\(\sigma _{P}^{2}=\frac{(0.4)^{2}(34.618)}{0.6}+\frac{(0.6)^{2}(106.07)}{0.6}+2(0.4)(0.6)(0.107)(5.88)\)

asked by guest
on Jul 31, 2025 at 10:06 am



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