COMPUTER-BASED TOEFL TEST

G+TiC=C a three-part argument map that uses deduction as a method of organization: G (general statement = opinion or premise) + TiC (body paragraph) = C (conclusion) G+2TiC=C a four-part argument map that uses deduction as a method of organization: G (general statement = opinion or premise) + 2[r]

PBT, the Structure and Written Expression section score is based on multiple-choice questions only. The essay score is reported separately from the total score.Using Section ScoresETS encourages institutions to set separate section scores in addition to, or instead of, a total score. Two appl[r]

TOEFL IBT: những điều cần biết Bạn chuẩn bị thi TOEFL IBT, bạn chưa nắm rõ về nội dung, cách làm bài và thời gian một vài lưu ý sau đay sẽ giúp bạn hiểu rõ hơn về kì thi này. TOEFL (theTest of English as a Foreign Language) là một kỳ thi chuẩn hóa có khả năng đánh giá sự thành thạo về tiếng Anh c[r]

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Sample Questions for the Computer-Based TestThe questions on the following pages are similar tothose that will be presented in the actual test.ListeningThe Listening section of the test measures the ability tounderstand conversations and talks in English. You willuse head[r]

www.candidates.CambridgeESOL.orgFlexible test optionsIn addition to the standard paper-based exam, CambridgeEnglish: First is available as a computer-based test.Advantages of computer-based testing include:• Additional exam sessionsMore exam dates are[r]

n = 0 (1)where a0≠ 0 and all ai′ s are real.For polynomials, if the coefficients are all real valued then the complex roots occurs inconjugate pair. Therefore we extract the quadratic factors that are the products of the pairs of92COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUEScomplex roo[r]

28561 20309855 32438416 2354 2812209 352 2050625 2706 814915 360 4065540 3066 4817981 408 4083521 3474 821455 416 20104976 3890 2825345 444130321 433429679160000y y y yyy→←134COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUESFrom this table we have the third differences are quite ir[r]

116COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUESExample 14. Evaluate the following:I. ∆2 (cos 2x)II. ∆2 (3ex)III. ∆ tan–1 xIV. ∆(x + cos x)the interval of differencing being h.Sol. I. We have ∆2 (cos 2x)= (E – 1)2 cos 2x because ∆= E – 1.=(E2 – 2 E + 1) cos 2x= E2 cos 2x – 2E cos[r]

,∆4y5 is as ∆2y10 = –73 and ∆4y5 = –1.108COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUESExample 5. Find f(6) given that f(0) = –3, f(1) = 6, f(2) = 8, f(3) = 12, the third differences beingconstant.Sol. For given data we construct the difference table:∆∆ ∆−−23() () () ()03916 72928[r]

when p and q are of required accuracy.98COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES11 6 18 24 161.9882 1.9882 7.97626 15.95299 0.0473431.9999 1.9999 8.023198 16.046871 4.0118 8.02384 0.023812 0.094213iab−−−−−−−−−−Thus b0 = 1, b1 = –4.0118, b2 = 8.02384 and thus the polynomial b[r]

Now f(1.375) = (1.375)3 – 1.375 – 1f(1.375) = 0.2246∴f(1.375) is positive.∴The required root lies between 1.25 and 1.375.Fourth iteration: The fourth approximation is given by 41.25 1.3751.3132x+==Now f(1.313) = (1.313)3 – 1.313 – 1f(1.313) = – 0.049438COMPUTER BASED NUMERICAL AND STAT[r]

So that f(2) = (2)3 – 2(2) – 5 = – 150COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUESand f(3) = (3)3 – 2(3) – 5 = 16Therefore, a root lies between 2 and 3.First approximation: Therefore taking x0 = 2, x1 = 3, f(x0) = – 1, f(x1)= 16, then by Regula-Falsi method, we get ()()()1020 01[r]

Step 4: Stop the process when the prescribed accuracy is obtained.80COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES2.8.2 Rate or Order of Convergence of Secant MethodOn substituting îå=+nnx, etc. in (1) we obtain the error equation as åå îåååîå îå−+−−+=−+−+111()()()( )nnfffnn nnn ([r]

0 be the initial approximation andh be a small corrections to x0, so thatf(x0 + h)= 0Expanding it by Taylor’s series, we getf(x0 + h) = f(x0) + hf ′(x0) + 22!h f ′(x0) + = 0Since h is small, we can neglect second and higher degree terms in h and therefore, we getf(x0) + hf ′ (x0)= 068COMPUTER[r]

In Second case, φ(x)= 15 (sin x + 2)∴φ’(x)= 15 cos x64COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUESHence, ()′φ≤15x for all x because cos1x ≤∴φ’(x) will succeed.Hence taking x = φ(x) = 15 (sin x + 2) and initial value x0 = 0.5, we have the first approximationx1

196COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUESExample 5. If f(x) is a polynomial of degree four find the value of f(5.8) using Gauss’s backwardformula from the following data:f(4) = 270, f(5) = 648, ∆f(5) = 682, ∆3f(4) = 132.Sol. Given (5) 682f∆=⇒ f (6) – f (5) = 682⇒ f (6) = 6[r]

200 220 260 ? 350 ? 430YearProduction[Ans. f(1964) = 306 f(1966 = 390)]5. Given, log 100 = 2, log 101 = 2.0043, log 103 = 2.0128, log 104 = 2.0170. find log 102.[Ans. log 102 = 2.0086]6. Estimate the missing term in the following:1234 5 6 72 4 8 ? 32 64 128xyExplain why the result differs from 16. [[r]

−−−−−0192039405960798099. 41 62 65 50 17Find no. of candidates who obtained fewer than 70 marks.Sol. First, we form the difference table.Marks less than x No of candidates y ∇∇ ∇ ∇−−−−234 () . ()19 416239 103 365 1859 168 15 050 1879 218 331799 235178COMPUTER BASED NUMERICAL AND STATIS[r]