Christmas Streak 21/88: Too Many Variables!

For a set of $n$ variables $X,$ $V(X)$ is its variance and $E(X)$ is its average.

$V(X) \\ \displaystyle =\sum_{k=1}^{n} \dfrac{(x_k-E(X))^2}{n} \\ \displaystyle = \sum_{k=1}^{n} \dfrac{{x_k}^2}{n}-2E(X)\sum_{k=1}^{n}\dfrac{x_k}{n} +\sum_{k=1}^n \dfrac{E(X)}{n} \\ =E(X^2)-2E(X)^2+E(X)^2 \\ =E(X^2)-E(X)^2.$

Same applies for $\sqrt{X},$ so plugging in, we get $V(\sqrt{X})=E(X)-E(\sqrt{X})^2.$

Since $V(X)=V,~E(X^2)=M,~E(X)=m,$ we get $V=M-m^2.$

And since $M-V+m=210,~m^2+m-210=0.$

So $m=14$ since $m>0.$

Then note that $v=V(\sqrt{X}).$ Therefore $v=m-E(\sqrt{X})^2.$

$E(\sqrt{X})^2=m-v=14-3=11.$

$\therefore~E(\sqrt{X})=\sqrt{11}\approx\boxed{3.317}.~(\because E(\sqrt{X})>0)$