Results 1 to 3 of 3

Thread: Help Hurry Pre-test 6a

  1. #1
    Jun 2006

    Help Hurry Pre-test 6a

    questions on attachment!!!
    Attached Files Attached Files
    Follow Math Help Forum on Facebook and Google+

  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:
    $\displaystyle g=10*\left(\log_{10}P_o-\log_{10}P_i\right)=10*\log_{10}\left(\frac{P_0}{P _i}\right)$

    Now plug in the given values:

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

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

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

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

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


    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

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

    4. The amount of power generated by a satellite's power supply is: $\displaystyle P\:=\:50e^{-\frac{t}{300}}$
    where $\displaystyle P$ is the power in watts and $\displaystyle t$ 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: $\displaystyle P = 22$

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

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

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

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

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

    5. The mathematical model for learning an aseembly line procedure needed for assembling
    one component of a manufactured item is: .$\displaystyle P\:=\:\frac{0.93}{1 + e^{-2n}}$
    where $\displaystyle P$ is the proportion of correctly assembled compnents after $\displaystyle n$ 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 $\displaystyle P = 0.75$
    The equation becomes: .$\displaystyle \frac{0.93}{1 + e^{-2n}}\:=\:0.75\quad\Rightarrow\quad0.93\:=\:0.75(1 + e^{-2n})$

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

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

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

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

    An assembler achieves a 75% proficiency level during the first session?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Ask for help on Poisson Distribution~~~ Hurry,hurry!!!
    Posted in the Advanced Statistics Forum
    Replies: 6
    Last Post: Apr 25th 2009, 11:00 PM
  2. in a hurry...thanks
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: Jan 21st 2008, 08:41 PM
  3. Replies: 8
    Last Post: Dec 7th 2006, 01:35 PM
  4. Pre-test 6b Help Hurry
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Jul 17th 2006, 04:08 PM
  5. help..hurry
    Posted in the Algebra Forum
    Replies: 4
    Last Post: May 4th 2005, 05:44 PM

Search Tags

/mathhelpforum @mathhelpforum