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Thread: General Equations - Word Problem

  1. #1
    Jan 2013

    General Equations - Word Problem

    Could someone help me figure this out please? I have attached the problem (jpg). If you would like to know the answer lmk. Thanks
    Attached Thumbnails Attached Thumbnails General Equations - Word Problem-dsc00163.jpg  
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  2. #2
    Super Member

    May 2006
    Lexington, MA (USA)

    Re: General Equations - Word Problem

    Hello, Sprinkledozer!

    A telephone company estimates that the number N of phone calls per day
    between two cities of population $\displaystyle P_1$ and $\displaystyle P_2$ that are $\displaystyle d$ miles apart is given by:

    . . $\displaystyle N \:=\:\frac{2.51P_1P_2}{d^2}$

    115. Estimate the population $\displaystyle (P_1)$ of a city given that
    . . . . . the population of a second city is: $\displaystyle P_2 = 48,\!000$,
    . . . . . the number of phone calls per day is: $\displaystyle N = 1,\!100,\!000$,
    . . . . . and the distance between the cities is: $\displaystyle d=75$ miles.
    . . . Round to the nearest thousand.

    First, solve for $\displaystyle P_1.$

    We have: .$\displaystyle \frac{2.51P_1P_2}{d^2} \:=\:N$

    Multiply by $\displaystyle d^2\!:\;\;2.51P_1P_2 \:=\:d^2N $

    Divide by $\displaystyle 2.51P_2\!:\;\;P_1 \;=\;\frac{d^2N}{2.51P_2}$

    Substitute: .$\displaystyle P_1 \;=\;\frac{75^2(1,\!100,\!000)}{2.51(48,\!000)} \;=\;51,357.07171$

    Therefore: .$\displaystyle P_1 \;\approx\;51,\!000$
    Thanks from Sprinkledozer
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