answer:Let f(t)= dfrac {t}{t^{2}+9}= dfrac {1}{t+ dfrac {9}{t}}, then f(t) is monotonically increasing for tin(0,2], thus we have f(t)_{max}=f(2)= dfrac {2}{13} Let g(t)= dfrac {t+2}{t^{2}}= dfrac {t+2}{(t+2)^{2}-4(t+2)+4}= dfrac {1}{(t+2)+ dfrac {4}{t+2}-4}, which is monotonically decreasing for (0,2], thus we have g(t)_{min}=g(2)=1 Since the inequality dfrac {t}{t^{2}+9}leqslant aleqslant dfrac {t+2}{t^{2}} holds for all tin(0,2], Therefore, f(t)_{max}leqslant aleqslant g(t)_{min} Thus, dfrac {2}{13}leqslant aleqslant 1 Hence, the correct choice is: boxed{B}. Let f(t)= dfrac {t}{t^{2}+9}= dfrac {1}{t+ dfrac {9}{t}}, g(t)= dfrac {t+2}{t^{2}}= dfrac {t+2}{(t+2)^{2}-4(t+2)+4}= dfrac {1}{(t+2)+ dfrac {4}{t+2}-4}, from the inequality dfrac {t}{t^{2}+9}leqslant aleqslant dfrac {t+2}{t^{2}} holding for all tin(0,2], we have f(t)_{max}leqslant aleqslant g(t)_{min}, by utilizing the monotonicity of the function y=x+ dfrac {k}{x}(k > 0) to solve for the function's extremum, and the transformation between the inequality always holding and the function's extremum.

answer:To determine the probability that a sum of selected numbers on a fair six-sided die equals 4, let's break it down step by step. 1. **Probability that a = 4**: Since the die has six faces, the probability of rolling a number that equals 4 is: [ P(a = 4) = frac{1}{6} ] 2. **Probability that a + b = 4**: We need to determine the ways in which the sum of two numbers equals 4. The valid pairings are: [ (1+3), (2+2), (3+1) ] Each pair has a probability of ( frac{1}{6} times frac{1}{6} ), and there are 3 pairings: [ P(a + b = 4) = 3 times left( frac{1}{6} times frac{1}{6} right) = 3 times frac{1}{36} = frac{3}{36} = frac{1}{12} ] 3. **Probability that a + b + c = 4**: We now consider the ways in which the sum of three numbers equals 4. The valid combinations are: [ (1+1+2), (1+2+1), (2+1+1) ] The probability for each combination is ( frac{1}{6} times frac{1}{6} times frac{1}{6} ), and there are 3 combinations: [ P(a + b + c = 4) = 3 times left( frac{1}{6} times frac{1}{6} times frac{1}{6} right) = 3 times frac{1}{216} = frac{3}{216} = frac{1}{72} ] 4. **Probability that a + b + c + d = 4**: Here, the only way this sum is possible is if each number is 1: [ (1+1+1+1) ] The probability for this combination is: [ P(a + b + c + d = 4) = left( frac{1}{6} times frac{1}{6} times frac{1}{6} times frac{1}{6} right) = frac{1}{1296} ] 5. **Summing up all probabilities**: Importantly, since a die roll outcome is unique, these events are mutually exclusive, meaning no two events can occur at the same time. Therefore, we sum the probabilities: [ P(a = 4 text{ or } a + b = 4 text{ or } a + b + c = 4 text{ or } a + b + c + d = 4) = frac{1}{6} + frac{3}{36} + frac{3}{216} + frac{1}{1296} ] Simplifying each term: [ frac{3}{36} = frac{1}{12} ] [ frac{3}{216} = frac{1}{72} ] Thus, we have: [ frac{1}{6} + frac{1}{12} + frac{1}{72} + frac{1}{1296} ] To combine these fractions, let's have a common denominator, 1296. [ frac{1}{6} = frac{216}{1296} ] [ frac{1}{12} = frac{108}{1296} ] [ frac{1}{72} = frac{18}{1296} ] Adding them up: [ frac{216}{1296} + frac{108}{1296} + frac{18}{1296} + frac{1}{1296} = frac{343}{1296} ] Conclusion: [ boxed{frac{343}{1296}} ]

answer:A is a print statement. B is an input statement. C is a conditional statement. Therefore, the correct answer is boxed{text{B}}.

answer:From the first condition, we have: - If a + b = 6, then g(a) = g(b). From the second condition, we have: - If c + d = 16, then g(c) = g(d). Let's consider any t: [ g(t) = g(6-t) = g(t+16), ] because t + (6-t) = 6 and (6-t) + (t+16) = 16. This indicates that if t is a root of g, then t+16 is also a root, and this pattern continues periodically every 16 units. Since t = 0 is a root, then t = 16n (where n is any integer) is also a root. Moreover, g(6) = g(0) = 0, so t = 16n + 6 is also a root. To count the roots in the interval -2000 leq x leq 2000: - Roots of the form 16n: The smallest n for which 16n geq -2000 is n = -125, and the largest n for which 16n leq 2000 is n = 125. This gives 125 - (-125) + 1 = 251 roots. - Roots of the form 16n + 6: The smallest n for which 16n + 6 geq -2000 is n = -125, and the largest n for which 16n + 6 leq 2000 is n = 124. This gives 124 - (-125) + 1 = 250 roots. Thus, the total number of roots is 251 + 250 = boxed{501}.