answer:(1) Since S_{m-1}=-4, S_{m}=0, and S_{m+2}=14, we have a_{m}=S_{m}-S_{m-1}=4 and a_{m+1}+a_{m+2}=S_{m+2}-S_{m}=14. Let d be the common difference of {a_n}, then 2a_{m}+3d=14, thus d=2. Since S_{m}=frac{a_{1}+a_{m}}{2}m=0, we have a_{1}=-a_{m}=-4. So, a_{m}=a_{1}+(m-1)d=-4+2(m-1)=4, which implies m=5. (2) From (1), we have a_{n}=2n-6. Since frac{a_{n}}{2}=log_{2}b_{n}, it follows that n-3=log_{2}b_{n}, so b_{n}=2^{n-3}. Thus, (a_{n}+6)·b_{n}=2n·2^{n-3}. Let T_n denote the sum of the first n terms of the sequence {(a_{n}+6)·b_{n}}. Then T_n=sum_{k=1}^{n}(a_{k}+6)·b_{k}=sum_{k=1}^{n}2k·2^{k-3}. We can compute T_n as: begin{align*} T_n&=sum_{k=1}^{n}2k·2^{k-3} &=2sum_{k=1}^{n}k·2^{k-3} &=2left((1cdot 2^{-2})+(2cdot 2^{-1})+(3cdot 2^{0})+cdots+(ncdot 2^{n-3})right) &=2left(frac{2^{-2}(1-2^{n})}{1-2}-ncdot 2^{n-2}right) &=2left(frac{2^{-2}}{1-2}-frac{(n-n2+1)2^{n-2}}{1-2}right) &=boxed{(n-1)2^{n-1}}. end{align*}

answer:Since x^2 + 1 geq 1, we know that lg(x^2 + 1) geq 0, so M = {y | y geq 0}. From 4x > 4, we know x > 1, so N = {x | x > 1}. Therefore, M cap N = {x | x > 1}. Hence, the correct option is boxed{C}.

answer:Given that the random variable X sim B(n,p), and EX= frac {5}{2}, DX= frac {5}{4}, Therefore, np= frac {5}{2}, np(1-p)= frac {5}{4}, Solving these equations, we get n=5, p= frac {1}{2}, Therefore, P(X=1)=C_{5}^{1}( frac {1}{2})^{1}( frac {1}{2})^{4}= frac {5}{2^{5}}= frac {5}{32}. Hence, the answer is boxed{frac {5}{32}}. This problem involves a random variable that follows a binomial distribution. By using the formulas for expectation and variance, we can solve for the values of n and p, and then calculate the probability for a given value of the random variable. This question tests the understanding of expectation and variance for a binomially distributed random variable, focusing on the memorization of formulas, making it a foundational question.

answer:To determine the average speed of the train over the entire journey, we need to calculate the total time taken for the journey and then divide the total distance by the total time. First, let's calculate the time taken for each segment: 1. Segment A: Distance = x km, Speed = 75 kmph Time = Distance / Speed = x / 75 hours 2. Segment B: Distance = 2x km, Speed = 25 kmph Time = Distance / Speed = 2x / 25 hours 3. Segment C: Distance = 1.5x km, Speed = 50 kmph Time = Distance / Speed = 1.5x / 50 hours 4. Segment D: Distance = x km, Speed = 60 kmph Time = Distance / Speed = x / 60 hours 5. Segment E: Distance = 3x km, Speed = 40 kmph (reduced by 10% due to wind resistance) Adjusted Speed = 40 kmph - (10% of 40 kmph) = 40 kmph - 4 kmph = 36 kmph Time = Distance / Speed = 3x / 36 hours Now, let's calculate the total time taken for the entire journey by adding the times for each segment: Total Time = (x / 75) + (2x / 25) + (1.5x / 50) + (x / 60) + (3x / 36) To simplify the calculation, let's find a common denominator for the fractions. The least common multiple (LCM) of the denominators (75, 25, 50, 60, 36) is 900. We'll express each time segment in terms of this common denominator: Total Time = (12x / 900) + (72x / 900) + (27x / 900) + (15x / 900) + (75x / 900) Now, let's add the numerators: Total Time = (12x + 72x + 27x + 15x + 75x) / 900 Total Time = (201x) / 900 hours The total distance covered is 7.5x km. Now we can calculate the average speed: Average Speed = Total Distance / Total Time Average Speed = 7.5x / (201x / 900) To simplify, we can cancel out the 'x' from the numerator and denominator: Average Speed = 7.5 / (201 / 900) Average Speed = 7.5 * (900 / 201) Now, let's calculate the value: Average Speed = 7.5 * (900 / 201) Average Speed = 7.5 * 4.4776 (approximately) Average Speed ≈ 33.582 kmph Therefore, the average speed of the train over the entire journey is approximately boxed{33.582} kmph.