Atlantic Posted February 8, 2004 Posted February 8, 2004 Sorry, it has been so long since i took math. If you have a value (lets say 8000) decreasing by 13% every hour (13% of its actual value not the original value). What would be the formula to find out the current value after 5 hours. Its like bonds. You invest money and it keeps increasing in accordance with the current value. thanks in advance.
hamzah Posted March 21, 2004 Posted March 21, 2004 I not sure what you mean by bonds, but with annual compound interest...to calculate how much you will have after a given number of years it is: (original value) * (increase)^(number of years) 15% increase would be 115/100 = 1.15 an example would be: £1000 in the bank...with a annual compound interest of 10%...how much is in the bank account after 3 years? 1000 * 1.1^3 = £1331 And in reply to your problem: Sorry, it has been so long since i took math.If you have a value (lets say 8000) decreasing by 13% every hour (13% of its actual value not the original value). What would be the formula to find out the current value after 5 hours. it would be: (initial value) * (decrease)^(number of hours) 8000 * (0.87)^5
xXxAuroraxXx Posted April 12, 2004 Posted April 12, 2004 y = 8000(1 - .13)^5 y = 8000(.87)^5 y = 8000(.498) y = 3987.86 Im not sure if "y" is the correct variable... If you wanted to see increase y = 8000(1 + .13)^5 I think.
Dave Posted April 12, 2004 Posted April 12, 2004 That's right. In general, for things like these interest values, you can use the formula: y = x*(interest)^n to calculate the value of x after n iterations of applying interest or whatever to it. Note that this only works for values of n which are in the natural (counting) numbers - it doesn't work for n = 12.32 for example, for that you need faf's equation.
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