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anonymous

  • 5 years ago

HELPP?? This problem is driving me nuts. Its Definite Integral Problem. Thanks a ton in advance!! If t is in years since 1990 , the population, P , of the world in billions can be modelled by P=5*3e^(0.014t) What does this model give for the world population in 1990 ? In 2000 ? Round answers to one decimal place. Use the Fundamental Theorem to find the average population of the world during the 1990's. Round the answer to one decimal place.

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  1. anonymous
    • 5 years ago
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    hmmm, looks like we should find the derivative and then do an integral. Is that the idea?

  2. anonymous
    • 5 years ago
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    the population P is already given by the equation, so i'm assuming you have no trouble in finding the first two answers, am I right?

  3. anonymous
    • 5 years ago
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    ill brb

  4. anonymous
    • 5 years ago
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    Okay, for the population of the world in 1990 and 2000, that's just plugging in 0 and 10 for t in the equation. I need these values for the next part, so I'll go ahead and answer those. P(0)=15, and P(10)=17.254 approximately. The average value of a function is \[[(1/(b-a)]\int\limits_{a}^{b}f(x)dx\] \[[(1/(10-0)]\int\limits_{0}^{10}(5*3e^{.014t})dt\] \[(1/10)*161.0076\] 16.101 billion is the average population during the 1990's.

  5. anonymous
    • 5 years ago
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    Although rounded to one decimal place it's 16.1 billion. Oops :p

  6. anonymous
    • 5 years ago
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    cant have done it better myself. how do you get the equations to work?

  7. anonymous
    • 5 years ago
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    What do you mean? How do I get them to show up correctly? If that's what you're asking, I just use the Equation button when I'm typing and make sure I use all the correct [] () and {} to make it look right.

  8. anonymous
    • 5 years ago
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    it's not showing up for me somehow, when i press the eqn button. guess im using an unsupported browser

  9. anonymous
    • 5 years ago
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    I guess, or your computer's old or something haha. Sorry!

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