Solution. Using the properties of mathematical expectation (formulas (31) - (33)), we get:

$$
\begin{aligned}
& M(Z)=M(2 X+4 Y+5)=2 M(X)+4 M(Y)+5= \\
& =2 \cdot 3+4 \cdot 5+5=31
\end{aligned}
$$
90. Find the mathematical expectation of the random variable $Z=2 X+4 Y+5$, if the mathematical expectations of $X$ and $Y$ are known: $M(X)=3, M(Y)=5$.