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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

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  1. anonymous
    • 5 years ago
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    Is this a first order reaction?

  2. Yuki
    • 5 years ago
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    these problems are easier than it looks like

  3. Yuki
    • 5 years ago
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    let's say N is the current amount and N_0 be the original amount

  4. Yuki
    • 5 years ago
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    in one year, N-0 becomes 99.75% of N_0 so \[N_1 = .9975*N_0\]

  5. Yuki
    • 5 years ago
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    next year, N_1 becomes 99.75% of N_1 so \[N_2 = .9975*N_1\]

  6. Yuki
    • 5 years ago
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    if you substitute N_1 with N_0, \[N_2 = (.9975)*(.9975N_0)\]

  7. Yuki
    • 5 years ago
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    as 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 \[N_t = N_0(.9975)^t\]

  8. Yuki
    • 5 years ago
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    I 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, \[.24 = (.9975)^t\]

  9. Yuki
    • 5 years ago
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    all you have to do is to solve for t :)

  10. Yuki
    • 5 years ago
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    oops, sorry not .24 it is actually .76 because the wording says "drops 24%" not "is 24%" so you are solving for \[.76 = (.9925)^t\]

  11. anonymous
    • 5 years ago
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    Is the answer 109.64 years?

  12. Yuki
    • 5 years ago
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    since\[t = {\ln(.76) \over \ln(.9975)} \] which is approximately 109.6 I'd use 110

  13. Yuki
    • 5 years ago
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    yep

  14. anonymous
    • 5 years ago
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    thanks!

  15. Yuki
    • 5 years ago
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    np :) most exponential decays and growths work the same way, so let me know if you need more help ok ?

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