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Thread: Help Hurry Pre-test 6a

  1. #1
    Jun 2006

    Help Hurry Pre-test 6a

    questions on attachment!!!
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  2. #2
    Super Member
    earboth's Avatar
    Jan 2006

    1,2,3 solved

    Quote Originally Posted by Lane
    questions on attachment!!!
    Hello, Lane,

    to 1) rearrange the expression:
    g=10*\left(\log_{10}P_o-\log_{10}P_i\right)=10*\log_{10}\left(\frac{P_0}{P  _i}\right)

    Now plug in the given values:

    g=10*\log_{10}\left(\frac{30 W}{.06 W}\right)=10*\log_{10}\left(500\right)\approx 26.9897...

    to 2) Use the substitution y=2^{x}. You'll get a quadratic equation:
    y^2-13y+42=0\quad\Longrightarrow\quad y=6\vee y=7

    Re-substitute: 2^x=6\ \Leftrightarrow\ x=\log_{2}(6)=\frac{\log_{10}(6)}{\log_{10}(2)} \approx 2.5849... or 2^x=7\ \Leftrightarrow\ x=\log_{2}7=\frac{\log_{10}7}{\log_{10}2}\approx 2.8073...

    to 3) Multiply the equation by the denominator. You'll get:
    900=25+25e^{-x} \Leftrightarrow\ 35=e^{-x}  \Leftrightarrow\ e^x=\frac{1}{35}

    Therefore: x=\ln\left(\frac{1}{35}\right)\ \Longrightarrow\ x\approx -3.555...


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  3. #3
    Super Member

    May 2006
    Lexington, MA (USA)
    Hello, Lane!

    4. The amount of power generated by a satellite's power supply is: P\:=\:50e^{-\frac{t}{300}}
    where P is the power in watts and t is the time in days.
    For how many days will 22 watts of power be available? .Round to the nearest whole day.

    We are given: P = 22

    The equation becomes: . 22\;=\;50e^{-\frac{t}{300}}

    Divide by 50: . 0.44\;=\;e^{-\frac{t}{300}}

    Multiply by e^{\frac{t}{300}}:\;\;0.44e^{\frac{t}{300}}\;=\;1 \quad\Rightarrow \quad e^{\frac{t}{300}}\:=\:\frac{1}{0.44}

    Rewrite in log form: . \frac{t}{300} \:=\:\ln\left(\frac{1}{0.44}\right)\:=\:-\ln(0.44)

    Therefore: . t\:=\:-300\cdot\ln(0.44) \:=\:246.2941656\:\approx\:246\text{ days}

    5. The mathematical model for learning an aseembly line procedure needed for assembling
    one component of a manufactured item is: . P\:=\:\frac{0.93}{1 + e^{-2n}}
    where P is the proportion of correctly assembled compnents after n practice sessions.

    How many practice sessions are required to have at last 75% of the components correctly
    assembled within the given time period? .Round any non-integer to the next higher whole number.

    Is there a typo? . . . I'm getting a strange answer.

    We want P = 0.75
    The equation becomes: . \frac{0.93}{1 + e^{-2n}}\:=\:0.75\quad\Rightarrow\quad0.93\:=\:0.75(1 + e^{-2n})

    . . . . 0.93\:=\:0.75 + 0.75e^{-2n}\quad\Rightarrow\quad 0.18\:=\:0.75e^{-2n}

    Multiply by e^{2n}:\;\;0.18e^{2n}\:=\:0.75\quad\Rightarrow \quad e^{2n}\;=\;\frac{0.75}{0.18}\:=\:\frac{25}{6}

    Log form: . 2n \:=\:\ln\left(\frac{25}{6}\right)\quad\Rightarrow \quad n \:=\:\frac{1}{2}\ln\left(\frac{25}{6}\right)

    Then we have: . n \:= \:0.713558178 \:=\:1

    An assembler achieves a 75% proficiency level during the first session?
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