} - 36 = 6.80$. $\frac{a^2+b^2+189}{5} = 42.80$. $a^2+b^2+189 = 214 \implies a^2+b^2 = 25$. Since $(a+b)^2 = a^2+b^2+2ab$,we have $7^2 = 25 + 2ab \implies 49 = 25 + 2ab \implies 2ab = 24 \implies ab = 12$. We need to find $\operatorname{Tan}^{-1} \frac{1}{a} + \operatorname{Tan}^{-1} \frac{1}{b} = \operatorname{Tan}^{-1} \left( \frac{\frac{1}{a} + \frac{1}{b}}{1 - \frac{1}{ab}} \right) = \operatorname{Tan}^{-1} \left( \frac{\frac{a+b}{ab}}{1 - \frac{1}{ab}} \right) = \operatorname{Tan}^{-1} \left( \frac{a+b}{ab-1} \right)$. Substituting the values: $\operatorname{Tan}^{-1} \left( \frac{7}{12-1} \right) = \operatorname{Tan}^{-1} \frac{7}{11}$.

The mean of numbers $a, b, 8, 5, 10$ is $6$ and their variance is $6.80$. Then $\operatorname{Tan}^{-1} \frac{1}{a} + \operatorname{Tan}^{-1} \frac{1}{b} =$

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