Please help with this interest rate problem. First correct answer gets 10 points.?
969001
The real interest rate is the nominal interest rate, less inflation.  In this case, we have RI = 0.1 - (-2.0) = 2.1%.
1153478
B is the answer.  If they pay before interest kicks in, no cost other than the goods themselves are involved. "0" percent APR.
26357
You don't need linear programming for this...\nYou want to maximize the investment in C, then in B, then in A.\n\nHe should in invest $6,000 in bond C.\nThen he can invest only $2,000 in bond B.\nThen he can invest only $8,000 in bond A.\n\nHe'll have $2,000 remaining.\n\n\nEdit to add:\nSorry, my LP skills are a bit rusty.  Here's all I remember:\nMaximize .05*A + .75*B + 1.05*C;\nsubject to   \nA + B + C <=18000\nA + B <=10000\nB + C <=8000\nC <=6000
487839
Need to know how much you invested at what interest rate first.
880662
I've got the first part.  I/rt = P\nSo, 93.08/(.065 times 1.33) = 1076.69 dollars\n.\nHere is the second one, I hope. I = Prt .\nSo, 56000times .12 times 1 = 6720\nNow divide 6720 by twelve to find the interest for each month.\nIt's 560 \nSubtract 560 from the monthly payment of 619.36 = 59.36. This is the amount coming off the principal. So after one month the principal is 56000- 59.36 = 55940.64\nHere is the third one.\nI = Prt\nSo 387000 times >147 times 15 = 853335 dollars to be paid back in 15 years. Divide that amount by 180 months(15 years) and you get a monthly payment of 4740.75\nI hope those folks have really good jobs. Ha
1126484
COMPOUND INTEREST\n---------------------------------\n\n S = P (1 + I) n \n\nwhere:\n\nS = the future value\n\nP = the principal amount\n\nI = the periodic rate of interest*\n\nn = the # of compounding periods for the term of the loan\n--------\n\nFor example,\n\nIf the principal amount is $10,000, the periodic rate of interest is 10% and the # of compounding periods is 6, what is the value at maturity?\n\nS = P (1 + I) n \n\nS = $10000 (1 + .10) 6\n\nS = $ 17715.61\n\nhttp://homepages.cambrianc.on.ca/tutorial/thetutorialcentre/mathscience/math/compound_interest.htm\n\n\nSIMPLE INTEREST \n------------------------------\n\nAt start:\n\nT = P\n\nAfter 1 period:\n\nT = P + Pr = P(1 + 1r)\n\nAfter 2 periods:\n\nT = P + Pr + Pr = P + 2Pr = P(1 + 2r)\n\nAfter 3 periods:\n\nT = P + Pr + Pr + Pr = P + 3Pr = P(1 + 3r)\n\nAfter n periods:\n\nT = P(1 + nr)\n\n--------------------\nWhen money is borrowed, interest is charged for the use of that money for a certain period of time. When the money is paid back, the principal (amount of money that was borrowed) and the interest is paid back. The amount to interest depends on the interest rate, the amount of money borrowed (principal) and the length of time that the money is borrowed. \nThe formula for finding simple interest is: Interest = Principal * Rate * Time. If $100 was borrowed for 2 years at a 10% interest rate, the interest would be $100*10/100*2 = $20. The total amount that would be due would be $100+$20=$120. \n\nSimple interest is generally charged for borrowing money for short periods of time. Compound interest is similar but the total amount due at the end of each period is calculated and further interest is charged against both the original principal but also the interest that was earned during that period. \nhttp://www.learningwave.com/lwonline/percent/interest.html\n\n\n\nhttp://www.aaamath.com/g84-simple-interest.html \na calculater ..
844754
I = Prt, \nwhere  I = interest, P = principal, r = interest rate and t = number of years\nP = 770, r = 3% = .03 and t = 9\nI = 770(.03)(9)\nI = 207.90
264272
to be able to solve this, first you have to determine the amount of money in the fund after 30 years of annual deposit. to solve this you'll use the formula for annuity, which is,\n\nP = A [1 - (1+i)^-n] / i\n\nP = is the pricipal, in this case this would be the amount of the fund after 30 years.\n\nA = is the annual withdrawal, which is 50,000\n\ni = is the interest rate, which is 10%\n\nn = is the number of years, in this case it would be 20 years\n\nsubstitute the values of A, i, and n\n\nP = 50,000 [1 - (1 + .1)^-20] / .1\n\nP = 425678.19, this would be the amount of the fund after 30 year\n\nfrom this you can get the annual deposit in the account, you'll be using this formula,\n\nF = A [(1+i)^n - 1] / i\n\nF = is the future amount, in this case it would be the amount of the fund after 30 years, which is 425678.19\n\nA = is the annual deposit, which we are looking for\n\ni = is the interest rate, 10%\n\nn = is the number of years, this time it would be 30 years\n\nsubstituting,\n\n425678.19 = A [(1+.1)^30 - 1] / .1\n\nA = 2587.80
819834
if interest is compounded daily, don't forget to...\n\n1.  divide 5.16% by 365 to get the daily rate\n2.  use 30 days as your period "n"\n3.  don't use the avg daily balance.  you must use the initial principal balance at the beginning of the 30 day period.\n\ni haven't worked it out, but if your formula is correct, the future value will work.\n\nif you have a business calculator (preferrably from HP)\n\nPV = initial principal balance\ni = 5.16% divided by 365 days\nn = 30 days\n\nsolve for FV
165316
So what if the rate in the 9th year is 5% and then the rate in the 15th year is 15% and all other 10%?\n\nUh oh......\n\nUse a spreadsheet program and you can investigate differences much more easily.
158774
if it is as i understood\nyou borrow =$10 000\nits profit during one year =$10 000* 0.09 (9%)= $900\nafter one year you must pay= $10 000+$ 900=$10 000 * (1.09)\n\nsimilary for the second year =$10 000*(1.09)^2\nsimilary for the third year =$10 000*(1.09)^3\n\nafter ten years=$10 000*(1.09)^10=$23 673.6367\n\nThe rule F=P*(1+i)^n\nF:Future Value Of Money\nP:Present Value Of Money\ni:Interest Rate During Certain Period (e.g. one year)\nn:Number Of Periods (e.g. ten years)
653412
(future value/present value)^(1/time)\n\nSo using your example, (200,000/$100,000)^(1/5) or the fifth root of 2, which is 1.148698 or 14.8698%.\n\nTo use your example, 1.148698 x 1.148698 x 1.148698 x 1.148698 x 1.148698 =2\n\nJust be careful with your units of time.  If you had found the monthly rate, you could not multiply it by twelve to find the annual rate, you would take it to the twelth power.\n\nOne simpler method, that is approximately correct is the rule of 72.  You can estimate most rates by dividing 72/time.  In this case it would give you 14.4.  It comes in handy when you don't have a calculator on hand.  See the wikipedia article on why it works.
267501
PRT/ 100 = interest P=principle R=rate T=time\n\nin your case\n\n10000(0.1)(10)= interest\n10000= interest\n\n(i'm not sure what your rate is? )
836682
Yr 1. 9000*.08= 720 + 9000\nYr 2. 9720*.08=777.60 +9720\nYr3. 10497.60*.08=839.81+10497.60\nYr 4. 11337.41*.08= 907 +11337.41\nYr 5. 12244.41*.08= 979.60 + 12244.41\nYr6. 13224.01*.08=1057.92+ 13224.01 = $14,281.93 total repaid\n5,281.93 total interest
289683
discounting factors for an interest rate of 8% \nat the end of  year 1 - 100/108 or 1/1.08 = 0.9259\nat the end of  year 2 - 1/1.1664 = 0.8573\nat the end of  year 3 - 1/1.2597 = 0.7938\n\n1.08, 1.1664, 1.2579 compound amounts at the end of year 1,2,3\n\nExpected Cash Flows\n\nDiscounting Factor.. 0.9259\nYear 1 Cash Flow .....$300...$500...$700\nYear 1 Dis Cash Flow..$278...$463...$648\nDiscounting Factor..0.8573.\nYear 2 Cash Flow .....$100...$200...$600\nYear 2 Dis Cash Flow..$86....$172...$514\nDiscounting Factor....0.7938\nYear 3 Cash Flow .....$200...$800...$900\nYear 3 Dis Cash Flow..$159...$635...$714\n\nYear 1\nDisc Cash Flow (x)..278...463...648\nProb P(x)...........0.3....0.4...0.3\nx.P(x) {px}.........83.4--185.2--194.4\n\nExpected Cash Flow in the first year = Sum(px)= $463\n\nYou should be able to calculate the rest..\n\nhttp://www.futureaccountant.com/probability/\n\nhttp://www.schoolingkids.com/
39368
Compound Interest= Principal Amount{1+(rate of intersest/100)^Numebr of years}\nFor you it will be\n21000=5000*(1+r/100)^5\n=>4.2=(1+r/100)^5\n=>1.33=1+r*0.01\n=>r=33%
393533
I don't feel like working it out, but you just discount back the cash flows assuming they're constant, and then do some simple algebra.\n\nIt looks kind of like\n\n(785000 - 137) = (x / 1.07) + (x / 1.07^2) + ... (x / 1.07 ^10)\n\nSolve for x.