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22 Jan 2011: 5.35 6.15 3.30 122.20 Tenerife 12.30 1.25 4.90 3.70 2.90 130.85 Florida 15.60 1.90 5.05 5.00 ... 30 Sicily 13.25 1.75 4.20 7.00 3.85 174.40 Madeira 10.25 0.70 5.10 6.85 6.85 153.70.

3 Mar 2011: dx log n for n N. Show that In =. 1/20. f (t/n) dt and hence that|In| < 1/9n2. By considering. ... 7. Let In = π/20. cosn x. Prove that nIn = (n 1)In2, and hence 2n2n1 I2n1/I2n 1.

28 Oct 2011: . 1 1 00 3 20 1 0. .  ,. . 1 1 10 3 20 1 0. .  ,. . 1 1 11 3 11 1 1. .  ,. .

3 Mar 2011: dx log n for n N. Show that In =. 1/20. f (t/n) dt and hence that|In| < 1/9n2. By considering. ... 7. Let In = π/20. cosn x. Prove that nIn = (n 1)In2, and hence 2n2n1 I2n1/I2n 1.

11 Feb 2011: By definition, X0 = 0. Plot a realisation of aPoisson process of rate 1 over the time interval t [0, 20]. ... Estimate by simulationthe probability that no more than 15 arrivals have occurred by t = 20.

9 Mar 2011: Estimate v from the following independentobservations made on 9 shells. α (deg) 5 10 15 20 25 30 35 40 45sin 2α 0.1736 0.3420 0.5 0.6428

18 Feb 2011: y 5.50 5.60 5.90 5.80 6.10 6.30 6.4 6.20. Exercise: Plot the data. ... 19 20 21. -0.17838665 -0.06180739 -0.27838665. > fitted(LinMod1). 4. 1 2 3 4 5 6 7 8.

16 Feb 2011: Improved No difference WorsePlacebo 18 17 15Half dose 20 10 20Full dose 25 13 12.

12 Oct 2011: 2. Between0 and 10 there are four primes. The same is true between10 and 20.

24 Jan 2011: Hence find the complete character table of S5. Repeat, replacing S4 by the subgroup 〈(12345), (2354)〉 of order 20 in S5.

14 Feb 2011: Of ourse, this is about 20 separate questions.)1(6) Show that OK = Z[(1 19)/2] is a PID, but is not Eu lidean.

4 Feb 2011: What do the functions summary and sample do? > X <- rnorm(20, mean = 1, sd = 2). >

12 Nov 2011: 3.20. Let ΦN Z[X] denote the N-th cyclotomic polynomial. Show that:(i) If N is odd and N ̸= 1 then Φ2N(X) = ΦN(X).(ii) If p

16 Oct 2011: We have written ϕ = 1 ϕ. 1 1 20 20 30 12 12 12 12I I ST ST S T T T 2 T 2. ... 8). D = ab(a10 11a5b5 b10)c4 = a. 20 228a15b5 494a10b10 228a5b15 b20.

16 Aug 2011: Compressions and Probably IntersectingFamilies. Paul A. Russell †. April 20, 2011.

21 Dec 2011: w13 0 0 1 1 2 0 1 1 2 3 8 w28 0 1 2 3 4 2 3 4 5 5 20. ... 2. 20 TOM FISHER. Lemma 7.9. Let φ X5(k) be a genus one model satisfying the conclusions ofLemma 7.8.

9 May 2011: So they can differ by at most a Möbius map. 20 T.A. ... It is shown in [AR, Corollary 23.28] that X(11) P4 has degree 20.

18 Jul 2011: See [23] for a precise statement.) Our work in the casen = 3 is a starting point both for the work on 6- and 12-descent in [20], andfor the work on ... 20 J.E. CREMONA, T.A. FISHER, C. O’NEIL, D. SIMON, AND M.

29 Jul 2011: Let. Mi = {m1 mi1 1,. ,m1 mi1 mi},. 20 J. ... 20 of [BB05]. The first part now follows froma direct calculation (alternatively, one could apply [Cas], Proposition1.3.3).

12 Jul 2011: 7.9) Example. G = S5. 1 15 20 24 10 20 30 |Cj |1 (12)(34) (123) (12345) (12) (123)(45) (1234) gj. ... Example. G = S5 (again). 1 15 20 24 10 20 30 |Cj |1 (12)(34) (123) (12345) (12) (123)(45) (1234) gj.