Mathematics · 2022
JEE Main · 27 June 2022, Shift 2 · Q16
The mean and variance of the data 4,5,6,6,7,8, x, y, where x<y, are 6 and 9/4 respectively. Then x^4+y^2 is equal to
The mean and variance of the data $\displaystyle 4,5,6,6,7,8, x, y$, where $\displaystyle x<y$, are $\displaystyle 6$ and $\displaystyle \frac{9}{4}$ respectively. Then $\displaystyle x^4+y^2$ is equal to
Official answer
(2)
$\displaystyle 320$
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.