Step-by-step solved example
BOD decays with k=0.23 /d. What fraction remains after 3 days?
1. Evaluate
N/N0=e^{−0.23×3}=e^{−0.69}≈0.50.
2. Read
About one half-life (ln2/0.23≈3.0 d).
Answer: ≈ 50% remains
BOD decays with k=0.23 /d. What fraction remains after 3 days?
1. Evaluate
N/N0=e^{−0.23×3}=e^{−0.69}≈0.50.
2. Read
About one half-life (ln2/0.23≈3.0 d).
Answer: ≈ 50% remains
1.If k>0 the quantity
2.Half-life for k=0.0693 /yr is
3.Doubling time is
4.e^{−0.693} is nearest
5.Continuous compounding F=Pe^{rt} matches this model with k=
6.After 2 half-lives the remainder is
7.k has units of
8.N=100 e^{−0.1 t} at t=0 is
9.Linearize by plotting
10.If remainder is 37% , kt is nearest