diff --git a/ex10/kn10.pdf b/ex10/kn10.pdf
index 653444d..7e773ae 100644
--- a/ex10/kn10.pdf
+++ b/ex10/kn10.pdf
Binary files differ
diff --git a/ex10/kn10.tex b/ex10/kn10.tex
index eb8e006..b1b24be 100644
--- a/ex10/kn10.tex
+++ b/ex10/kn10.tex
@@ -130,9 +130,9 @@
     %TODO: Quellen
 \end{enumerate}
 \Aufgabe{Wi-Fi - CSMA/CA, RTS/CTS}{10+10+10}
-%TODO: labled figure
+%TODO: labeled figure
 \begin{enumerate}
-    \item Time to transmit one packet and receiving the acknowledgment:
+    \item Time to transmit one packet and receiving the acknowledgement:
     \begin{align*}
       T_1 &= \frac{l_d}{N_d} + \frac{l_a}{N_a} \\
          &= \frac{12000~bit}{11 \cdot 10^{6} ~bit/s} + \frac{112~bit}{1 \cdot 10^{6} ~bit/s} \\
@@ -141,9 +141,9 @@
     \end{align*}
     The maximum throuput $D_1$ in $bit/s$ is therefore:
     \begin{align*}
-      D_1 = (1 / 1.203 ms) * (12000~bit + 112~bit) = 10.07 \cdot 10^{3} bit/s
+      D_1 = (1 / 1.203 ms) * (12000~bit + 112~bit) = 10.07 \cdot 10^{6} bit/s
     \end{align*}
-    \item Time to transmit one packet and receving the acknowledgment:
+    \item Time to transmit one packet and receiving the acknowledgement:
     \begin{align*}
       T_2 &= t_{difs} + \frac{\sum_{X=0}^{31}X \cdot t_{slot}}{32} + t_{sifs} + T_1 \\
           &= 50 \mu s + 310 \mu s + 10 \mu s + 1203 \mu s \\
@@ -152,18 +152,18 @@
     \end{align*}
     The average throughput $D_2$ in $bit/s$ is therefore:
     \begin{align*}
-      D_2 = (1000 ms / 1.2485 ms) * (12000 + 112 bit) = 9.701 \cdot 10^{6} bit/s
+      D_2 = (1 / 1.573 ms) * (12000 + 112 bit) = 7.70 \cdot 10^{6} bit/s
     \end{align*}
-    \item Time to transmit one packet and receiving the acknowledgment:
+    \item Time to transmit one packet and receiving the acknowledgement:
     \begin{align*}
       T_3 &=  \frac{l_{rts}}{N_a} + \frac{l_{cts}}{N_a} + t_{sifs} + t_{sifs} + T_2 \\
-          &= 160 \mu s + 112 \mu s + 20 \mu s + 1285 ms\\
-          &= 292 \mu s + 1285 \mu s \\
-          &= 1.577ms
+          &= 160 \mu s + 112 \mu s + 20 \mu s + 1573 \mu s\\
+          &= 292 \mu s + 1573 \mu s \\
+          &= 1.865 ms
     \end{align*}
-     The maximum throughput $D_@$ in $bit/s$ is therefore:
+     The maximum throughput $D_3$ in $bit/s$ is therefore:
      \begin{align*}
-       D_3 = (1000 ms / 1.577 ms) * (12000 + 112 + 160 + 112) = 7.852 \cdot 10^{6} bit/s
+       D_3 = (1 / 1.865 ms) * (12000 + 112 + 160 + 112) = 6.640 \cdot 10^{6} bit/s
      \end{align*}
 \end{enumerate}
 \Aufgabe{Broadband Internet Access}{3+3+3+3+3+3+3+3+3+3}