Exponential distribution?
924716
has it not occurred to you that pretty much anyone who posts on here is trying to get help with their homework? i don't see a problem with helping, even the solution as long as it can be useful in understanding problems
144487
the answer is exponents.
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.
264938
The exponential equation a^x = b is equivalent to the log equation log_a(b) = x (where log_a is the log base a). So if you've got variables in the exponent the general procedure is to put the equation in the form a^x = b (ie isolate the exponential) and convert it into log form.\n\nYou can combine the exponentials in the first term into a single exponential 2^(3x). So you've got\n2^(3x) + 3 = 39\n2^(3x) = 36\n\na^x=b ---> log_a(b) = x\nSo 2^(3x) = 36 is equivalent to:\nlog_2(36) = 3x\n(1/3)log_2(36) = x\n\nYou can use the change of base formula to write the log base 2 in terms of logs base 10 (which is what the log button on your calculator will compute):\nlog_2(36) = log_10(36)/log_10(2).
770136
P = Ae^rt\nP = 100e^(.2*52)\nP = 100 * 32824.21\nP = 3282421
925143
This is an exponential increase function so it has the form\n\nN(t) = N0 e^(- λ * t)\n\nwhere N is the number of infected, N0 is the initial number of people infected, λ is the decay rate and t is the amount of time.\n\nIt is given that N(0) = 1 person and N(3) = 25 people\n\nUsing these values we can solve this for λ. These two equations are\n\n1 person = (1 person) e^(- λ * 0)   [a]\n\nand\n\n25 people= (1 person) e^(- λ * 3)  [b]\n\nSolving [a] will not let us determine λ so we must use [b] to find it. Taking the natural log of both sides gives us\n\nln[(25 people)/(1 person)] = ln(e^(- 3 days * λ))\nln25 = -3 days * λ\n(ln25)/-3 days = λ\nλ = -1.07/day\n\nSince we used days to determine the 'decay' rate we must now convert one week to days. So one week is 7 days. Now to find the number of infected we just plug in the values into the formula to find N(7)\n\nN(7) = (1 person) e^[-(-1.07/day) * (7 days)]\n\nnotice here that the units in the exponential cancel out, as they must, so that out units for out answer is people, as we would expect.\n\nEvaluating this expression gives \n\nN(7) = (1 person) e^(7.51)\n= 1826.21 people\n\nSince there are only 500 people everyone is infected. Because you referenced carbon decay that is why I used an exponential function. I'm not sure if this is correct, but it's what I came up with. This seems to be a particularly nasty virus.
272822
One cool way that I have seen it derived is based on Jayne's Formalism.\n\nAssume we sample process, and measure the mean and standard devation of the samples. There are many possible distributions that will produce those exact mean and standard deviation. \n\nJayne's Formalism chooses the distribution which maximizes the entropy of the distribuiton. This is viewed mathematically as the distribution that makes the minimum number of assumptions regarding the distribution.\n\nI always thought that was kinda cool.
428474
m(e) = (2/3) log(e / 10^4.4) energy release = 2 * 10^17?\n\nWhat is the question?  Do you want to know what value of e makes (2/3) log(e / 10^4.4) = 2 * 10^17 ?\n\nThat would mean\nlog(e / 10^4.4) = (3/2) * 2 * 10^17 = 3 * 10^17\ne / 10^4.4 = exp(3 * 10^17)\ne = 10^4.4 exp(3 * 10^17)\n\nwhere exp(x) is just e^x for e = 2.718281828459... the base of natural logarithms, but I write exp(x) because you are already using "e" for something else.
648708
I know this! I did a problem like this a while ago! It's an exponential equation. It can be used to predict exponential increases or decreases, like bacteria or double/triple/quadrupled information.
152635
You program the data in to a calculator (like a TI-83), graph, and model the data points using an exponential profile.  It will then give you the form of the equation that best fits your data points in the form of a*e^(b*x), which you can then graph over your data points.  In your case, "b" should be negative to indicate decay.
773699
You have said that they aren't independent; however, you have no mention of how they depend on each other. Therefore there is no way to calculate the answer.\nIf they *were* independent, then E[X*Y] = E[X]*E[Y]. Each one is an exponential distribution with lambda = 1, so the mean is just 1 (could also integrate e^(-x) from 0 to infinity, you also get 1.)
979515
the formula for sd.dev of binomial distribution is "\nsqrt(n*p*q).\ncan't understand the table so use the fomula urself.hope this helps
458039
In the experiment of testing for the number of errors in a chapter...\n\nLet the event an error appearing in a chapter be a success..\n\nLet "x" represent the number of success.\nThe number of errors in a chapter can be either 0,1,2,...\n==> "x" can carry the values 0,1,2,3,...\n==> "x" is a discrete random variable with range = {0,1,2,..} [Countably infinite]\n\nThe probability distribution of "x" is a poisson distribution. \n[Since there are a large number of trials all of which are identical and independent with only two possible outcomes in each trial, either a success (there being an error) or a failure (there not being an error), with the probability of success in all the trials remaining the same all throught ]\n\nThe probability function of a poisson distribution\n==> P(X = x) = {e^(-m) * m^(x)}/x!\n \nParameter of the distribution\n==> Mean of the distribution\n==> Mean number of errors per chapter =0.8 \n==> m = 0.8\n\nTherefore P(X = x) = {e^(-0.8) * (0.8)^(x)}/x!\n\nProbability that there are less than 2 errors in a chapter \n==> P(x < 2) \n= P(x=0) + P(x=1)\n= {e^(-0.8) * (0.8)^(0)}/0! + {e^(-0.8) * (0.8)^(1)}/1!\n= {e^(-0.8) * 1)/1 + {e^(-0.8) * (0.8)}/1\n= e^(-0.8) + {e^(-0.8) * (0.8)}\n= e^(-0.8) [ 1 + 0.8]\n= e^(-0.8) [1.8]\n\n\nhttp://www.futureaccountant.com/probability/
729281
The  exponent becomes a coefficient\n-1+2x lne\nln e=1\n-1+2x
86293
I suppose you are asking about the exponential function e, which is approximately equal to 2.71828. Its the base number for natural logarithms.\n\nIt is computed by using binomial theorem. The formula used to find e is an infinite series but easy to find by substituting the value of x with different number. If you substitute x with "1" then you get 2.71828. \nI found the below equation in math forum website.\ne^x = 1 + x + x^2/2! + x^3/3! + ... to infinity
1135408
OH NOES THIS IS LIKE A MATH PROBLEM!!!! i can't handle any more variables :( sry.\n\n\nhahahaha.
979650
For any random variable X, if a is a positive number, it is clear that the probability that X will be larger than a goes to 0 as a grows without limit. This is because a probability distribution has integral 1, and therefore the area under the probability density curve (continuous distribution) to the right of a, or the sum of the probabilities for values of X larger than a (discrete distribution) have to go to 0 as a goes to infinity.\nHope this helps -