2007-12-27 15:26:59 · 5 answers · asked by Steve 1 in Science & Mathematics ➔ Mathematics
1 - e^(- 35 / θ ) = 0.2
0.8 = e^(- 35 / θ )
ln 0.8 = - 35 / θ
θ = - 35 / ln 0.8
θ = 156.8
2007-12-27 22:26:59 · answer #1 · answered by Como 7 · 1⤊ 1⤋
1- e^(-35/theta) = 0.20
-e^(-35/theta) = 0.20 -1
= -0.80
e^(-35/theta) = 0.8
take the logarithm of both sides:
ln(e^(-35/theta)) = ln(0.8)
-35/theta = -0.22314
-35 = -0.22314 * theta
theta= -35/-0.22314
theta = 35/0.22314
theta = 156.85
2007-12-27 15:37:53 · answer #2 · answered by pico t 2 · 0⤊ 0⤋
1 - e^(-35/theta) = 0.20
Solution:
1 - 0.20 = e^(-35/θ)
0.80 = e^(-35/θ)
ln(0.80) = lne^(-35/θ)
ln(0.80) = (-35/θ)ln(e)
ln(0.80) = (-35/θ)1
θ = -35/ln0.80
θ = 156.85 ANS
Hop I help you.
teddy boy
2007-12-27 15:43:02 · answer #3 · answered by teddy boy 6 · 0⤊ 0⤋
theta=156.85
2007-12-27 16:21:20 · answer #4 · answered by someone else 7 · 0⤊ 0⤋
e^(-35/x) = 0.8
-35/x = ln 0.8 = -0.22
x = -35/-0.22 = 156.85
2007-12-27 15:34:23 · answer #5 · answered by norman 7 · 1⤊ 0⤋