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anonymous

  • 5 years ago

If 7,500 bacteria, with a growth constant k=0.4, are present 3hrs after beginning the experiment, how many are present 8 hrs later? a) 24,849 bacteria b) 2,259 bacteria c) 610,500 bacteria d) 184,000 bacteria

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  1. amistre64
    • 5 years ago
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    i assume the k refers to 'percentage'

  2. amistre64
    • 5 years ago
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    7500 = P(1.4)^3

  3. amistre64
    • 5 years ago
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    P= 7500/(1.4)^3 X = [7500/(1.4)^3](1.4)^11

  4. amistre64
    • 5 years ago
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    the closest number i get to the choices is d, but i cant say for sure that I was right

  5. amistre64
    • 5 years ago
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    i see it; e^kt ....

  6. amistre64
    • 5 years ago
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    7500 = P.e^(.4*3)

  7. amistre64
    • 5 years ago
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    183 ....... so its still D but with more confidence lol

  8. anonymous
    • 5 years ago
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    okay thank you

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