document.write( "Question 76870: PLEASE SOMEONE HELP ME?
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document.write( "Business and finance. 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? \n" );
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Algebra.Com's Answer #55110 by tutorcecilia(2152)![]() ![]() You can put this solution on YOUR website! Use the formula for simple interest: \n" ); document.write( "Interest = (Principle)(rate)(time) \n" ); document.write( ".\r \n" ); document.write( "\n" ); document.write( "Plug-in the values: \n" ); document.write( "$165=(1500)(rate)(1 year) \n" ); document.write( "165=1500(r)(1) [solve for rate (r)] \n" ); document.write( "165=1500r \n" ); document.write( "165/1500=1500r/1500 \n" ); document.write( ".11=r \n" ); document.write( ".11 X 100% = 11% = rate \n" ); document.write( ". \n" ); document.write( "Use the rate of 11% for the second part of the problem: \n" ); document.write( "I=Prt \n" ); document.write( "Interest =($2500)(.11)(1year) [solve for interest (I)- \n" ); document.write( "I=(2500)(.11)(1) \n" ); document.write( "I=275.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |