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Mathematik Klausurübungen Lösung
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Sommersemester Mathematik Lösungen Übungen für Klausur
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Lösungen Probeklausur IIILösung 1
P1 P1 P3
2 2 10 6a
6 8 3 18a
1 2 3 3a
1 2 3 3a
6 8 3 18a
1 1 5 3a
1 2 3 3a
0 4 15 0
0 1 2 0
1 2 3 3a
0 1 2 0
0 4 15 0
1 2 3 3a
0 1 2 0
0 0 23 0
Daraus ergibt sich: P3 = 0, P2 = 0, P1 = 3a.
Es können keine 3 Einheiten von P3 produziert werden.
Lösung 2
0 1
Überweg T-Kreuzung Normale Kreuzung
a) MRZ = @ Fuß 2 6 8 A und
Auto 2 3 4
0 1
Wohngebiet Verkehrsgebiet
B Überweg 8 4 C
MZE = B @ T-Kreuzung
C und
A
6 7
Normale Kreuzung 10 14
0 1
Wohngebiet Verkehrsgebiet
MRE = @ Fuß 3 3 A.
Auto 2 5
0 1
✓ ◆ 8 4 ✓ ◆
2 6 8 @ A 3 3
M = MRZ MZE + MRZ = 6 7 + =
2 3 4 2 5
10 14
✓ ◆ ✓ ◆ ✓ ◆
132 162 3 3 135 165
= + =
74 85 2 5 76 90
✓ ◆
~ 20
b) E =
6
✓ ◆✓ ◆ ✓ ◆
~ ~ 135 165 20 3690
R = ME = =
76 90 6 2060
Damit werden für den Stadtbezirk 3690 Fußgängerampeln und 2060 Autoampeln ge-
braucht.
~ · (b a). es ist b = 1000 · t + a, t bezeichnet die Tausender-
c) Formal sieht das so aus: ~sT2 R
stelle. (Also die Gesamtsumme aller Ampeln mal den b a Watt Leistungseinsparung.
(3690 + 2060) · (b a) = 5750 · 1000 · t = 5750000 · t. Also 5750000 · t Watt Einsparung.
Lösung 3
a) Modell 1: geometrisch
2012: a0 = a, 2020: a8 = a0 q 8 . Es ist somit, a8 = b = 1000 · t + a, wobei t die
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