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Math Help - Exam tomorrow need to check answers on growth and decay

  1. #1
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    Exam tomorrow need to check answers on growth and decay

    6) The radioactive isotope uranium 235 has a half-life of 8.8x10^8 years.
    A) How much of a 1-gram sample will decay after 1000 years?
    B) How long will it take for 90% of the mass to decay?

    7) A bacteria culture is known to grow at a rate proportional to the amount present. After 1 hour, 1000 bacteria are present and after 4 hours, 3000 bacteria are present.
    A) find an expression for the number of bacteria at any time T.
    B) Find the number of bacteria that will be present after 5 hours.
    C) When will the population reach 20,000?
    D) How long does it take for the population to triple?

    I did these out and i want to check my answers with anyone here
    My answers:
    6a) .99 or ~1
    6b) 2.9x10^9 years

    7a) Q(t)=1000e^(ln3/3)t
    b) 6240 bacterias after 5 hours
    c) 8.18 hours
    d) 3 hours


    thank you a lot
    Last edited by mr fantastic; December 10th 2008 at 11:41 PM. Reason: Deleted admission of double posting
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  2. #2
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    Quote Originally Posted by tjnguyen View Post
    Ok i made a post about this on the pre-calc forum but I am in a hurry to get these checked because i have a exam tomorrow.



    6) The radioactive isotope uranium 235 has a half-life of 8.8x10^8 years.

    A) How much of a 1-gram sample will decay after 1000 years?

    B) How long will it take for 90% of the mass to decay?



    7) A bacteria culture is known to grow at a rate proportional to the amount present. After 1 hour, 1000 bacteria are present and after 4 hours, 3000 bacteria are present.

    A) find an expression for the number of bacteria at any time T.

    B) Find the number of bacteria that will be present after 5 hours.

    C) When will the population reach 20,000?

    D) How long does it take for the population to triple?



    I did these out and i want to check my answers with anyone here

    My answers:

    6a) .99 or ~1

    6b) 2.9x10^9 years



    7a) Q(t)=1000e^(ln3/3)t

    b) 6240 bacterias after 5 hours

    c) 8.18 hours

    d) 3 hours



    thank you a lot
    6(a) ... look carefully at what the question is asking.

    6(b) is fine.


    7. your equation is fine if the population is 1000 at t = 0 ... however, the problem statement says P = 1000 at t = 1. what can you easily do to correct your equation? (think horizontal shift)
    Last edited by mr fantastic; December 10th 2008 at 11:35 PM. Reason: Added the quote to make for easier reading when I shift to original thread
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  3. #3
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    Follow up questions

    would i have to convert 1gram to kg? so .001kg?

    would it be .0001 for 6a?

    and for 7 im still stumped, can you help me?

    ok idk if this is right for 7a) but is the initial amnt bacteria would be 693? because i found K first its (ln3)/3 through and i found k through (((1000/(e^k)) * e^(4k)), and 1000/e^k = initial, plugged in 1000/e^((ln3)/3)) and got 693
    Last edited by mr fantastic; December 10th 2008 at 11:37 PM. Reason: The double posting of this question and the subsequent replies is making a mess of things.
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