current population= 257 growth rate=5% no. of yrs=10 find initial population?
369524
Initial Population was 157.\n\nUsing  (Initial)*e^(rate*years)=current
943178
A = A_(0)*exp(k*t) \n\nA represents the amount at a given time t.\n\nAo represents the initial amount of the growing entity.  Note that this is the amount when t = 0.\n\n\nk is a constant that represents the growth rate.  It is POSITIVE when talking in terms of exponential GROWTH.\n\n\nt is the amount of time that has past.  If the information for time is given in dates, you need to convert it to how much time has past since the initial time.  For example, if the model is set up at an initial year of 2000 and you need to find out what the value is in the year 2010, t would be 2010 - 2000 = 10 years.\n\nYou can use this formula to find any of its variables, depending on the information given and what is being asked in a problem.  For example, you may be given the values for Ao and t and you need to find the amount A after the given time.  Or, you may be given the final amount A and the initial amount Ao and you need to find the time t.\n\nSome examples of exponential growth are population growth and financial growth.  The information found, can help predict what a population for a city or colony would be in the future or what the value of your house is in ten years.  Or you can use it to find out how long it would take to get to a certain population or value on your house.
699905
mass initial = Mi\nmass final = Mf\n\nMass growth = Mf - Mi\nPercent growth = Mf/Mi * 100
977857
The population can be modelled using the following equation.\nP(t) = P(initial) * 2^(t/10)\n\nYou know P(90) = 90000:\nSo plug this in:\nP(90) = 90000 = P(initial) * 2^(90/10)\n\n90000 = P(initial) * 2^9\n\nSolve for P(initial):\nP(initial) = 90000 / 512\nP(initial) = 175.78125\n\nHere are the details:\nP(0) = 175.78125\nP(10) = 351.5625\nP(20) = 703.125\nP(30) = 1,406.25\nP(40) = 2,812.5\nP(50) = 5,625\nP(60) = 11,250\nP(70) = 22,500\nP(80) = 45,000\nP(90) = 90,000\n\nYou can also do this working backwards, just dividing by 2.  So at time t = 80, you have half the population (45,000) at t = 70 you have half of that (22,500), etc.  Until you get to t = 0.\n\nSo the answer is:\nInitial population = 175.78125 ≈ 176 bacteria
523621
(10+15+23+37+45) / 5 = 26
903062
Revenue = Price x Quantity\nR = P*Q\n\nLet i be the increase in Price\nP = 40 + i\nQ = 5,000 - 100 * i\n\nR = (40 + i) * (5,000 - 100 * i)\n   = 200,000 + 5,000i - 4,000i -100i^2\n   = 200,000 + 1,000i - 100i^2\n\nDifferentiate R and set = 0 to find the maximum\n\ndR/di = 1,000 - 200i = 0\n 200i = 1,000\n      i = 5\n\nPlug this back in to the equation for Price = $40 + i\nP = 40 + 5 = $45\nQ = 5,000 - 100 *5 = 4,500\nR = P * Q = 45 * 4,500 = $202,500\n\nYou can calculate R at other prices ( try i = 4 and i = 6 ).  Then, you can verify that i = 5 is the MAX!\n\nThe important thing to learn is how to set-up the equations.  In this case we needed to create a new variable "i" in order to describe the price - quantity relationship.\n\nGood Luck\nm
1043517
Use the compound interest formula:\n\nA(t) = A (1+r/n)^(nt)  where A(t) is final value, A is starting value, r is interest rate, n is number of times interest is counted in a year, t is number of years. In this case we want to figure r.  Thus,\n1690 = 1255 (1+ r/4)^(2.75)(4)\n1690/1255 = (1+ r/4)^(11)\n1.3466 = (1+ r/4)^(11)\n(1.3466)^(1/11) = (1+ r/4)\n1.0274 = (1+ r/4)\n.0274 = r/4\nr = .1096 or 10.96%
724452
23*0.6/30000*0.05\n=13.8/1500\n=0.0091
915560
let\nx = amount in growth fund,\ny = amount in income fund,\nz = amount in money market fund\n\n\nx + y + z = $3,200\n0.07x + 0.10y + 0.05z = $250\nx = 2z
774657
Total = P * (1 + r)^t = $1762.34\n\nGeneral formulation:\nP * (1 + r/c)^(t*c)\nP = initial principle\nr = rate/yr in decimal.\nt = # of years\nc = # of periods that is compounded in a year
1085819
10 - 15% = 8.5\n\nor\n\n10 - (10*0.15) = 8.5\n\nturn % to decimal and multiply it by the number from which you are subtracting the percent. Then subtract that number from the original number\n\nremember each percent is 2 decimal places left - 8% = 0.08, 25% = 0.25, 125% = 1.25
753269
300000*8%=?\nthen subtract that amount from the 300000
32878
200\n\n(1,000,000/5,000=200)
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To find a rate of change, you need to find out how much the average has changed from one point to the next, then divide it by how many units that change is spread across:\n\n1. ((78/10) - (48/4))/(10-4) = (7.8 - 12)/6 = -4.2/6 = -,7\nThe price decreased by $.70 per person\n\n2. (120/4 - 30/1)/(4-1) = (30 - 30)/3 = 0\nThe rate did not change\n\nSlopes:\nThe easiest way to do a point-slope problem is to use the slope-intercept form of the linear equation, y = mx + b, where m is the slope and b is the y-intercept (the place where x = 0, and the line crosses the y-axis).  Plug in your known values and solve:\n\n1. 5 = -2(3) + b\n5 = -6 + b\n11 = b\ny = -2x + 11\n\n2. -3 = (-4/3)(2) + b\n-3 = -8/3 + b\n-1/3 = b\ny = -4/3x - 1/3
625377
% increase=difference/original value\n\ntherefore % inc. =(21.0-20.7)/20.7\n                 =0.3/20.7\n                 =0.014492\n                 =0.01% to nearest hundredth
98834
The rule of 70 states that to find the doubling time for any exponential growth, divide 70 by the percent of annual increase. So, if the population grows 10% per year, it will double in 7 years.
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The way you solve this problem is by\n\n25,000 *.057\n1,425 a year, but need only 9 months so divide by 12 and get\n118.75 a month and multiply by 9 to recieve your answer\n1068.75