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Monitor and range from % of HRmax and 4060% of HRmax during the high and lowintensity portions, respectively (Table 2) Intervals will consist of a high/lowintensity ratio of 11 (eg, 30 seconds high followed by 30 seconds low) ranging from 1560 seconds in durationU P v Ç E µ } } v î ì î í o } v Southwest Florida Chapter 125 8 8 640% Volusia/Flagler Chapter 90 4 4 444% West Coast Florida Chapter 417 21 21 504% Georgia State Council 1,376 68 68 494% Chattahoochee Chapter 56 6 6 1071%The mean µ and that attains its maximum value of √1 2πσ ' 0399 σ at x = µ as represented in Figure 11 for µ = 2 and σ 2= 15 The Gaussian pdf N(µ,σ2)is completely characterized by the two parameters µ and σ2, the first and second order moments, respectively, obtainable from the pdf as µ = EX = Z ∞ −∞ xf(x)dx, (12

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µ 40 an d σ 4Answer 416 µ 40 and σ 8Answer 432 µ 40 a nd σ 16Answer 464 µ 40 from EN 111 at Caldwell College^ } o o } Z } } u } ( Z P v o z } µ v P W } ( } v o v À Z v P î X K v Ç } µ µ } Z KW Z v } v Z µ } v Young Professionals enjoy a savings of more than 40% per year for the first five years after graduation Bu not just the savings that are amazing;2 Intuitively, if the evidence (data) supports H1, then the likelihood function fn(X1;¢¢¢;jµ1) should be large, therefore the likelihood ratio is small Thus, we reject the null hypothesis if the likelihood ratio is small, ie LR • k, where k is a constant such that P(LR • k) = fi under the null hypothesis (µ = µ0)To flnd what kind of test results from this criterion, we expand

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For the capacitors in parallel = 15 5 40 = 60 µ F eq C Hence 10 1 60 1 30 1 1 C eq = = C eq = 10 µ Chapter 6, Solution 19 We combine 10, , and 30 µ F capacitors in parallel to get 60 µ F The 60 µ capacitor in series with another 60 µ F capacitor gives 30 µ 30 50 = 80 µ F, 80 40 = 1 µ The circuit is reduced to 35 ₹ /crores total revenue 123 97 150 370 34 53 87 175 apr may jun q1 cy py total revenue ebitda pbt total revenue pat ihcl consolidatedP ÷m ñÍó ö ö ó ) û þ ö1Võ A ò)ò Rûà P ûÍ ó ñ ÿ bõhû ò ö ü ý~ m h ùö~ óAò¾ó ö õ Rø_ö~ ) v ÷ óAñ ý~ g V b÷ ñ ó ÷ û ò)ó bö~VûÕ ) ~ ¾ P ò ö~û þbòAó V A ÷

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2 The curve has a peak at x = µ 3 The curve is symmetric about the line x = µ 4 The curve always lies above the xaxis but approaches the xaxis as x extends indefinitely in either direction 5 The area under the curve is 1 6 The Standard Normal distribution has µ =0and = 1 We denote the standard normal random variable with Z = z#è Ê>K>3> >4>>k %40°6 "> 7g M ;µ £ å Ð4E m ± 0°6 "> 7g M '¨ G ô ¾X(x) and mean µ X gives a quantitative measure of how much spread or dispersion there is in the distribution of xvalues The variance is calculated as σ2 X = VX = Z ∞ −∞ (x−µ X)2 f X(x) dx = = = = The standard deviation (sd) of Xis σ X = p VX The coefficient of variation (cov) of Xis defined as the ratio of the standard

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The sample proportion ˆp is anX −µ σ = X − 63 8 ∼ N(0,1) (a) Using the table with cumulative probabilities for the N(0,1) we find that P({student obtains a I}) = P(X ≥ 70) = P Z ≥ 70− 63 8 = P(Z ≥ ) = 1−P(Z ≤ ) = 1−F() = 1−8106 = 1840 (b) We want to find P(X < 40) Using the table and the symmetry of the N(0,1) distribution (draw aA 9 È a 9 È a P$ È a P$ È 9 (© EÃ é ?

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Ie E(X) = µ As Hays notes, the idea of the expectation of a random variable began with probability theory in games of chance Gamblers wanted to know their expected longrun 40/36 1 5/36 9 4/36 36/36 4 16/36 10 3/36 30/36 9 27/36 11 2/36 22/36 16 32/36 12 1/36 12/36 25(µ∗I)(1) = µ(1)I(1) = 1·1 = 1 = e(1) Now suppose n > 1 Then by (d), µ∗I = Dµ, so (µ∗I)(n) = (Dµ)(n) = 0 = e(n) Therefore, the formula holds for all n The next result is very powerful, but the proof will look easy with all the machinery I've collected Theorem (Mo¨bius Inversion Formula) If f is an arithmetic function, thenD^W µ Z } Ì ^ P v µ D^W W v E u Gayle Lowery ^ P v Ç WD u v } ( , µ u v ^ À Z v P K } Exhibit A1 W ì í s v } E u W u W } v P ID# Katie Belange ( 0840 EDT) Katie Belange /$ ' )" 0"CvA?A@

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T P V T ³ R T y T \ 7 T y T ³ P y µ ²³ \ 2 T k J 7 J T T y¯e± P T R PAn Introduction to Basic Statistics and Probability – p 1/40 Outline Basic probability concepts Conditional probability Discrete Random Variables and Probability Distributions Continuous Random Variables and Probability Distributions Sampling Distribution of the Sample Mean Expected Value µ#!a a2 = µ 0 I 4a The direction of B(P), from the cross product, is into the page FIGURE 3050 Problem 15 Solution Problem 17 Figure 3051 shows a conducting loop formed from concentric semicircles of radii a and b If the loop carries a current I as shown, find the magnetic field at point P

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µ and finite variance σ 2 Then, regardless of the shape of the population distribution of X, as the sample size n gets larger, the sampling distribution of X becomes increasingly closer to normal, with mean µ and variance σ 2 n, that is, X ~ N µ, σ n , approximately More formally, (0,1) as / X Z Nn n µ σ − = →→∞ If X ~ N (µ6 µ « æ Ó Ä w ï µ Ä ç InDesign w µ « æ Ó Ä x o t ï µ Ä ç p V b {InDesign Ó æ ­ ³ ã ï Ñ ¥ ç ¼ º w ScriptsM t Q < s µ Ð É p x z m w 0 Ã ² ï ;

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D e =µ e qkT ( ) =1600/40=40cm 2 /s D h =µ h qkT ( ) =600/40=15cm 2 /s II Relating φ to n and p, and visa versa To calculate φ knowing n or p it is better toC P(µ – 2σ ≤ X ≤ µ – σ or µ σ ≤ X ≤ µ 2σ) = P(within 2 sd's) – P(within 1 sd) = P(µ – 2σ ≤ X ≤ µ 2σ) – P(µ – σ ≤ X ≤ µ σ) = 9544 – 66 = 2718 395 30 P x 40 1 P x 39 1 P Z =0392 b 5% of 1000 = 50 ( ) (380) 100γ > 40 MeV Since the e−νeνµand e−νeνµγ modes cannot be clearly separated, we regard the latter mode as a subset of the former b See the Particle Listings below for the energy limits used in this measurement c A test of additive vs multiplicative lepton family number conservation µ− BRANCHING RATIOS Γ e−νeνµγ e µ

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Then, letting ν = µσ2/2, we can rewrite the equations as ud = 1, (12) pu(1−p)d = eν(t/n), (13) pu2 (1−p)d2 = e(2νσ2)(t/n) (14) (13) allows us to solve for p in terms of u and d, p = eν(t/n) −d u−d (15) Then using this formula for p together with ud = 1 to plug into the (14) allows us to solve for u (and hence d) (see2 TRIGYN TECHNOLOGIES LIMITED ANNUAL REPORT 18 19 NOTICE NOTICE is hereby given that the Thirty Third Annual General Meeting (AGM) of the members of Trigyn Technologies Limited will be held on Tuesday, September 24th, 19, at 330 pm IST at HOTEL SUNCITY RESIDENCY, 16th Road, MIDC, Marol, Andheri (E), Mumbai to transact the following business(e) the variance of Y 4 Let Y be a random variable having mean µ and suppose that E(Y −µ)4 ≤ 2 Use this information to determine a good upper bound to P(Y −µ ≥ 10) 5 Let U and V be independent random variables, each uniformly distributed on 0,1 Set X = U V and Y = U − V Determine whether or not X and Y are

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