how effective are the items

... from the mean, i.e., subtract the mean from the mark Square each of the deviation Sum the squares Average the sum of squares by dividing it with N, that is, the number of students Take the positive ... has the lowest range (0) and the lowest SD (0.00) On the other hand, Group C has the greatest range (55) and the greatest SD (22.9) The other two groups have the “in-between” ranges and the “in-between” ... from the mean.” Look at Table 3.1 Take Group A All marks are the same as the average, and therefore, none of them deviate from the mean So, the sum of deviations is zero, and the SD is therefore...

Ngày tải lên: 13/08/2016, 09:04

... If the first two letters are the same as an entry in the prefix dictionary, a note is made of the 76 prefix, these two letters are dropped from the word, and the third and fourth letters are ... prefixes, the remainder of the word being regarded as the stem The stem dictionary is now searched for this stem If it is found, the associated meaning, and the meanings of the recognized prefixes, are ... search for the stem toto, and then check the prefix for number In the case of verb forms, however, the active voice, in unmodified form, occurs far more frequently than any of the other forms,...

Ngày tải lên: 23/03/2014, 13:20

... using the Digamma formulation (7) are the same) Dashed lines show the values given by the Antoniak approximation (4) (the line for α = 100 falls below the bottom of the graph) Stars show the mean ... as in the unigram LM Circles show the mean of empirical table counts when P1 is inferred, as in the bigram LM Standard errors in both cases are no larger than the marker size All plots are based ... in the corpus All the bigram restaurants share a common base distribution P1 over unigrams, which must be inferred Predictions in this model are as follows: these kinds of models, the counts are...

Ngày tải lên: 31/03/2014, 00:20

Ngày tải lên: 05/08/2014, 15:21

... to obtain the lower bound let us observe that there are exactly ϕ maximal (k, )-sum-free arithmetic progressions with the difference ρ Precisely ϕr of them have length n/ρ and ϕ − ϕr are of length ... Furthermore we have less than n2 possible choices for the pair (d , ρ), hence there are at most 2n2 2n/(ρ+1) such (k, )-sum-free sets Thus, the number of (k, )-sum-free sets satisfying neither ... them have length n/ρ and ϕ − ϕr are of length n/ρ Since these progressions are pairwise disjoint, there are at least (ϕ + ϕr )2 n/ρ the electronic journal of combinatorics (2000), #R30 (k, )-sum-free...

Ngày tải lên: 07/08/2014, 06:20

... Unfortunately, the condition (k − t) = in the hypothesis of our theorem precludes a new proof of Wilson’s theorem via our recursive formula In the special case when the characteristic p of F is larger ... does apply, with the same conclusion and proof as the above corollary In conclusion, we raise the question as to whether there is a q-analogue of formula (1), (q) i.e., for the (0,1)-inclusion ... seen by fixing x in X and classifying t-sets and k-sets according to whether x belongs to them or not Further, there is the elementary product formula Wt,k (v)Wk,l (v) = l−t k−t Wt,l (v) (4) whose...

Ngày tải lên: 07/08/2014, 06:22

... degree” we mean the ratio between its outdegree and the cardinality of the opposite vertex class), then there exists a directed cycle of length at most four that contains v Thus, there is also ... αmn In order to prove the upper bound in Theorem 1.1 we need to show that if α > 2/3 then G has a directed C4 For v ∈ V let dv denote the indegree and outdegree of v (it is the same by assumption) ... M2 Denote by M3 the block consisting of the rows i = |A+ | + 1, , m and the columns j = 1, , ρn Denote by M4 the block consisting of the rows i = |A+ | + 1, , m and the columns j = 2ρn+...

Ngày tải lên: 07/08/2014, 06:23

... apply the induction hypothesis using the paths P and Q to obtain a path R satisfying the lemma Next, we repeat the above argument with the portion of R beginning at um−1 and the vertex um Theorem ... such that → v1 and all other arcs are of the form vi → vj with i < j In other words, reversing the arc → v1 gives the transitive tournament of order n As noted by Moon [1], there is a bijection ... allow the special case where m = 0; in this case the path Q is a path on vertices, and R = P satisfies the lemma trivially The remainder of the proof is by induction on m For m = 1, let i be the...

Ngày tải lên: 07/08/2014, 13:21

... two of them there are either all possible edges or none of them The factorial layer is substantially richer It contains most of the unitary classes mentioned above (the unique exception in the above ... consists of classes with the same index Moreover, the classes Ei,j with the same value of i + j are the only minimal classes in the respective subset In particular, for k = 2, there are exactly three ... listed in the order of their appearances in the linear interval representation of G Fuzzy linear interval strips are defined analogously, provided that if a, b are the endpoints of the strip then φ(a),...

Ngày tải lên: 08/08/2014, 14:23

... fact, in the article the authors say that the methodology is to calculate the risk of the cost of capital, although at the end they say it is to define the risk for the equity cost The way the methodology ... before starting the work in the spreadsheet or when Excel declares the presence of circularity After these instructions are done, then, the WACC can be calculated as the sum of the debt and equity ... the finance textbooks teach that the discount rate for the TS should be the cost of debt However, the tax savings depend on the firm profits Hence, the risk associated to the tax savings is the...

Ngày tải lên: 07/03/2014, 00:20

... completes the proof of Theorem 1.1 Competing interests The author declares that she has no competing interests Acknowledgments The author was very grateful to the referees and the Editors for their ... also removed the condition nP(|X1 | > ηn ) ≤ c(log n)ε0 , < ε0 < in theorem 1.1 of [12] Remark 1.4 If EX < ∞, then X is in the domain of attraction of the normal law Therefore, the class of random ... interest to determine the optimal weights On the other hand, by the Theorem of Schatte [3], Equation (3) fails for weight dk = The optimal weight sequence remains unknown The purpose of this article...

Ngày tải lên: 20/06/2014, 21:20

... (2010GXNSFA013120) The authors are very grateful to the referees and the editors for their valuable comments and some helpful suggestions that improved the clarity and readability of the article Author ... ,P(A)>0 m n+1 |P(B|A) − P(B)| When these are compared with the corresponding results of independent random variable sequences, there still remains much to be desired The concept of complete convergence ... out the design of the study and performed the analysis QZ participated in its design and coordination All authors read and approved the final manuscript Competing interests The authors declare...

Ngày tải lên: 20/06/2014, 22:20

... can be proved Theorem 1.6 by using Theorem 2.4 But it is not easy to prove Theorem 2.4 by using Theorem 1.6 So Theorem 2.4 is more general than Theorem 1.6 Proof of Theorem 1.6 The set of all ... and then they proved the case of mean zero by using these results However, this approach is not good For the case of negatively associated mean zero, an additional condition is assumed in Theorem ... (i) Hence, the result is proved ■ It can be proved Theorem 1.4 by using Theorem 2.6 But it is not easy to prove Theorem 2.6 by using Theorem 1.4 So Theorem 2.6 is more general than Theorem 1.4...

Ngày tải lên: 20/06/2014, 22:20

... satisfies the hypotheses of the Theorem and that any of the equivalent conditions holds, then for r = |z| log |σ (z)| = o log log 1−r We also provide the following restatement of the hypotheses of Theorem ... proceeding, we prove the following lemma Lemma Under the hypotheses of the theorem, w and gw tend to the same prime end at ∞, and ∂gΩ ⊂ Ω Proof The first assertion follows from the fact that the closure ... to show that if lim f (w) = lim w→∞ w→∞ − log δ (w) =∞ log |w| then holds To complete the proof, we require the following lemma whose proof we merely sketch Lemma Under the hypotheses of the theorem,...

Ngày tải lên: 21/06/2014, 01:20

... symmetric properties of them Furthermore, using these symmetric properties of them, we can obtain some relationships between the twisted q-Bernoulli numbers and polynomials and between the twisted generalized ... u2 u1 ⎞ ⎠ 2.17 n t n! By comparing the coefficients of tn /n! on both sides of 2.16 and 2.17 , we obtain the following theorem Theorem 2.1 Let u1 , u2 , n ∈ N Then for all x ∈ Zp , n i where n qu1 ... un−i , 2.18 i is the binomial coefficient From Theorem 2.1, if we take u2 1, then we have the following corollary Corollary 2.2 For m ≥ 0, one we has q Bi,w u1 x where n i is the binomial coefficient...

Ngày tải lên: 21/06/2014, 16:20

... decomposition theory If fixed τ > 0, there are some efforts on the global existence of smooth solutions to the system (1.1)-(1.2) for the isentropic gas or the general hyperbolic system, the interested ... of the multidimensional isothermal Euler equations,” Transactions of the American Mathematical Society, vol 359, no 2, pp 637–648, 2007 [2] S Junca and M Rascle, “Strong relaxation of the isothermal ... Problems To be concerned with the small relaxation-time analysis, we define the scaled variables nτ ,uτ (x,s) = (n,u) x, s τ (1.3) Then the new variables satisfy the following equations: nτ +...

Ngày tải lên: 22/06/2014, 19:20

... Since the dimension of the row-span of a matrix is equal to the dimension of its column-span, we can equally well study the dimension of the space spanned by the columns of the table By the preceeding ... that fλ (q) is the generating function for the entries of the column corresponding to the λ-cycles The dimension of the column-span of our table is therefore equal to dim Fn , and the proposition ... to be the complex vector space spanned by all of these fλ (q)’s We have: Proposition 1.2 dim Fn = D(n) Proof Consider the table of the characters χN ; we are interested in the dimension of the...

Ngày tải lên: 07/08/2014, 06:20

... using the product of the probabilities of each of the events as an upper bound In our approach, a key step to further improving the upper bound obtained in [2] is the exact computation of the probability ... points close to the point at which downhill simplex finds the maximum of ln E and we confirmed that the value of ln E is not above the value returned by the method The above discussion shows that -0.00000998 ... involving the conjunction of the events using the occupancy problem for random placements of balls into bins We make use of the sharp estimates obtained by Kamath et al in [8] for the probability of the...

Ngày tải lên: 07/08/2014, 06:20

... [17]) are very useful In these applications, the parameter b is quite large, and may depend on n There are other applications of b-tries in computer science, communications and biology Among these ... Results We establish the five parts of Theorem Since the analysis involves a routine use of the saddle point method (cf [1, 12]), we only give the main points of the calculations The distribution ... We next consider the limit b → ∞ We now find that there are five cases of (n, k) to consider, and we summarize our final results below The necessity of treating the five cases in Theorem is better...

Ngày tải lên: 07/08/2014, 06:20

... depend on the language, but not the specific input Furthermore, the entire conversation must have length polynomial in the length of the input In the above model, Arthur’s coin flips are public– ... calculation shows that 1 Pr[∀v, χ(v) = χ(φ(v))] = ( )
(φ) ≤ ( )
(G) χ d d Collecting together these bad events, we have Pr[∃φ = id, ∀v, χ(v) = χ(φ(v))] ≤ |Aut(G)|( )
(G) χ d The hypothesis of the theorem ... interesting to notice that the above theorem is actually tight in the case of the dihedral groups D3, D4, mentioned in the introduction (and in [1]) (The answers are
(D3) = 3,
(D4) = 4,
(D5)...

Ngày tải lên: 07/08/2014, 06:22