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anonymous
 5 years ago
Suppose the amount of ozone in the atmosphere is decaying exponentially, losing 0.25% each year. How many years does it take for the ozone level to drop 24%?
A) 78 years
B) 86 years
C) 94 years
D) 102 years
E) 110 years
anonymous
 5 years ago
Suppose the amount of ozone in the atmosphere is decaying exponentially, losing 0.25% each year. How many years does it take for the ozone level to drop 24%? A) 78 years B) 86 years C) 94 years D) 102 years E) 110 years

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is this a first order reaction?

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0these problems are easier than it looks like

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0let's say N is the current amount and N_0 be the original amount

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0in one year, N0 becomes 99.75% of N_0 so \[N_1 = .9975*N_0\]

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0next year, N_1 becomes 99.75% of N_1 so \[N_2 = .9975*N_1\]

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0if you substitute N_1 with N_0, \[N_2 = (.9975)*(.9975N_0)\]

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0as you can see, the number of years will just tell you how many times the initial amount decreased by .25%, so the general eqn after t years is \[N_t = N_0(.9975)^t\]

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0I f we used this eqn, we can say that N_t = .24N_0, because it is 24% of the original state. or in other words, \[.24 = (.9975)^t\]

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0all you have to do is to solve for t :)

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0oops, sorry not .24 it is actually .76 because the wording says "drops 24%" not "is 24%" so you are solving for \[.76 = (.9925)^t\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is the answer 109.64 years?

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0since\[t = {\ln(.76) \over \ln(.9975)} \] which is approximately 109.6 I'd use 110

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0np :) most exponential decays and growths work the same way, so let me know if you need more help ok ?
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