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+# Gegeben
+* Messwert: $f = \qty{145}{\mega\hertz}$
+* Genauigkeit: $\qty{10}{\ppm}$
+
+# Lösungsweg
+$\unit{ppm}$ steht für *parts per million*, $\qty{10}{\ppm}$ bedeutet also $10$ pro $10^6$.  
+
+$\Delta f = \qty{145}{\mega\hertz} \cdot \frac{10}{10^6} = 145\cdot \cancel{10^6}\unit{\hertz}\cdot\frac{10}{\cancel{10^6}} = \qty{1450}{\hertz}$
+
+$f_\text{min} = f - \Delta f = \qty{145}{\mega\hertz} - \qty{1450}{\mega\hertz} = \qty{144,99855}{\mega\hertz}$
+$f_\text{max} = f + \Delta f = \qty{145}{\mega\hertz} + \qty{1450}{\mega\hertz} = \qty{145,00145}{\mega\hertz}$
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