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anonymous

  • 5 years ago

8. A certain strain of virus grows in numbers at the rate of 50% per hour. If its present population is 611,000 what will be its population count in 4 hours?

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  1. anonymous
    • 5 years ago
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    Nt=No*e^(r*t) Nt = Future Population No = Starting population r = growth rate t= number of years.

  2. anonymous
    • 5 years ago
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    "t" in this case i in hours not in years

  3. anonymous
    • 5 years ago
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    Can you please explain a little better. I am so confused.

  4. dumbcow
    • 5 years ago
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    the rate of growth is 50% per hour not continuously, i dont think we use "e" here it will grow too quickly \[P _{t}=611*1.5^{t}\] each hour it will be 150% or 1.5X the prev population

  5. anonymous
    • 5 years ago
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    ok. Let's do with V variables for Virus:\[V(t) = (Vo) e^{rt}\] V(t) = Future Population of virus Vo = Present/initial population of virus r = growth rate (grows in numbers at the rate of ) t = how much time (in this case hours) From the question we know that: V(t) = we are trying to find this Vo = 611,000 r = 50% = 0.50 t = 4 hours hence, \[V(4) = 611,000*e^{0.5*4}\]

  6. anonymous
    • 5 years ago
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    hmm... i might be wrong thou

  7. dumbcow
    • 5 years ago
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    your math is good, but i dont think it applies to this problem e^rt is usually used when modeling a rate compounded continuously

  8. anonymous
    • 5 years ago
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    It would be something like P*(1+r/100)^t where, P -> current population r -> rate of growth t -> time period

  9. anonymous
    • 5 years ago
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    so it would come up to 3093187.5

  10. anonymous
    • 5 years ago
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    ooh... i see.. you are right, it's just a basic form P = Po*(1+r)^t

  11. anonymous
    • 5 years ago
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    thanks everyone. this is so confusing I think I am may be seeing the light with these problems.

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