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Umwandlung von keilschriftzahlen in Dezimalzahl und Dezimalbruch, Zufallswerte.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Die Baylonischen Keilschriftzahlen wurden sowohl für natürliche Zahlen als auch für Bruchzahlen verwendet, dabei ergeben sich Mehrdeutigkeiten in der Zahldarstellung.

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{image('resources/question-resources/'+chosenimage2)}

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Zur Vereinfachung gehen wir davon aus, dass zwischen den hier dargestellten drei Zifferngruppen keine leeren Stellenwerte vorkommen.

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", "advice": "

a) Dargestellt sind die Ziffern $\\var{chosenfact[0]},\\var{chosenfact[1]},\\var{chosenfact[2]}$.

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b) Es ergibt sich $\\var{chosenfact[0]}\\cdot 60^2+\\var{chosenfact[1]}\\cdot 60^1+\\var{chosenfact[2]}\\cdot 60^0=\\var{whole}$.

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c) Es ergibt sich $\\var{chosenfact[0]}\\cdot 60^0+\\var{chosenfact[1]}\\cdot 60^{-1}+\\var{chosenfact[2]}\\cdot 60^{-2}\\approx\\var{fraction}$

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Ermitteln Sie die in den drei dargestellten Zifferngruppen dargestellten Ziffern ($0<z_n<60$).

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1. Zifferngruppe: $z_1=$ [[0]]
2. Zifferngruppe: $z_2=$[[1]]
3. Zifferngruppe: $z_3=$[[2]]

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Welche ganze Zahl (im Dezimalsystem) ergibt sich, wenn die letzte Zifferngruppe Einer darstellt? 

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Welcher Dezimalbruch ergibt sich, wenn die erste Zifferngruppe Einer darstellt? 

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Bitte Beistrich (,) als Dezimaltrennzeichen verwenden!

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