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V = (g Dh)05 = (g y)05 Q = (g y)05 (b y) y = Q / b g052/3 = 137 m 339 upstream downstream Example #15 (Cont) Specific Energy 340 A 137 m depth in the upstream channel (Section 1, blue) corresponds to specific energy E = 155 m, compared with E = 5 m in the control section The upstream E2 must exceed 5 m, which requires anP(a ≤ X ≤ b) = ∫f(x) dx =Area under f(X) from a to b b a That is, the probability of an interval is the same as the area cut off by that interval under the curve for the probability densities, when the random variable is continuous and the total area is equal to 100 C For a continuous variable, the cumulative distribution function isThen f g(b) = f(g(b)) = f(a) = b, ie f g = idB 119 Show that for an injective function f A !

E) 102 p 6 19;Subset If every element of a set A is also an element of a set B,thenwesaythatA is a subset of B and we write A B Note If we write A ⇢ B,thenthismeansthatA is a proper subset of B,withoutthepossibility of equality Therefore, for any set A, A is NOT a proper subset of itselfStructural Design II My = the maximum moment that brings the beam to the point of yielding For plastic analysis, the bending stress everywhere in the section is Fy , the plastic moment is a F Z A M F p y ⎟ = y 2 Mp = plastic moment A = total crosssectional area a = distance between the resultant tension and compression forces on the crosssection a A

^ _ g b x ^ _ e d e Z k k Z;V 8 = 0 V = 800 kip 6 ft 6 x 10 ft M = {x2 300x 216} kip # ft ©M NA = 0;NCERT Solutions for Class 12 Maths Chapter 13 – Free PDF Download The NCERT Solutions for Class 12 Maths Chapter 13 Probability are provided here to help the students in their term – II studies Probability is the chance of occurrence of an event In this chapter, students learn about Probability in detail

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N R s q ` s s n q h ` r h ͂ O V Y ̓ X g c ƒ B Ƒ A 劽 } I C y ɂ p B X g c Ǝ l ̂ m 点 N ̉Ă̓ ` c Ƃ p A 肪 Ƃ ܂ B { ďI Ă ܂ BC) P(Y XjY 025) SOLUTION These problems are most easily solved by drawing pictures and computing the areas of the triangular regions that you see For part (a) the answer is 1 12(2)(3 4)(3 4) = 7 16 (we did this as an example in class) For part (b) the answer is 1 1 2 (1)(4 5) 1 2 (1)(3 4) = 9 40 For part (c), P(Y X;Y 025) = P(Y X) P(Y XAn example I obtain an F ratio of 396 with (2, 24) degrees of freedom I go along 2 columns and down 24 rows The critical value of F is 340 My obtained Fratio is larger than this, and so I conclude that my obtained Fratio is likely to occur by chance with a p

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(a) Find PX > Y and PX Y ≤ 1 (b) Find Pmin(X,Y ) ≥ 1 (c) Find Pmax(X,Y ) ≤ 1 Problem 443 Solution The joint PDF of X and Y is fX,Y (x,y) = ˆ 6e−(2x3y) x ≥ 0,y ≥ 0, 0 otherwise (1) (a) The probability that X ≥ Y is Y X X Y³ P X ≥ Y = Z∞ 0 Zx 0 6e−(2x3y) dydx (2) = Z∞ 0 2e−2x −e−3y y=x y=0 dx (3Problem Ten (1738) Let Aj = { 2, 1, 0, 1, , j} Find n a) ∪ Aj j=1 Each Aj is the set { j}, so every Aj fully contains the sets Aj1 Aj2 etc as subsets Therefore, the union of the sets A1 through An is exactly AnWe can take this one step further and say that, since n) x p y x y;

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7 Given that P(A) = 02, P(B) = 07, and P(AB) = 015, find P(A0 ∩B0) Solution By De Morgan's Law, P(A0 ∩ B0) = P((A ∪ B)0) = 1 − P(A ∪ B) = 1−P(A)−P(B)P(A∩B) Using the given values of P(A) and P(B) and P(A∩B) = P(AB)P(B) = 015 07 = 0105 (the multiplication formula), we get P(A0 ∩ B0) = 1−02−A B and the 40 T 7 Y I 4 W and a F 28 D L 3 M in a B A P is W 1000 W 2 F 2 F 13 S on the U S of A F S W and the 7 D 19 x 1080 P in F H D R 6 B in the H B 4 H of the A 10 T I H A Y 8 P in the S SB there is a left inverse g B !

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P F B P F B P F B P F A P F A P F A P A P B C C C C a) 9 5 P(A) = ≈ b) Para calcular P(A/F) se aplica el teorema de Bayes, con los sucesos A y B como sistema completo de sucesos 37 25 37 25 25 12 25 9 4 03 9 5 05 9 5 05 ( / ) ( ) ( / ) ( ) ( / ) ( ) ( / ) = = ≈ = = = P F A P A P F B P B P F A P A P A F 10 DadosP −1, then a= b=0 30 if a ib= x iy,wherei= p −1, then a= xand b= y 31 The roots of the quadratic equationax2bxc=0;a6= 0 are −b p b2 −4ac 2a The solution set of the equation is (−b p 2a;M 8(10 x) 40 = 0 c©F y = 0;

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P 6 f) 1 p x p y Answers a) pa b41;D) xy =5 y;R i R j P i P j Q i Q j S i S j x x x y y y x y x y x y x y • Resolve each force into rectangular components, then add the components in each direction x x x x x F R P Q S • The scalar components of the resultant vector are equal to the sum of the corresponding scalar components of the given forces y y y y y F R P Q S x y x y R R

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Page 2 c) d) wx x′y BD B′D′ A′B 312) Simplify the following Boolean functions in products of sums a) F(w, x, y, z) = ∑ (0, 2, 5, 6, 7, 8, 10)3For any two numbers aand bwith aNot Mutually Exclusive because P(F ∩ B) = 0077 ≠ 0 Independent because P(FB) = P(F) = 064 9 The following table shows the breakdown of sex and degree among a university's faculty What is the probability that a randomly selected professor is a Male and has a Doctorate P(M ∩ D) = 28 66 b Male or has a Doctorate P(M U

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6 ft 4 ft 300 2x V = 0 0 x 6 6 ft 6–18 Draw the shear and moment diagrams for the beam, and determine the shear and moment throughout the beamGeometrically the probability that X is between a and b, ie, P(a X b), is then represented by the area shown shaded, in Fig 22 The distribution function F(x) P(X x) is a monotonically increasing function which increases from 0 to 1 and is represented by a curve as in Fig 23 d dx3 x a f (u) du f (x) F(x) μ 0 x 1 x 2 1 x 2 1 x 2 dF(x) dxMinimum SOP and POS •f= xyz x`yz xy`z Is called sum of products The is sum operator which is an OR gate The product such as xy is an AND gate for the two inputs x and y

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Answers a) p 10 10;If not, find Cov (X, Y) The support of (X, Y) NOT independentis NOT a rectangle ⇒ X and Y are OR(a) (4 points) Describe the events B ∩F and (B ∪T)∩F in words B ∩F is the event that Bob is at the grill, and the grill is on fire (B ∪T)∩F is the even that Bob or Tina are at the grill, and the grill is not

The X Component Of Vector A Is 25 0 M And The Y Component Is 40 0 M A What Is The Magnitude Of A B What Is The Angle Between The

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 P C h a n e l B r a y s B ayou S i m s Bayo u G r e e n s B a y o u 8 22 81 4 26 23 18 36 40 29 45 Eastside Special Operations Polic eH adQu rt s William OP Hobby Airport Leija EastL Freeway B ro adw y Squ e Melcher Central Stanaker Mancuso V ParkA such that g f = idA Solution For each b 2 B such that b = f(a) for some a 2 A, we set g(b) = a This is wellde ned since for each b 2 B there is at most one such a Now pick some element 2 A and for each b 2

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F(x) = (1 P(y= 1jx) P(y= 0jx) 0 P(y= 0jx) > P(y= 1jx) 2 3 pts Suppose that the class y= 0 is extremely uncommon (ie, P(y= 0) is small) This means that the classi er f(x) = 1 for all xwill have good risk We may try to put the two classes on even footing by considering the risk R= P(f(x) = 1jy= 0) P(f(x) = 0jy= 1)X 0 10 30 40 y 7 18 32 48 85 Solution Here x 0 = 0, x 1 = 10, x 2 = , x 3 = 30, x 4 = 40, x 1 x 0 = 10 = x 2 x 1 = x 3 x 2 = x 4 x 3 The given data is equispaced As x= 5 lies between 0 and 10 and at the start of the table and data is equispaced, we−b− p 2a) where = discriminant = b2 −4ac 32 The roots are real and distinct if >0 33 The roots are real and coincident if = 0 34 The

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Ap,i~w f nm ~r 5 ?/at ,ta duf~ Attached are copies of the above documents which I offer as proof of delivery and evidence of my claim I submit the above claim and state that it is true under penalty of law Date Y/J/~ / ' zip 5763 Phone # dw 43 7 2 dm &U 3~7/@ Check box if email onlv service is acceptable Email address(a) P(Y = 3) = (i) 0 (ii) 025 (iii) 050 (iv) 075 (b) P(Y 3) = F(3) = R 3 2 x 6 dx= x 2 12 i x=3 x=2 = 3 12 2 12 = 5 12 requires (i) summation (ii) integration and is a value of a (i) probability density function (ii) cumulative distribution function which is a (i) stepwise (ii) smooth increasing function (c) E(Y) = (i) P yf(y) (ii) R yf(y)dyAnswer A Explanation a=5,b=3 , as there are only two format specifiers for printing

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C) 2 3x2 x;The cumulative distribution function (CDF) of random variable X is defined as F X ( x) = P ( X ≤ x), for all x ∈ R Note that the subscript X indicates that this is the CDF of the random variable X Also, note that the CDF is defined for all x ∈ R Let us look at an example ExampleB) 4 p 23;

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R2 Statistic (1) R2 is a measure of how well the fit function follows the trend in the data 0 ≤ R2 ≤ 1 Define yˆ is the value of the fit function at the known data points For a line fit yˆ i = c1x i c2 y¯ is the average of the y values y¯ = 1 m X y i Then R2 = X (ˆy i − y¯) 2 X (yi − y¯) 2 =1− r 2 P 2 (yi − y¯)2 When R2 ≈ 1 the fit function follows the trendThe following properties hold for all events A, B • P(∅) = 0 • 0 ≤ P(A) ≤ 1 • Complement P(A) = 1−P(A) • Probability of a union P(A∪B) = P(A)P(B)− P(A∩ B) For three events A, B, C P(A∪B∪C) = P(A)P(B)P−P(A∩B)−P(A∩C)−P(B∩C)P(A∩B∩C) If nd B are mutually exclusive, then P(A∪B) = P(A)P(B)B) p 15 6;

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B a p 1 (f0(t))2 dt Therefore, Z b a f(t)dt= Z b a p 1 (f0(t))2 dt for all nonnegative aand b In particular, we can write Z x 0 f(t)dt= Z x 0 p 1 (f0(t))2 dt for all nonnegative x Both sides of the above equation de ne a function of x, and since they areM 216 2xa x 2 b 300x = 0 V = {300 2x} kip c©F y = 0;Disjoint P(A and B) = 0 If two events are mutually exclusive, then the probability of either occurring is the sum of the probabilities of each occurring Specific Addition Rule Only valid when the events are mutually exclusive P(A or B) = P(A) P(B) Example 1 Given P(A) = 0, P(B) = 070, A and B are disjoint

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