document.write( "Question 591829: If $3000 is invested at an interest rate of 5% each year, find the amount of the investment at the end of 4 years for the following compounding methods.\r
\n" ); document.write( "\n" ); document.write( "B)semiannually, which is six months, but it is also half a year, so I'm not sure which variable this is, as well as what number it is. \r
\n" ); document.write( "\n" ); document.write( "Note: I know that the formula I need to use is this: A(t)= P[1+(r/n)]^nt
\n" ); document.write( "Initially I tried plugging in the values: A(t)= 3000(1+0.05/4)^4, this was just a guess (this is one is for Annually though). I couldn't figure that one out, but since I think B) might be harder I tried that one. I had no success getting the right answer. I'd appreciate your help on this one, Thank you so much.
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Algebra.Com's Answer #375743 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
\"A=P%281%2Br%2Fn%29%5E%28n%2At%29\" Start with the compound interest formula\r
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\n" ); document.write( "\n" ); document.write( "\"A=3000%281%2B0.05%2F2%29%5E%282%2A4%29\" Plug in \"P=3000\", \"r=0.05\", \"n=2\" (since we're compounding twice a year) and \"t=4\".\r
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\n" ); document.write( "\n" ); document.write( "\"A=3000%281%2B0.025%29%5E%282%2A4%29\" Evaluate \"0.05%2F2\" to get \"0.025\"\r
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\n" ); document.write( "\n" ); document.write( "\"A=3000%281.025%29%5E%282%2A4%29\" Add \"1\" to \"0.025\" to get \"1.025\"\r
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\n" ); document.write( "\n" ); document.write( "\"A=3000%281.025%29%5E%288%29\" Multiply \"2\" and \"4\" to get \"8\".\r
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\n" ); document.write( "\n" ); document.write( "\"A=3000%281.21840289750992%29\" Evaluate \"%281.025%29%5E%288%29\" to get \"1.21840289750992\".\r
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\n" ); document.write( "\n" ); document.write( "\"A=3655.20869252975\" Multiply \"3000\" and \"1.21840289750992\" to get \"3655.20869252975\".\r
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\n" ); document.write( "\n" ); document.write( "\"A=3655.21\" Round to the nearest hundredth (ie to the nearest penny).\r
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\n" ); document.write( "\n" ); document.write( "So there is $3655.21 in the account after 4 years (where $3000 is invested at an interest rate of 5% each year and it's compounded semiannually).
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