Turtle population=> P(t)= 175(1.145)^t. how long does it take for the population to double.?
813587
2x=x(1.145)^t\nlog2+logx=logx+t log1.145\nt log 1.145t=log2\nt=log2/log1.145\n=0.3010/0.0588\n=5.2 years approx
639100
A = Pe^(rt)\nP = 25000, r = .08 and A = 50000 (since you are doubling and 25000*2 = 50000)\n50000 = 25000e^(.08t)\nDivide both sides by 25000\n2 = e^(.08t)\nTake the ln of each side\nln(2) = .08t\nDivide each side by .08\nln(2)/.08 = t\nt = 8.664339757
331500
(m)\n\nU.S. 298,594,235\nWorld 6,511,912,875\n10:24 GMT (EST+5) Apr 25, 2006
526116
Basically, all you need to know is the initial amount (8.0 grams I assume), and the half life (46 days).  Then you can build an exponential model of the form:\n\nA(t)=A*e^(k*t)  \nA is initial amount\nk is decay constant\nt is time\n\nKnowing that after 46 days, there will be half as much present:\n\n4=8*e^(k*46)\n\nThen solve for k using logarithms.  \n\nln(0.5)=46k\nk=ln(0.5)/46=-0.01507\n\nSubstitute into equation to get time dependent equation:\n\nA(t)=8*e^(-0.01507*t)\n\nNow you want to know how much is left after t=184:\n\nA(184)=8*e^(-0.01507*184)\n\nA(184)=0.5 [grams]\n\nViola....
778184
PV=nRT\n\nn is doubled.\nT, V same.\nR is constant\n\n=>P has to be doubled to validate the eqn.
511329
Start by multiplying both sides by D this gives you\n(D*P) = (2*s*t)\n\nthe divide both sides by 2t which gives you\n(D*P)/(2*t) = s\n\nor\n\ns = (D*P)/(2*t)
912617
P=P(0)*exp(-1.1e-4*H)\n\nP(0) is the pressure at sea level. Note it's temperature dependent and it assumes a stable Boltzmann type distribution.
625531
The formula is\nA = Pe^(rt), where A is the amount returned, P is the original amount, r is the interest rate, and t is the number of years.\nP = 2000\nr = 2% = .02\nA = 2000e^(.02t)\n"How long will it take?" means you are looking for t.\nIn order to find t, you will need to know A.
266644
I've just answered this in your previous question: "Statistics: Variability & SRS Samples of Population?"\n \nIf you look there, you'll find the formula for the s.d. of p_hat. So that increasind n by multiplying it by "m", reduces (devides) sd (or se) by sqrt(m).\n \nFor example, lets say the sd is 0.2 and the sample size is 100. If you increase the sample to 400 (multiply it by 4), the sd is devided by sqrt(4) = 2, i.e., sd = 0.1.
15480
Let me give you the most exact figure and accurate figure of world population at the time I'm answering the question (As on 26th December, 2006 at 6:00 GMT):\n\n6579803290\n6579803293 (1 Second Later\n6579803296 (2 Second Later) \n6579803299 (3 Second Later)\n6579803302 (4 Second Later)\n...\nand counting\nYou can see the time answered the question\n\nThe following clock shows the accurate population at current time : http://www.ibiblio.org/lunarbin/worldpop
32878
200\n\n(1,000,000/5,000=200)
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time to the minus one.\n\nDo a dimensional analaysis on a rate equation to confirm this.
279724
come on. all these questions are almost EXACTLY the same. \np1v1/t1 = p2v2/t2 = nr = constant.\n(0.998atm)(2.1L) / (36+273K) = (0.9atm)v2 / (28+273K)\nand then solve for v2. and u get\nv2 = 2.268L\n\n3. p1v1 = p2v2 = nr/t = constant\n(1.08atm)(14.57mL) = (1.43atm)v2\nv2 = 11.0039mL.
365048
1. Sorry, but I can't draw a graph on Yahoo! Answers.\n\n2. t = time, 600 (subtract 200 from 800), (subtract 2.9 from 5.1), -200 (subract 800 from 200).\n\n3. # of foxes = t (number of foxes added each year)\n\n4. Divide 5.1/2.9, then multiply the total number of foxes added per year and multiply it by 7, 0, 9.5, 2, or whatever number to get the fox population for that year.\n\nGood luck! :)
208940
theres some  decay equation, i think. \nyou know that   the element   will have half the original after 40 days. so its like F = I (1/2)^d/40    i think. so d is the number of dayz and I is the initial.  so if d=40 dayz  youll have half  the original. \nanyways i might be wrong so sorry if i am but if im right, then u would put 100 for I (100%) and then put d=48 and find F.\nF = 100 (1/2)^(48/40)\nF = 100 (1/2)^1.2\nF = 43.52........   % of the original left after 48 days
1132233
this question will require calcs using logarithm.\nhence will take time
840090
P1V1 = nRT = P2V2\nP1/P2 = V2/V1