if a clean up crew reduces pollutants at 8% a week, how long until only 1/5 th of the original concentration remains??
I believe the half-life is 70/.08 = 8.75 weeks
I can't figure out the time!!?? PLEASE HELP ME
Thanks so much :)
2007-08-07 16:50:18 · 5 answers · asked by princesssweetsarah 1 in Science & Mathematics ➔ Mathematics
(1 - 0.08)^w = 0.20, where w is weeks
w * ln(0.92) = ln(0.20)
w = ln(0.20) / ln(0.92)
w = 19.30 weeks
The half-life, by the way is:
(1 - 0.08)^w = 0.50
w * ln(0.92) = ln(0.50)
w = ln(0.50) / ln(0.92)
w = 8.313 weeks
2007-08-07 16:56:46 · answer #1 · answered by McFate 7 · 0⤊ 0⤋
Use the formula n=(1-r)^t, with n=the amount left, r=the rate of cleaning and t=number of weeks.
We plug in n=.2 (because only 1/5 is left) and r=.08, for 8%. This gives us .2=.92^t. We solve for t.
ln(.2)=ln(.92^t)
ln(.2)=t*ln(.92)
t=ln(.2)/ln(.92)=19.3 weeks
2007-08-07 17:00:22 · answer #2 · answered by Matt 4 · 0⤊ 0⤋
1/5 = (.92)^w
w = log base .92 of 1/5 = log(1/5)/log(.92) = 19.3021 weeks
2007-08-07 16:58:20 · answer #3 · answered by pki15 4 · 0⤊ 0⤋
(1-.08)^t = 1/5
t = -ln5 / ln.92 = 19.3 weeks
2007-08-07 16:56:11 · answer #4 · answered by sahsjing 7 · 1⤊ 0⤋
missiing info, how much pollutants at t=0?
2007-08-07 16:55:53 · answer #5 · answered by vlee1225 6 · 0⤊ 0⤋