$\begin{array}[t]{rll} p(h) &=& 787 \\[5pt] 1.013\cdot \mathrm e^{-0,126\cdot h} &=& 787 &\quad \scriptsize \mid\; :1.013\\[5pt] \mathrm e^{-0,126\cdot h}&=& \frac{787}{1.013}&\quad \scriptsize \mid\; \ln\\[5pt] -0,126\cdot h &=& \ln \frac{787}{1.013} &\quad \scriptsize \mid\; :(-0,126) \\[5pt] h &\approx& 2,00 \end{array}$

$\begin{array}[t]{rll} p(h)&=& 1.013\cdot \mathrm e^{-0,126\cdot h} \\[10pt] p(h+1)&=& 1.013\cdot \mathrm e^{-0,126\cdot (h+1)} \\[5pt] &=& 1.013\cdot \mathrm e^{-0,126\cdot h - 0,126} \\[5pt] &=& 1.013\cdot \mathrm e^{-0,126\cdot h}\cdot \mathrm e^{- 0,126} \\[5pt] &=& \mathrm e^{- 0,126} \cdot p(h) \end{array}$

$\begin{array}[t]{rll} \overline{m} &=& \frac{1}{11-0} \cdot \displaystyle\int_{0}^{11}p(h)\;\mathrm dh \\[5pt] &=& \frac{1}{11} \cdot \displaystyle\int_{0}^{11} 1.013\cdot \mathrm e^{-0,126\cdot h}\;\mathrm dh \\[5pt] &=& \frac{1}{11} \cdot \left[-\frac{1.013}{0,126}\cdot \mathrm e^{-0,126\cdot h}\right]_0^{11} \\[5pt] &=& \frac{1}{11} \cdot \left(-\frac{1.013}{0,126}\cdot \mathrm e^{-0,126\cdot 11}- \left(-\frac{1.013}{0,126}\cdot \mathrm e^{-0,126\cdot 0} \right)\right) \\[5pt] &=& \frac{1}{11} \cdot \left(-\frac{1.013}{0,126}\cdot \mathrm e^{-0,126\cdot 11}+ \frac{1.013}{0,126} \right) \\[5pt] &\approx& 548,11 \end{array}$