answer:Solution: (1) Given S_n=(2^n-1)a_n, we can derive S_{n-1}=(2^{n-1}-1)a_{n-1} for ngeqslant 2, Subtracting these two equations, we get S_n-S_{n-1}=(2^n-1)a_n-(2^{n-1}-1)a_{n-1}, (2^n-2)a_n=(2^{n-1}-1)a_{n-1}, which implies dfrac{a_n}{a_{n-1}}=dfrac{1}{2} for ngeqslant 2, Therefore, {a_n} is a geometric sequence with the first term 1 and common ratio dfrac{1}{2}, So, a_n=left(dfrac{1}{2}right)^{n-1} for ninmathbb{N}^*; (2) b_n=na_n=ncdotleft(dfrac{1}{2}right)^{n-1}. T_n=b_1+b_2+b_3+ldots+b_n=1timesleft(dfrac{1}{2}right)^0+2timesleft(dfrac{1}{2}right)^1+3timesleft(dfrac{1}{2}right)^2+ldots+nleft(dfrac{1}{2}right)^{n-1}, dfrac{1}{2}T_n=1timesleft(dfrac{1}{2}right)^1+2timesleft(dfrac{1}{2}right)^2+ldots+(n-1)left(dfrac{1}{2}right)^{n-1}+nleft(dfrac{1}{2}right)^n, Subtracting the second equation from the first, we get dfrac{1}{2}T_n=1+left(dfrac{1}{2}right)^1+left(dfrac{1}{2}right)^2+ldots+left(dfrac{1}{2}right)^{n-1}-nleft(dfrac{1}{2}right)^n=2-dfrac{n+2}{2^n}, Therefore, T_n=4-dfrac{n+2}{2^{n-1}}. The final answers are: (1) a_n=boxed{left(dfrac{1}{2}right)^{n-1}} (2) T_n=boxed{4-dfrac{n+2}{2^{n-1}}}

answer:The possible values of a and b remain the same, which are 20 values as listed before. Let x = cos(api) and y = sin(bpi). The expression becomes: [(x + yi)^6 = x^6 - 15x^4y^2 + 15x^2y^4 - y^6 + i(6x^5y - 20x^3y^3 + 6xy^5).] For this to be real, the imaginary part must equal zero: [6x^5y - 20x^3y^3 + 6xy^5 = 0.] Factoring out common terms gives: [6xy(x^4 - 10x^2y^2 + y^4) = 0.] The solutions are x = 0, y = 0, or x^4 - 10x^2y^2 + y^4 = 0. The third equation simplifies further into a quadratic in terms of (x/y)^2: [(x/y)^4 - 10(x/y)^2 + 1 = 0,] which can be solved using a substitution u = (x/y)^2. This quadratic gives two solutions: u = 5 pm 2sqrt{6}, which implies conditions on a and b that might not be solvable in the set S. Counting the simple cases where x=0 or y=0: - If x = 0 (i.e., cos(api) = 0), then a = frac{1}{2}, frac{3}{2}, and b can be any of the 20 values, giving 40 pairs. - If y = 0 (i.e., sin(bpi) = 0), then b = 0, 1, and a can be any of the 20 values, giving 40 pairs. However, overlaps (4 pairs) must be subtracted, resulting in 36 additional pairs. The complex solutions from the quadratic may not yield additional pairs due to the specific values required, not found in the set S. Thus, the number of pairs that make (cos(api) + i sin(bpi))^6 real is 40 + 36 = 76. The total number of ways to choose (a, b) is 20^2 = 400. The probability is frac{76}{400} = boxed{frac{19}{100}}.

answer:To find the man's rowing speed in still water, we need to determine his speed going against the stream and with the stream, then calculate the average of the two. First, let's find his speed against the stream: Distance against the stream = 750 m Time against the stream = 675 seconds Speed against the stream = Distance / Time Speed against the stream = 750 m / 675 s Speed against the stream = 1.111 m/s Now, let's find his speed with the stream: Distance with the stream = 750 m Time with the stream = 7.5 minutes (we need to convert this to seconds to match the units) Time with the stream = 7.5 minutes * 60 seconds/minute = 450 seconds Speed with the stream = Distance / Time Speed with the stream = 750 m / 450 s Speed with the stream = 1.667 m/s The man's rowing speed in still water (V_s) is the average of his speed against the stream (V_a) and his speed with the stream (V_w). However, we need to account for the stream's speed (V_c) as well. The formula to find the rowing speed in still water is: V_s = (V_a + V_w) / 2 But we don't have the stream's speed yet. We can find it by subtracting the speed against the stream from the speed with the stream and then dividing by 2: V_c = (V_w - V_a) / 2 Now we can find the man's rowing speed in still water: V_s = V_a + V_c V_s = 1.111 m/s + (1.667 m/s - 1.111 m/s) / 2 V_s = 1.111 m/s + 0.556 m/s / 2 V_s = 1.111 m/s + 0.278 m/s V_s = 1.389 m/s Therefore, the man's rowing speed in still water is approximately boxed{1.389} meters per second.

answer:The question requires us to correctly understand several concepts in the sampling process. Commonly, the math scores of 400 students in their first mock college entrance examination are the population, the math scores of 12 students in each class are the sample, 400 is the population size, and 96 is the sample size. Therefore, we select the answer as boxed{C}.