document.write( "Question 44689: Kevin earned $165 interest for 1 year on an investment of
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document.write( "$1500. At the same rate, what amount of interest would be earned by an investment of $2500?\r
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document.write( "Thank You!!! \n" );
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Algebra.Com's Answer #29560 by tutorcecilia(2152) You can put this solution on YOUR website! Use the formula for interest: I = prt \n" ); document.write( "I=interest \n" ); document.write( "p=principle \n" ); document.write( "r=rate or percent \n" ); document.write( "t=time in years (annually) \n" ); document.write( ". \n" ); document.write( "Plug-in the known values into the formula for interest and solve for the unknown variable: \n" ); document.write( "I = prt \n" ); document.write( "$165=(1500)(r)(1) \n" ); document.write( "165=1500r \n" ); document.write( "165/1500=4 \n" ); document.write( ".11=r or 11% is the rate. \n" ); document.write( ". \n" ); document.write( "For the second part, plug-in the known values into the formula for interest and solve for the unknown variable: \n" ); document.write( "I = prt \n" ); document.write( "I=(2500)(.11)(1) \n" ); document.write( "I=$275 \n" ); document.write( " \n" ); document.write( " |