How did they compute the value of e?
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
863606
E indicates that the precedding number is multiplied by 10 and then the result to the next number in power. \n\nYour example: \n6E15 can be also expressed as\n6X10^15\nor 60,000,000,000,000,000\n\nyou already know 6X10 is 60, so you take that amount and add the next number in zeros to make the expresion in standard form.\n\nSo: 6E15 is the scientific form of 60,000,000,000,000,000
808176
Correct answer should be E
889920
Yes, if it's converted to energy.  E=mc^2
524536
A = Pe^(rt)\nYou are given everything you need.  All you have to do is evaluate it on the calculator.\nA = 20000e^(.08*3)\nA = 20000e^.24\nA = 20000(1.27124915032)\nA = 25424.9830064\nSince we are talking about money, you would round it to the nearest cent.\nA = 25424.98
83972
R*exp(i*theta) = R [ cos(theta) +i*sin(theta) ]\nthis is the polar representation (raduis R ,angle theta) transformed to the cartisienne form (X=R cos(theta), Y=R sin(theta))  where i is imaginary to matk that y is perpendic to x\n\nR=1 theta=pi/2\nexp(i*pi/2) = cos(pi/2) + i sin(pi/2)\nexp(i*pi/2) = 0 +i *1 =i\n\nso i=exp(i*pi/2)=e^(i*pi/2)    where e=2.7818281828\ni^i=e^(i*pi/2)^i\ni^i=e^(i^2*pi/2)\ni^i=exp(-pi/2)\n=0.207879576
775240
Just apply the formulas you gave and the definitions.\nNote that the mean of the measurements (usually mu) is L here.\n\nThe expected value of S(1) is, expanding and using the independence of L(1) and L(2), \n\nES(1) = E[0.5(L(1)+ L(2))]^2\n= (1/4){E[L(1)^2] + 2E[L(1)L(2)] + E[L(2)^2]}\n= (1/4){...............+ 2E[L(1)]E[L(2)] +  ........}\n= (1/4)[(sigma^2+L^2) + 2L^2 + (sigma^2+L^2)]\n= L^2 + sigma^2/2 = A + (1/2)sigma^2\n\nso S(1) is biased upwards (as an estimator of A).\n\nSimilarly \n\nES(2) = E[0.5(L^2(1) + L^2(2))] = (1/2){E[L(1)^2] + E[L(2)^2]}\n= (1/2)[(sigma^2+L^2) + (sigma^2+L^2)] = L^2 + sigma^2\n= A + sigma^2\n\nwhich is more biased tahn S(1).\nHow about S(3) = L(1)L(2)? Using the independence of L(1) and L(2), \n\nE[L(1)L(2)] = E[L(1)] E[L(2)] = L^2\n\nso S(3) is unbiased.
934968
Okay, assuming you mean:\n∫e^(2x)/√(1-e^(2x)) dx\nLet u=1-e^(2x), du=-2e^(2x) dx, dx = -1/2 e^(-2x)\n-1/2 ∫1/√u du\n-1/2 ∫u^(-1/2) du\n-u^(1/2) + C\n-√(1-e^(2x)) +C
550920
5^x=.125\nTaking log (base e) on both sides\n\n-> xlog5=log(0.125)\n-> x= log(0.125)/log5 = -1.291099302
408697
For experimental data it is 100 * (average of the set - measured member of the set) / (average of the set).\n\nIf you are measuring something, how have you established an "Expected Value"? Why are you conducting an experiment/measurement if you already know what to expect?
955889
Alright.\n\nE(S1) = E{(L1/2 + L2/2)^2}\n= Var(L1/2 + L2/2) + E^2(L1/2+L2/2) (Uses the fact that Var(X) = E(X^2) - E^2(X))\n= 0.25(sigma^2) + 0.25(sigma^2) + (L/2 + L/2)^2 (Since L1 and L2 are independent)\n= 0.5(sigma^2) + L^2.\n\nSo the bias of S1 is 0.5(sigma^2)\n\nE(S2) = E{0.5(L1^2 + L2^2)}\n= 0.5E(L1^2) + 0.5E(L2^2)\n= 0.5(Var(L1) + E^2(L1)) + 0.5(Var(L2) + E^2(L2))\n= 0.5(sigma^2) + 0.5L^2 + 0.5(sigma^2) + 0.5L^2\n= sigma^2 + L^2\n\nSo the bias of S2 is sigma^2.\n\nSo that means that S1 has the lower bias.\n\nPerhaps S3 = L1L2 is unbiased?\n\nE(L1L2) = E(L1)E(L2) (Since both are independent)\n= L*L\n= L^2\n\nBias is zero.  There you go!
864021
[a b]\n[c d]\n\n1. E=[2 -2]\n ____[-1 2]= 2*2-(-1)(-2)=4-2=2\nE^-1=1/2 [2 2]  = [1   1]\n________[1 2] = [1/2 1]   \nYou multiply 1 divided by ad-cb and switch a and d and make b and c negative.\n\n2.  E= [1 -1]= 1+1=2\n______[1 1]\nE^-1=1/2 [1 1]   =   [1/2 1/2]\n________[-1 1] =  [-1/2 1/2]\n\nHope this helped!
29143
It is called a caret.  You can use it to indicate an exponent or a superscript if the software you are using does not have the ability to make superscripts. \nEx: E=mc^2 (Einstein's famous equation relating mass and energy)
755844
1 correct \n\n2 ∫x/e^x dx \n= ∫x*e(-x) dx\n= -xe^(-x) - ∫-e^(-x). 1.dx\n= -xe^(-x) - e^(-x) + c\n\nSo∫[x = 1 to ∞] x/e^x dx = lim t →∞ (-te^(-t) - e^(-t)) - (-1/e - 1/e)\n= 2/e (NB lim t →∞ (-te^(-t)) = o by d'Hopitals rule)\n\n3 correct\n\n4. Use (n² + 1)/[e^n * (n + 1)²] \n=  [(n² + 1)² - 2n]/[e^n * (n + 1)²]\n= 1/e^n - 2n/e^n(n + 1)² \nNow 1/e^n < 1/3^n and  2n/e^n(n + 1)² < 2n/3^n*n² \n= 2/n*3^n \nSo both parts converge so the difference converges\n\nNOTE: It is many, many years since I have done these BUT I am confident that I am correct ... Maybe you can tell me if that is so!
313019
This is a simple application of Coulomb's Law.\nThe Electric Field is simply given by the charge on the nucleus (26e) divided by 4*Pi*EpsilonNought*distance^2\nWhere e is 1.6*10^-19 Coulombs
739921
Not too hard.\n1. plug in what you know\n\nP/2 = Pe^(-0.0027x)\n\n2. cancel the Ps.\n1/2 = e^(-0.0027x)\n\n3. use some tricks with the ln and e to get x by itself\nln(1/2) = -0.0027x\n\n4. solve for x\nln(1/2) / (-0.0027) = x\n\n5. calculator, most definitely\nx = 256.72\n\n\nOn step 3, try to remember this:\na = e^b, therefore ln(a) = b. That's usually the make-or-break with exponential decays.
967786
I'm not going to tell you but look use Saha's equation to find out the energy in ev of each level. Then you can work out the energy value of the jump. It's just a simple bit of maths. But you are wrong so have a think