answer:Let Aleft(a, frac{1}{a}right), Bleft(b, frac{1}{b}right)(a, b>0), then vec{m}=left(1, sqrt{a^{2}+frac{1}{a^{2}}}right), overrightarrow{O B}=left(b, frac{1}{b}right). Thus, vec{m} cdot overrightarrow{O B}=b+frac{1}{b} sqrt{a^{2}+frac{1}{a^{2}}} geqslant b+frac{sqrt{2}}{b} geqslant 2 sqrt[4]{2}, equality holds when (a, b)=(1, sqrt[4]{2}). Therefore, the minimum value of vec{m} cdot overrightarrow{O B} is 2 sqrt[4]{2}.

answer:First, calculate the mean of x as follows:overline{x} = frac{3+5+7+12+13}{5} = 8Next, substitute the mean of x into the given linear regression equation:hat{y} = frac{1}{2}x + 20Now, find overline{y} by substituting overline{x} into the equation:overline{y} = frac{1}{2} times 8 + 20 = 24Therefore, the answer is boxed{D}.The reasoning behind this solution is that the regression line always passes through the center of the sample points. This question tests your understanding of the linear regression equation and its relationship with the sample data center.

answer:Since tan left( frac {pi}{4}+alpha right)= frac {tan frac {pi}{4}+tan alpha}{1-tan frac {pi}{4}tan alpha}= frac {1+tan alpha}{1-tan alpha}= frac {1}{2},then 2(1+tan alpha)=1-tan alpha, solving this gives: tan alpha= - frac {1}{3}.Therefore, the correct answer is: boxed{A}.This problem involves using the values of trigonometric functions for special angles and the formula for the tangent of the sum of two angles to simplify and solve. It tests the application of these concepts in simplifying and solving trigonometric functions, focusing on transformational thinking, and is considered a basic question.

answer:## Solution.By subtracting the given equations, we obtain the equationw-w^{2}=z-z^{2}which is equivalent to the equation(w-z)(w+z-1)=0Since w-z neq 0, it follows thatw+z=1Squaring this equality, we getw^{2}+z^{2}=1-2 w zand squaring again,w^{4}+z^{4}=1-4 w z+2(w z)^{2}The first equation of the given system is equivalent to the equationfrac{w^{5}-z^{5}}{w-z}=-1It can be written in the formbegin{aligned}frac{(w-z)left(w^{4}+w^{3} z+w^{2} z^{2}+w z^{3}+z^{4}right)}{w-z} & =-1 w^{4}+w^{3} z+w^{2} z^{2}+w z^{3}+z^{4}+1 & =0 w^{4}+z^{4}+w zleft(w^{2}+z^{2}+w zright)+1 & =0end{aligned}Substituting (2) and (3) and simplifying, we obtain the equation(w z)^{2}-3 w z+2=0Solving this, we find that w z=1 or w z=2.Substituting z=frac{1}{w} and z=frac{2}{w} into (1), we get the quadratic equations w^{2}-w+1=0 and w^{2}-w+2=0, from which we finally obtain the four solutions(w, z)=left(frac{1 pm sqrt{3} i}{2}, frac{1 mp sqrt{3} i}{2}right) quad text { and } quad(w, z)=left(frac{1 pm sqrt{7} i}{2}, frac{1 mp sqrt{7} i}{2}right)

answer:AnalysisThis problem involves understanding basic events and applying the classical probability formula. We will determine the total number of basic events, identify the cases where the average of the drawn labels is 3, and then use the classical probability formula to calculate the probability.Step-by-step solution1. Determine the total number of possible outcomes. When drawing 3 labels from 5 distinct labels, the number of possible combinations is {5 choose 3} = 10.2. Identify the favorable outcomes. We are looking for combinations of 3 labels that have an average of 3. The possible combinations are: - (1, 3, 5) with an average of frac{1+3+5}{3} = 3. - (2, 3, 4) with an average of frac{2+3+4}{3} = 3.3. Apply the classical probability formula. The probability of an event is defined as the number of favorable outcomes divided by the total number of possible outcomes. In this case, p = frac{2}{10} = boxed{frac{1}{5}}.

answer:: 2/7. For i = 1, 2, 3, 4, 5, let p i (n) = prob that the final score is n+i if the player stops when his total score is at least n. We note that p(n) is also the probability that the player's total equals n on some throw if he throws repeatedly. Now we can see that p 5 (n) = 1/6 p(n-1), because the only way to achieve a final score of n+5 without passing through n, n+1, n+2, n+3, n+4 is to reach n-1 and then throw a 6. Similarly, p 4 (n) = 1/6 p(n-1) + 1/6 p(n-2), because to reach n+4 without passing through n, n+1, n+2, n+3 you must either to through n-1, which requires reaching n-1 and then throwing a 5, or not, in which case you must reach n-2 and then throw a 6. Similarly, p 3 (n) = 1/6 p(n-1) + 1/6 p(n-2) + 1/6 p(n-3), p 2 = 1/6 p(n-1) + 1/6 p(n-2) + 1/6 p(n-3) + 1/6 p(n-4), and p 1 (n) = 1/6 p(n-1) + 1/6 p(n-2) + 1/6 p(n-3) + 1/6 p(n-4) + 1/6 p(n-5). Adding, we get: 1 = p(n) + p 1 (n) + p 2 (n) + p 3 (n) + p 4 (n) + p 5 (n) = p(n) + 5/6 p(n-1) + 4/6 p(n-2) + 3/6 p(n-3) + 2/6 p(n-4) + 1/6 p(n-5). In the limit, p(n-5) = p(n-4) = ... = p(n). Hence we get: p 1 (n) = 5/6 p(n), p 2 (n) = 4/6 p(n), p 3 (n) = 3/6 p(n), p 4 (n) = 2/6 p(n), p 5 (n) = 1/6 p(n). But they sum to 1, so p(n) = 2/7. The objection to the above is that it fails to establish that lim p(n) exists. So one should arguably add the following. We have (as above) p(n) = 1/6 p(n-1) + 1/6 p(n-2) + 1/6 p(n-3) + 1/6 p(n-4) + 1/6 p(n-5) + 1/6 p(n-6) (*). Let m(n) = min{ p(n-1), p(n-2), p(n-3), p(n-4), p(n-5), p(n-6) }. Then (*) establishes that p(n) ≥ m(n), and so m(n+1) ≥ m(n). Similarly, let M(n) = max{ p(n-1), p(n-2), p(n-3), p(n-4), p(n-5), p(n-6) }. Then (*) shows that p(n) ≤ M(n), so M(n+1) ≤ M(n). Thus m(n) is a monotonic increasing sequence and M(n) is a monotonic decreasing sequence. But m(n) is obviously bounded above by any M(m), and M(n) is bounded below by any m(m). So both sequences converge. Suppose they converged to different limits. So m(n) converges to m and M(n) converges to M with M - m > 36k > 0. Take n sufficiently large that m(n) > m - 6k. At least one of the terms on the rhs of (*) must equal M(n) and the others are at least m(n), so p(n) ≥ 5/6 m(n) + 1/6 M(n) > 5/6 (m - 6k) + 1/6 M > 5/6 m - 5k + 1/6 (m + 36k) = m+ k. But that means that m(n) > m+k for all sufficiently large n. Contradiction. Hence M and m are the same and p(n) must have the same limit. 21st Putnam 1960 © John Scholes [email protected] 15 Feb 2002