how to calculate an expectation of a variable?
612882
You first need the probablity function of the basic variable... Lets call it P(x) where x is the variable... Then to get the epectation take the integral of x*P(x). The limits of integration should be the min and max values that x can be.
781591
P(wrong size) = 0.2\nP(defective) = 0.35\nP(not defective and right size) = 0.60\n\nThus P(defective or wrong size) = 1 - P(not defective and right size) (These two events are opposites.  Look at the Venn diagram for help with this, it's DeMorgan's law if you're curious)\n= 1 - 0.60 = 0.40\n\nP(defective or wrong size) = P(defective) + P(wrong size) - P(defective and wrong size)\n\nSo 0.40 = 0.2 + 0.35 - P(defective and wrong size)\n0.40 = 0.55 - P(defective and wrong size)\nP(defective and wrong size) = 0.55 - 0.40 = 0.15
1116823
Isolate your variable by dividing by -3 to get -5 for y.\n\nsubtract 8 and divide by -1 to get -4 for r.
950045
Correct formula should be as follows:\n\nE = 1/2*I*w^2
345069
This can be shown for X as either continuous or discrete.  Let mu = E(X).\n\nVar(Y) = E(Y^2) - E(Y)^2\n= E[(X-mu)^2] - E(X-mu)^2\n= E(X^2 - 2Xmu + mu^2) - (E(X)-mu)^2\n= E(X^2) - 2muE(X) + mu^2 - (mu-mu)^2\n= E(X^2) - 2mu^2 + mu^2 - 0\n= E(X^2) - mu^2\n= E(X^2) - E(X)^2\n= Var(X)\n\nIf you need to be more specific than this, the replace E(X) = int(-inf to inf) xf(x)dx and E(X^2) = int(-inf to inf) x^2f(x)dx.  Everything should come out the same way.\n\nRemember that using substitution in the integral, let x = y + mu, the dx = dy, and the limits of integration would remain -inf to inf.
323008
By calculationg it's mean. Add all the values of observations and then divide by the number of observations.\nThat's all!
50664
ln(e) = 1 is a basic rule\nwhen you have ln(x^y)\nyou can rewrite it as yln(x)\nso for that specific problem bring the 4 to the front to make 2*4*ln(e)\nthen ln(e)=1\nso 2*4*1 = 8
407618
k e =1/2  * m *v *v\n    =1/2 *3 * (3.5*10) * (3.5*10)\n    =0.18j (ans)
1166133
Multiply both sides by R+r\nI(R+r) = E\nDistribute the I\nIR + Rr = E\nSubtract IR form each side\nRr = E - IR\nDivide both sides by R\nr = (E-IR)/R
758848
let y = x^x\nw = ln y = xlnx\nthen y = e^w\ndy/dx = e^w * dw/dx\ndw/dx = x(1/x) + lnx = 1+lnx\ndy/dx = (1+lnx)*x^x
1181156
It looks like that first 1 should be an i. In this case,\ni^z=exp(i*pi/2*z),\nwhere I am using the principal branch of the logarithm to define the exponential.\nThen the derivative will be \n(i*pi/2)*exp(i*pi/2*z)=\npi/2*i^(z+1)\n\nEasy.
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.
158690
substitute x=e^y\nthen use a table of integration formulas?
353406
You will want to isolate the variable (many times it is 'x'). First you will add or subtract from each side. Then divide. For example:\n5x+4=19, subtract 4 from each side giving 5x=15. Then divide by 5 giving x=3. I hope this helps.
496124
Calculate the electricity.
913689
Add additional details to your question and I may be able to answer it.  I assume you are looking for the expected value, but some of you data is missing.\nE(X) = .7(175000) + .3(-amount of the loss)
764897
use derivative rule for exponential as\n\nd/dx(A exp(bx))=(A*b)*exp(bx)\n\nyou got hint. No more Spoon Feeding