document.write( "Question 982861: the fish population is decreasing at a rate of 3% per year. in 2002 there were about 1900 fish. find the fish population in 2010 \n" ); document.write( "

Algebra.Com's Answer #603641 by josgarithmetic(39618) $\"\"$ $\"About$

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One way is to say population decreases to 97% each year.
\n" ); document.write( "and then form:
\n" ); document.write( " $\"p%28t%29=p%2A%280.97%29%5Et\"$ , where t is time since year 2002.\r
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\n" ); document.write( "\n" ); document.write( "p=1900,
\n" ); document.write( "t=0 for year 2002;\r
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\n" ); document.write( "\n" ); document.write( "Year 2010 is then t=8.\r
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\n" ); document.write( "\n" ); document.write( " $\"p%288%29=1900%280.97%29%5E8\"$ \r
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\n" ); document.write( "\n" ); document.write( " $\"p%288%29=1900%2A0.783743\"$ \r
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\n" ); document.write( "\n" ); document.write( " $\"highlight%28p%288%29=1490%29\"$ , which could be said as 1500, just for less accuracy if the 1900 were meant as two significant figures. \n" ); document.write( "

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