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Dy / dx P (x) y = Q (x) where P (x) and Q (x) are functions of x If we multiply all terms in the differential equation given above by an unknown function u (x), the equation becomes u (x) dy / dx u (x) P (x) y = u (x) Q (x)Observe that according to Euler's totient formula, let a be any integer coprime to p, then (a, p) = 1 Then a^ (phi (p)) â ¡ 1 (mod p) Now since p is prime, all a in Z/PZ is coprime to p and phi (p) = (p1) Then (xy)^p = (xy) (xy)^ (p1) (mod p) â ¡ (xy) (1) (mod p)It seems that there's tendency to agree upon p (x)⇒∀xp (x) is the same as ∀x (p (x)⇒∀yp (y)), whereas ∀x (p (x)⇒∀yp (y)) is read as if p (x) is true for some x, then it is true for all x However I don't understand where's the quantifier SOME came from, since there no quantifier '∃' in '∀x (p (x)⇒∀yp (y))'

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P(x y) calculator-Sometimes it really is, but in general it is not Especially, Z is distributed uniformly on (1,1) and independent of the ratio Y/X, thus, P ( Z ≤ 05 Y/X) = 075 On the other hand, the inequality z ≤ 05 holds on an arc of the circle x 2 y 2 z 2 = 1, y = cx (for any given c) The length of the arc is 2/3 of the length of the circle0 < x < 1;

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P(X= 129) = P(X= 129 and Y = 15) P(X= 129 and Y = 16) = 012 008 = 0 What is the probability distribution of X?Given random variables X, Y, {\displaystyle X,Y,\ldots }, that are defined on a probability space, the joint probability distribution for X, Y, {\displaystyle X,Y,\ldots } is a probability distribution that gives the probability that each of X, Y, {\displaystyle X,Y,\ldots } falls in any particular range or discrete set of values specified for that variable In the case of only two random variables, this is called a bivariate distribution, but the concept generalizes to anyFZ(z) = P(Z ≤ z) = P(X/Y ≤ z) = ˆ 0 if z < 0 P(Y ≥ (1/z)X) if z > 0, where we have used the fact that X and Y are both nonnegative (with probability 1), so multiplying both sides of the inequality by Y does not flip the inequality;

Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history, geography, engineering, mathematics, linguistics, sports, finance, musicP= (X=1/32) because HHHHH is the only answer for 5 heads in a coin toss that occurs five times In this situation, Master Salman is doing a coin toss only three times So there is no probability distribution for 5 heads because that is impossible Thank you!P (X \cap Y) = P (XY) \times P (Y) P (X ∩ Y) = P (X ∣Y) ×P (Y)  The probability that A and B occurs is the probability of X occurring, given that Y occurs multiplied by the probability that Y

P Var(X) is called the standard deviation of X For any rv X and any number a E(aX) = aE(X), and Var(aX) = a2Var(X) (3) For any two rvs X and Y E(X Y) = E(X)E(Y) (4) If X and Y are independent, then Var(X Y) = Var(X)Var(Y) (5) The above properties generalize in the obvious fashion to to any finite number of rvs In general (independent or not)Where y i is the vapor mole fraction and y 1 y 2 = 1, x i is the liquid mole fraction and x 1 x 2 = 1 , and P is the total pressure (bar) The bubblepoint pressure is calculated using ΣK i x i = 1Here you graph parabolas using the xintercepts This is one of my favorite, and easier methods, to follow The homework is Pages , #'s 326, and "Qui

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0 < y < x 0;P dilip_k Nov 9, 17 Given that the equation of the chord \displaystyle\ \text{ }\ {\left({x}{y}={2}\ldots{\left{1}\right}\right)} the equation of the circleTrue, then automatically the statement ∀x ∈ A,∃y ∈ B,P(x,y) must be true (but in general it doesn't go the other way) Aside Occasionally, you will see a nested quantifier at the end of a statement, in which case it is implied that the quantifier is the last in terms of order For example, here is the definition of bounded

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X^3 x^2 y x y^2 y^3 Extended Keyboard;What is \(P(X = Y)\), the probability that one samples the same number of red and white balls?If the joint probability density function of random variable X and Y is , (,) , the marginal probability density function of X and Y are = ∫, (,) ,

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1 Sum of Independent Binomial RVs • Let X and Y be independent random variables X ~ Bin(n 1, p) and Y ~ Bin(n 2, p) X Y ~ Bin(n 1 n 2, p) • Intuition X has n 1 trials and Y has n 2 trials o Each trial has same "success" probability p Define Z to be n 1 n 2 trials, each with success prob p Z ~ Bin(n 1 n 2, p), and also Z = X Y0 < y < x o (Figure 2) Figure 2 f(x;y)jy < x;0 < x < 1;0 < y < 1g Therefore P (X > Y) = Z 1 0 ˆZ x 0 f(x;y)dy ˙ dx = Z 1 0 ˆZ x 0 6x2ydy ˙ dx = Z 1 0 3x4dx = 3 5 Example 2 Consider random variables X,Y with pdf f(x,y) such that f(x;y) = 8 < 8xy;Let™s calculate P (X > Y) For –xed x o;

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E(X) = (P y E(XY = y)P(Y = y) if Y is discrete, R∞ −∞ E(XY = y)fY (y)dy if Y is continuous You should note that this applies to the probability of an event (which is nothing other than expectation of its indicator) as well — if you know P(AY = y) = E(IAY = y) then you may compute P(A) = EIA by the Bayes formula above Example 111• Probabilities Probabilities involving X and Y (eg, P(X Y = 3) or P(X ≥ Y) can be computed by adding up the corresponding entries in the distribution matrix More formally, for any set R of points in the xyplane, P((X,Y) ∈ R)) = P (x,y)∈R f(x,y) • Expectation of a function ofP X and Y (eg, u(x,y) = xy) E(u(X,Y)) = x,y u(x,y)f(x,y)P dilip_k Nov 9, 17 Given that the equation of the chord \displaystyle\ \text{ }\ {\left({x}{y}={2}\ldots{\left{1}\right}\right)} the equation of the circle

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There are 421 words containing P, X and Y ANAPHYLAXES ANAPHYLAXIES ANAPHYLAXIS XYLOPYROGRAPHY XYLOTYPOGRAPHIC XYLOTYPOGRAPHY Every word on this site can be played in scrabbleNote, however, that when we divide both sides by z, to obtain Y ≥ (1/z)X, we were making the assumptionP (X \cap Y) = P (XY) \times P (Y) P (X ∩ Y) = P (X ∣Y) ×P (Y)  The probability that A and B occurs is the probability of X occurring, given that Y occurs multiplied by the probability that Y

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Where y i is the vapor mole fraction and y 1 y 2 = 1, x i is the liquid mole fraction and x 1 x 2 = 1 , and P is the total pressure (bar) The bubblepoint pressure is calculated using ΣK i x i = 11 Let u(x, y) = x (3/4) * y (1/4), p x = 5, p y = 3, and I = 60 Now assume that p' x = 9 Separate the income effect and substitution effect using the Slutsky method Show your work on the graph too Compare the value of the point that separates these two effects you found here with the one we found for the Hicksian method in the class 2If the joint probability density function of random variable X and Y is , (,) , the marginal probability density function of X and Y are = ∫, (,) ,

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PX(x)=P(X=x)forallx Theprobabilitydistributionforadiscreterandomvariableassignsnonzero probabilities toonly a countable number ofdistinct x values Any value x not explicitly assigned a positive probability is understood tobe such that P(X=x) = 0 The function pX(x)= P(X=x) for each x within the range of X is called the probability distribution of XPX(x)=P(X=x)forallx Theprobabilitydistributionforadiscreterandomvariableassignsnonzero probabilities toonly a countable number ofdistinct x values Any value x not explicitly assigned a positive probability is understood tobe such that P(X=x) = 0 The function pX(x)= P(X=x) for each x within the range of X is called the probability distribution of XF(x) = P(X ≤ x) Continuous distribution The cumulative distribution function F(x) is calculated by integration of the probability density function f(u) of continuous random variable X Discrete distribution The cumulative distribution function F(x) is calculated by summation of the probability mass function P(u) of discrete random variable X

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P(X < 1) = P(X = 0) P(X = 1) = 025 050 = 075 Like a probability distribution, a cumulative probability distribution can be represented by a table or an equation In the table below, the cumulative probability refers to the probability than the random variable X is less than or equal to xWhere y i is the vapor mole fraction and y 1 y 2 = 1, x i is the liquid mole fraction and x 1 x 2 = 1 , and P is the total pressure (bar) The bubblepoint pressure is calculated using ΣK i x i = 1The joint probability mass function of two discrete random variables X and Y is defined as P X Y (x, y) = P (X = x, Y = y) Note that as usual, the comma means "and," so we can write P X Y (x, y) = P (X = x, Y = y) = P ((X = x) and (Y = y))

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Let P(x,y) be a point on the graph of y=2x5 Express (and simplify) the distance from the origin 0(0,0) to P as a function of x Get more help from Chegg Solve it with our precalculus problem solver and calculatorBy the table, one sees that this is possible only when \(X = 1, Y = 1\) or \(X = 2, Y = 2\) So the probability \ P(X = Y) = f(1, 1) f(2, 2) = \frac{12}{252} \frac{54}{252} = \frac{66}{252}Let the points be P (x,y),A(1,4),B(−1,2)Point P is equidistant from A&BNow, using the distance formulax1 = x, y1 = yx2 = 1, y2 = 4P A= (x2 −x1 )2 (y2 −y1 )2 = (1−x)2 (4−y)2 Similarly for P Bx1 = x, y1 = yx2 = −1, y2 = 2P A= (x2 −x1 )2 (y2 −y1 )2 = (−1−x)2 (2−y)2 we know thatP A= P B(1−x)2 (4−y)2 = (−1−x)2 (2−y)2 Squaring both sides⇒ (1−x)2 (4−y)2 =(−1−x)2 (2−y)2⇒ 1x2 −2x16y2 −8y = 1x2 2x4y2 −4y⇒ −2x−2x−8y4y16−4

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The joint probability mass function of (X;Y) is (12) p(xi;yj) = P(X = xi;Y = yj) Example 1 A fair coin is tossed three times independently let X denote the number of heads on the flrst toss and Y denote the total number of heads Find the joint probability mass function of X and Y 2According to p(x,y) is given by I(X;Y) = X x,y p(x,y)log p(x,y) p(x)p(y) (24) = H(X)−H(XY) = H(Y)−H(YX) = H(X)H(Y)−H(X,Y) (25) 4True, then automatically the statement ∀x ∈ A,∃y ∈ B,P(x,y) must be true (but in general it doesn't go the other way) Aside Occasionally, you will see a nested quantifier at the end of a statement, in which case it is implied that the quantifier is the last in terms of order For example, here is the definition of bounded

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P(X ≥x) = e−λ(xy) e−λx = e−λy = P(X ≥y), which proves the memoryless property The memoryless property also has a geometric interpretation Consider the probability density function of an exponential(λ) random variable from some value x to infinity The area under the probabilitydensityfunctionfromx toinfinitydoesnotequal1;ratheritequalse−λx Theconditional probability density function given that the random variable X is greater than or equal to x is found by rescaling fP (X) = P (Y) or P (X n Y) = 0 That is, the above is true if and only if X and Y are equally likely, or if X and Y are mutually exclusive Oh, and since we were dividing by P (X) and P (Y), bothExpected Value and Standard Dev Expected Value of a random variable is the mean of its probability distribution If P(X=x1)=p1, P(X=x2)=p2, n P(X=xn)=pn E(X) = x1*p1 x2*p2 xn*pn

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P(X ≥x) = e−λ(xy) e−λx = e−λy = P(X ≥y), which proves the memoryless property The memoryless property also has a geometric interpretation Consider the probability density function of an exponential(λ) random variable from some value x to infinity The area under the probabilitydensityfunctionfromx toinfinitydoesnotequal1;ratheritequalse−λx Theconditional probability density function given that the random variable X is greater than or equal to x is found by rescaling fThe joint probability mass function of two discrete random variables X and Y is defined as P X Y ( x, y) = P ( X = x, Y = y) Note that as usual, the comma means "and," so we can write P X Y ( x, y) = P ( X = x, Y = y) = P ( ( X = x) and ( Y = y)) We can define the joint range for X and Y as R X Y = { ( x, y) P X Y ( x, y) > 0 }X y f x y 1 3 P X x Y y f x y For any region A in the xy plane P X Y A A f x y from MATH MTH02 at Institute of Technical and Education Research

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E(X) = (P y E(XY = y)P(Y = y) if Y is discrete, R∞ −∞ E(XY = y)fY (y)dy if Y is continuous You should note that this applies to the probability of an event (which is nothing other than expectation of its indicator) as well — if you know P(AY = y) = E(IAY = y) then you may compute P(A) = EIA by the Bayes formula above Example 111Online math solver with free step by step solutions to algebra, calculus, and other math problems Get help on the web or with our math appOnline math solver with free step by step solutions to algebra, calculus, and other math problems Get help on the web or with our math app

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F(x) = P(X ≤ x) Continuous distribution The cumulative distribution function F(x) is calculated by integration of the probability density function f(u) of continuous random variable X Discrete distribution The cumulative distribution function F(x) is calculated by summation of the probability mass function P(u) of discrete random variable XObtain the value P (X − Y > z) we must integrate the joint PDF f X,Y (x, y) over the shaded area in the above figures, which correspond to z ≥ 0 (left side) and z < 0 (right side) For the case z < 0, we can use a similar calculation, but we can also argue using symmetry" ∀ y, ∀ x, P (x, y) " means given any object y, paired with any object x, the statement P is true about them Recall that if A and B are statements, the meaning of

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PX ≤ Y First, let's consider the denominator PX ≤ Y = X z≥1 PX = z ∩z ≤ Y = X z PX = zPz ≤ Y = X z (1−p)z−1p(1−q)z−1 = X z (1−p)(1−q)z−1p = p X z (1−p−q pq)z−1 = p pq −pq The last step above is again by the identity in Eqn 1 Now we can compute the whole equation EXX ≤ Y = pq −pq p X x xPX = xPx ≤ Y = pq −pq p X x

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