document.write( "Question 157665: the population of glenbrook is 375000 and is increasing at the rae of 2.25% per year. predict when the population will be 1 million.\r
\n" ); document.write( "\n" ); document.write( "i neeed help. please answer this.... \n" ); document.write( "

Algebra.Com's Answer #116166 by jojo14344(1513) $\"\"$ $\"About$

You can put this solution on YOUR website!
$\"Population=P=375000\"$
\n" ); document.write( " $\"Rate=R=%280.0225%2A375000%29%2Fyear=8437.50%2Fyear\"$
\n" ); document.write( " $\"ProjectedPopulation=PP=1000000\"$ ------------> in $\"n\"$ years
\n" ); document.write( "Putting into equation:
\n" ); document.write( " $\"PP=P%2BR\"$ -----------------------------> working eqn
\n" ); document.write( " $\"1000000=375000%2B%288437.50%2Fcross%28year%29%29%28n%28cross%28years%29%29%29\"$
\n" ); document.write( " $\"%281000000-375000%29%2F8437.50=n\"$
\n" ); document.write( " $\"n=74.07407407years\"$ ---------> no. of years when popul'n will be 1,000,000
\n" ); document.write( "Or,
\n" ); document.write( " $\"74years\"$ and $\"%280.07407407%29%28cross%28year%29%29%2A%2812months%2Fcross%28year%29%29=0.8888months\"$
\n" ); document.write( "Or,
\n" ); document.write( " $\"74years\"$ and $\"%280.8888888%29%28cross%28month%29%29%2A%2830days%2Fcross%28month%29%29=26.66666days\"$
\n" ); document.write( "Or,
\n" ); document.write( " $\"74years\"$ , $\"26days\"$ and $\"%280.66666%29%28cross%28day%29%29%2A%2824hours%2Fcross%28day%29%29=16hours\"$
\n" ); document.write( "In doubt? Go back working eqn:
\n" ); document.write( "

\n" ); document.write( " $\"1000000=375000%2B625000\"$
\n" ); document.write( " $\"1000000=1000000\"$
\n" ); document.write( "thank you,
\n" ); document.write( "Jojo\r
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