feature: solution for AI507
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# Gegeben
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* Anzeigefrequenz: $f_\text{Anzeige} = \qty{14100}{\kilo\hertz}$
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* Genauigkeit: $\pm \qty{0,00001}{\percent}$
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# Lösungsweg
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$\qty{1}{\percent}$ entspricht $1 \cdot {10^{-2}}$;
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$\qty{0,00001}{\percent}$ entspricht $0,1 \cdot {10^{-6}}$.
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$\Delta f = \qty{14100}{\kilo\hertz} \cdot \qty{0,00001}{\percent} = \qty{14,1 \cdot \cancel{10^6}}{\hertz} \cdot 0,1\cdot \cancel{10^{-6}} = \qty{1,41}{\hertz}$
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