document.write( "Question 991355: The amount of annual interest earned by $8,000 invested at a certain rate is $270 less than $13,000 would earn at a 1% lower rate. At what rate is the $8,000 invested? \n" ); document.write( "
Algebra.Com's Answer #611244 by macston(5194)\"\" \"About 
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\n" ); document.write( "R=rate for $8000 in decimal form
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\n" ); document.write( "$8000R+$270=(R-0.01)$13000
\n" ); document.write( "$8000R+$270=$13000R-$130
\n" ); document.write( "$400=$5000R
\n" ); document.write( "0.08=R
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\n" ); document.write( "ANSWER: The $8000 was invested at 8%.
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\n" ); document.write( "CHECK:
\n" ); document.write( "$8000R+$270=(R-0.01)$13000
\n" ); document.write( "$8000(0.08)+$270=(0.08-0.01)$13000
\n" ); document.write( "$640+$270=(0.07)$13000
\n" ); document.write( "$910=$910\r
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