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Math Help - Pre-test 6b Help Hurry

  1. #1
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    Pre-test 6b Help Hurry

    questions are on attachment!!!
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  2. #2
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    Hey Brookester:

    Let's do #4. The bacteria.

    What you need to do is use the given data and solve for k.

    Once you know k, you can use that to solve the problem.

    Given:

    125=50e^{3k}

    Solve for k:

    \frac{5}{2}=e^{3k}

    ln(\frac{5}{2})=3k

    k=\frac{ln(\frac{5}{2})}{3}

    Now use that in the problem to solve. The initial population is 50. Triple that is 150.

    150=50e^{\frac{ln(\frac{5}{2})}{3}t}

    3=e^{\frac{ln(\frac{5}{2})}{3}t}

    ln(3)=\frac{ln(\frac{5}{2})}{3}t

    t=\frac{3ln(3)}{ln(\frac{5}{2})}\approx{3.60}
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  3. #3
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    Hello, Brooke!

    I believe we're suppose to use: A \:= \:P\left(1 + \frac{i}{n}\right)^{nt}\quad\Rightarrow\quad P \;=\;\frac{A}{\left(1 + \frac{i}{n}\right)^{nt}}

    where P is the principal investment,
    i is the annual interest rate,
    n is the number of compounding periods per year,
    t is the number of years.


    2. An investment with an interest rate of 7% per annum yields $4000 in 3.5 years.
    What is the present value for each compounding period?

    . . A: annually . . B: semi-annually . . C: quarterly . . D: monthly


    (A) annually: n = 1

    P \;= \;\frac{4000}{(1 + 0.07)^{3.5}} \;=\;3156.579774 \:\approx\; \$3,156.58


    (B) semi-annually:  n = 2

    P \;= \;\frac{4000}{\left(1 + \frac{0.07}{2}\right)^{(2)(3.5)}} \;= \; 3143.963843 \:\approx\:\$3143.96


    You can do the other two . . .

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