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diff --git a/mr/ub6/UB6.tex b/mr/ub6/UB6.tex
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 	\end{enumerate}
 	\Aufgabe{Stereo Vision}{8}
 	\begin{enumerate}
+		\item  As higher $d_{max}$ gets as smaller $z_b$ can be. It has no influnce on the highest observable $z_b$\\
+		$z_{p,min}=\frac{f\cdot b}{d_{max}}=\frac{512px\cdot10cm}{64px}=80cm$	
 		\item 
+		$  d \sim N(d_{true},\sigma_d)$\\
+		 $z_{p}(d)=\frac{f\cdot b}{d}$\\
+		TaylorApprox:\\$z_p(d)=z(d_{true})+z'(d_{true})(d-d_{true})=\frac{f\cdot b}{d_{true}}- \frac{f\cdot b}{d_{true}^2}(d-d_{true})= \frac{2f\cdot b}{d_{true}}-\frac{f\cdot b}{d_{true}^2}\cdot d\\
+		\sigma_z= a\cdot\sigma_d\cdot a = a^2\cdot \sigma_d =  - \frac{f^2\cdot b^2\cdot \sigma_d}{d_{true}^4}= - \frac{f^2\cdot b^2\cdot \sigma_d}{\frac{f^4\cdot b^4}{z_p^4}}= \frac{z_p^4}{f^2\cdot b^2}\sigma_d  \\
+		$schneller: $\sigma = f'(a)^2\cdot \sigma_{alt}
+		$
+		\item siehe b)
+		\item $f=512px \\
+		b=5cm\\
+		o_d=0.5cm\\
+		z_p=5m\\
+		\sigma_z= \frac{5^4}{512^2\cdot5^2}\cdot 0.5=\frac{5^2}{512^2\cdot 2}=0.0000477=4.77*10^{-5}cm
+		$
+		\item $o_{z}=5\\
+		b=\sqrt{\frac{z_p^4}{f^2\cdot o_z}\sigma_d}	= \sqrt{\dfrac{5^3}{512^2\cdot2}}= 0.0154 cm	
+		$
 	\end{enumerate}
 \end{document}