document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1135854>Question 1135854</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Can someone please help:\r<BR>\n" );
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document.write( "Consider this scenario: A town's population has been decreasing at a constant rate. In 2010 the population was 5,700. By 2013 the population had dropped 5,400. Assume this trend continues. \r<BR>\n" );
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document.write( "Predict the population in 2016.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #753557 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=753557\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> in 2010 the population was 5700.<BR>\n" );
document.write( "in 2013 the population was 5400.\r<BR>\n" );
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document.write( "that's a loss of 300 in 3 years which becomes an average loss of 300 / 3 = 100 per year.\r<BR>\n" );
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document.write( "if this trend continues, the population in 2016 is predicted to be 5400 - (2016 - 2013) * 100 = 5400 - 3 * 100 = 5400 - 300 = 5100.\r<BR>\n" );
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document.write( "this assumes a straight line loss which is a constant amount per year.\r<BR>\n" );
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document.write( "the equation for the straight line loss would be y = 5700 - 100 * x, where x is the number of years since 2010.\r<BR>\n" );
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document.write( "this can be graphed and looks like this.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" src = \"http://theo.x10hosting.com/2019/030201.jpg\" alt=\"$$$\" >\r<BR>\n" );
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document.write( "when x = 3, the year is 2010 + 3 = 2013.<BR>\n" );
document.write( "when x = 6,the year is 2010 + 6 = 2016.\r<BR>\n" );
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document.write( "the above formuls is used if the drop in the population is at a constant amount per year.\r<BR>\n" );
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document.write( "if the drop in the population is at a constant rate per year, then the formula is different.\r<BR>\n" );
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document.write( "to calculate the average rate of loss, use the following formula.\r<BR>\n" );
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document.write( "f = p * (1 + r) ^ n\r<BR>\n" );
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document.write( "f is the future value<BR>\n" );
document.write( "p is the present value<BR>\n" );
document.write( "r is the interest rate per time period (assumed to be years in this problem).<BR>\n" );
document.write( "n is the number of time periods (assumed to be years in this problem.\r<BR>\n" );
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document.write( "since p = 5700 and f = 5400, and n = 3 (from 2010 to 2013), the formula becomes:\r<BR>\n" );
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document.write( "5400 = 5700 * (1 + r) ^ 3\r<BR>\n" );
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document.write( "divide both sides of this equation by 5700 to get 5400 / 5700 = (1 + r) ^ 3\r<BR>\n" );
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document.write( "take the third root of both sides of this equation to get (5400 / 5700) ^ (1/3) = 1 + r.\r<BR>\n" );
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document.write( "subtract 1 from both sides of this equation to get (5400 / 5700) ^ (1/3) - 1 = r\r<BR>\n" );
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document.write( "solve for r to get r = -.0178609748.\r<BR>\n" );
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document.write( "that's your average interest rate per year.\r<BR>\n" );
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document.write( "confirm by using that rate in the formula of f = p * (1 + r) ^ 3\r<BR>\n" );
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document.write( "when p = 5700 and r =-.0178609748, the formula becomes f = 5700 * (1 - .0178609748) ^ 3 = 5400, which is correct, so the interest rate per year is good.\r<BR>\n" );
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document.write( "to find the population in s016, use the formula of f = p * (1 + r) ^ n, where:\r<BR>\n" );
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document.write( "p = 5700<BR>\n" );
document.write( "r =-.0178609748<BR>\n" );
document.write( "n = (2016 - 2010) = 6\r<BR>\n" );
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document.write( "to get f = 5700 * (1 - .0178609748) ^ 6 = 5115.789474 which rounds to 5116.\r<BR>\n" );
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document.write( "this formula can be graphed and looks like this.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" src = \"http://theo.x10hosting.com/2019/030202.jpg\" alt=\"$$$\" >\r<BR>\n" );
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document.write( "the difference between the methods is not so pronounced in only 6 years.\r<BR>\n" );
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document.write( "if you go out more years, the difference will be more pronounced.\r<BR>\n" );
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document.write( "going out 20 years, or even 30 years, you can see the difference in the following graph that shows both methods on the same graph.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" src = \"http://theo.x10hosting.com/2019/030203.jpg\" alt=\"$$$\" >\r<BR>\n" );
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document.write( "the red line is the constant amount per year loss.<BR>\n" );
document.write( "formula is y = 5700 - 100 * x.\r<BR>\n" );
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document.write( "the blue line is the constant rate per year loss.<BR>\n" );
document.write( "formula is y = 5700 * (1 - .0178609748) ^ x.\r<BR>\n" );
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document.write( "the values for the values of x indicated are:\r<BR>\n" );
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document.write( "year     x    red line     blue line    blue line minus red line\r\n" );
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document.write( "2010     0    5700         5700         0\r\n" );
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document.write( "2013     3    5400         5400         0\r\n" );
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document.write( "2016     6    5100         5116         16\r\n" );
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document.write( "2030     20   3700         3975         275\r\n" );
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document.write( "2040     30   2700         3319         619\r\n" );
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document.write( "</PRE>\r<BR>\n" );
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document.write( " </SPAN>\n" );
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