Gegeben
f(5) = 0.7
f(6) = 0.6
Exponentielle Abnahme angenommen.
f(5) = 0.7 = a*q^5 |*q (I)
f(6) = 0.6 = a*q^6 (II)
==> 0.7 * q = 0.6
==> q = 0.6/0.7 = 6/7 in (I)
0.7 = a*(6/7)^5
0.7 / (6/7)^5 = a
Also ca. 1.51 Promille war der Anfangswert a.
f(5) = 0.7
f(6) = 0.6
Lineare Abnahme angenommen.
f(5) = 0.7 = 5m + q | +m (I)
f(6) = 0.6 = 6m+q (II)
==> m = -0.1 in (I)
0.7 = -0.1*5 + q
1.2 = q
Also ca. 1.2 Promille Anfangswert.