document.write( "Question 44689: Kevin earned $165 interest for 1 year on an investment of
\n" ); document.write( "$1500. At the same rate, what amount of interest would be earned by an investment of $2500?\r
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Algebra.Com's Answer #29560 by tutorcecilia(2152)\"\" \"About 
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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)
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\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.
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\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
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