diff --git a/is/kaggle/1st.py b/is/kaggle/1st.py
index d4ace7e..3129032 100644
--- a/is/kaggle/1st.py
+++ b/is/kaggle/1st.py
@@ -22,3 +22,4 @@
 predictions = test.join(data,on='ID')
 
 predictions[['ID','WeightInter']].to_csv('sampleSubmission.csv', header = ['ID','Weight'],index_label=False,index=False)
+
diff --git a/mr/ub6/UB6.pdf b/mr/ub6/UB6.pdf
index 1d9b25c..27f9c10 100644
--- a/mr/ub6/UB6.pdf
+++ b/mr/ub6/UB6.pdf
Binary files differ
diff --git a/mr/ub6/UB6.tex b/mr/ub6/UB6.tex
index 8c66a69..9d6f652 100644
--- a/mr/ub6/UB6.tex
+++ b/mr/ub6/UB6.tex
@@ -144,12 +144,12 @@
 	 \end{pmatrix}$
 			\end{itemize}
 			werte falsch ... finde rechenfehler nicht.
-		
+
 	\item $(K^{-1} p_R)^T\cdot E\cdot K^{-1}p_L = p_R^{T}\cdot K^{T^{-1}}\cdot E\cdot K^{-1} \cdot p_L\\
 	E= K^T\cdot F\cdot K	\\
 	F= K^{T^{-1}}\cdot E\cdot K^{-1}$\\
 	K = calibration matrix of the cameras
-	\item 
+	\item
 	$
 	K=\begin{pmatrix}
 	500&0&320\\
@@ -190,9 +190,9 @@
 		s_1=4, s_2=6, s_3=7\\
 		\begin{tabular}{c||llllllllll}
 			 outlier ratio  & 0.1  &0.2  &0.3  &0.4  &0.5  &0.6  &0.7  &0.8  &0.9    \\ \hline
-			s1=4& 5 & 9 & 17 & 34 & 72 & 178 & 566  & 2876  & 4650     \\
+			s1=4& 5 & 9 & 17 & 34 & 72 & 178 & 567  & 2876  & 4650     \\
 			s2=5 & 6 & 12 &26  &57  &146  &448  &1893  &14389  &460515    \\
-			s3=7 & 8 &20  & 54 &163  &588  &2809  &21055  &359777  &46051700  
+			s3=7 & 8 &20  & 54 &163  &588  &2809  &21055  &359777  &46051700
 		\end{tabular}
 		$
 		\item A: 4 datapoints 150ns a iteration\\
@@ -201,16 +201,16 @@
 		\eta = 0.7\\
 		k_A= 567\\
 		k_B=1893\\
-		140/40= 3.5$ B is 3.5 times faster than A\\
-		$1893/3.5= 541$\\
-		So we choose B although it needs more points.
-		
+		k_A\cdot t_A=567\cdot 140ns=79380ns\\
+        k_B\cdot t_B=1893\cdot 40ns= 75720ns$\\
+        So the time for computation of B is smaller even with more points needed. So we choose B because of faster computationtime. If we are short for points however we should use A even though it takes more time.
+
 	\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 
+		$z_{p,min}=\frac{f\cdot b}{d_{max}}=\frac{512px\cdot10cm}{64px}=80cm$
+		\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\\
@@ -225,7 +225,7 @@
 		\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	
+		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}