answer:1. Let's start by setting up the given problem and define the involved distances and velocities. - At 17:00, let the Audi (A) be at a distance of (x) from the intersection (P) and BMW (B) at a distance of (2x). - Both vehicles travel at a constant speed (v). 2. Considering the times 17:00 and 18:00 and the given distances, we analyze the possible scenarios: - If both A and B had already passed the intersection (P) by 17:00, at 18:00 they would be at distances (x + v) and (2x + v) respectively from P. This does not satisfy the given condition that at 18:00, BMW is twice as far as Audi from the intersection. - Similarly, if both A and B were moving towards P at 18:00, their distances would not match the given condition. Thus, one of them has to cross the intersection between 17:00 and 18:00. 3. Next, we analyze which car crossed the intersection: - Suppose Audi (A) crosses P between 17:00 and 18:00: - At 18:00, Audi (A) would be at a distance (z - x) from P, and BMW (B) would be at a distance (2x - v) (if moving towards P) or (2x + v) from P. - Comparing the distances, for the moving towards case ((2x - v) is twice (z - x)): [ 2x - v = 2(z - x) ] [ 2x - v = 2v - 2x ] [ 3v = 4x ] - For the moving away case ((2x + v) is twice (z - x)): [ 2x + v = 2(z - x) ] [ 2x + v = 2x ] [ v = 4x ] 4. Calculating the times for Audi (A) to be at certain distances: - If (v = 4x): [ frac{x}{v} = frac{1/4x}{x} = frac{1}{4} ] Audi (A) was at the intersection at 17:15 because: [ 17:00 + 0.25 = 17:15 ] - If (v = 3x): [ frac{4x}{3x} = frac{4}{3} ] Audi (A) was at the intersection at 17:45 because: [ 17:00 + 0.75 = 17:45 ] 5. Therefore, based on the assumptions, Audi (A) crossed the intersection either at 17:15 or 17:45. # Conclusion: [ boxed{17:15 text{ or } 17:45} ]

answer:Solution: (1) Let vec{c} cdot vec{d}=0, then (3vec{a}+5vec{b}) cdot (mvec{a}-3vec{b})=0, which gives 3m|vec{a}|^2-15|vec{b}|^2+(5m-9)vec{a} cdot vec{b}=0 Solving this, we get m= frac{29}{14}. Therefore, when m= frac{29}{14}, vec{c} perp vec{d}. (2) Let vec{c}=lambda vec{d}, then 3vec{a}+5vec{b}=lambda(mvec{a}-3vec{b}) This implies (3-lambda m)vec{a}+(5+3lambda)vec{b}=0, Since vec{a} and vec{b} are not collinear, We have begin{cases} 3-lambda m=0 5+3lambda =0 end{cases}, solving this gives begin{cases} lambda =- frac{5}{3} m=- frac{9}{5} end{cases} Therefore, when m=- frac{9}{5}, vec{c} and vec{d} are collinear. Thus, the answers are (1) m= boxed{frac{29}{14}} for vec{c} perp vec{d}, and (2) m= boxed{-frac{9}{5}} for vec{c} and vec{d} being collinear.

answer:Let's denote the number of students taking history as H, the number of students taking statistics as S, and the number of students taking both history and statistics as B. We are given the following information: H = 58 (students taking history) S = 42 (students taking statistics) H ∪ S = 95 (students taking history or statistics or both) We want to find the number of students taking history but not statistics, which can be represented as H - B (since B represents the students taking both). We can use the principle of inclusion-exclusion to find B, which is the number of students taking both history and statistics: B = H + S - (H ∪ S) Plugging in the given values: B = 58 + 42 - 95 B = 100 - 95 B = 5 So, there are 5 students taking both history and statistics. Now, to find the number of students taking history but not statistics, we subtract the number of students taking both from the total number of students taking history: H - B = 58 - 5 H - B = 53 Therefore, boxed{53} students are taking history but not statistics.

answer:(I) Finding the Probability Distribution and Mathematical Expectation of xi 1. **Definition of xi**: Let xi be the number of correctly answered questions by person 甲 (Jia) in a randomly chosen set of 3 questions out of 10. 2. **Possible Values of xi**: xi can take values 0, 1, 2, or 3. 3. **Probability Calculation for xi**: - **P(xi = 0)**: Probability that Jia answers all three questions incorrectly. [ P(xi = 0) = frac{C_4^3}{C_{10}^3} = frac{binom{4}{3}}{binom{10}{3}} = frac{4}{120} = frac{1}{30} ] (Jia can choose any 3 from the 4 questions he answers incorrectly). - **P(xi = 1)**: Probability that Jia answers exactly one question correctly. [ P(xi = 1) = frac{C_6^1 cdot C_4^2}{C_{10}^3} = frac{binom{6}{1} cdot binom{4}{2}}{binom{10}{3}} = frac{6 cdot 6}{120} = frac{36}{120} = frac{3}{10} ] - **P(xi = 2)**: Probability that Jia answers exactly two questions correctly. [ P(xi = 2) = frac{C_6^2 cdot C_4^1}{C_{10}^3} = frac{binom{6}{2} cdot binom{4}{1}}{binom{10}{3}} = frac{15 cdot 4}{120} = frac{60}{120} = frac{1}{2} ] - **P(xi = 3)**: Probability that Jia answers all three questions correctly. [ P(xi = 3) = frac{C_6^3}{C_{10}^3} = frac{binom{6}{3}}{binom{10}{3}} = frac{20}{120} = frac{1}{6} ] 4. **Probability Distribution Table**: [ begin{array}{|c|c|c|c|c|} hline xi & 0 & 1 & 2 & 3 hline P & frac{1}{30} & frac{3}{10} & frac{1}{2} & frac{1}{6} hline end{array} ] 5. **Mathematical Expectation of xi**: Using the definition of expectation E(xi) = sum (text{value}) times (text{probability}), [ E(xi) = 0 cdot frac{1}{30} + 1 cdot frac{3}{10} + 2 cdot frac{1}{2} + 3 cdot frac{1}{6} ] [ E(xi) = 0 + frac{3}{10} + 2 cdot frac{1}{2} + 3 cdot frac{1}{6} ] [ E(xi) = frac{3}{10} + 1 + frac{1}{2} = frac{3}{10} + frac{5}{10} + frac{6}{10} = frac{14}{10} = frac{9}{5} ] 6. **Conclusion for Part (I)**: The probability distribution is given by the table above, and text{E}(xi) = frac{9}{5}. (II) Finding the Probability that at Least One Person Passes the Exam 1. **Defining Events**: - Let (A) be the event that Jia passes the exam. - Let (B) be the event that 乙 (Yi) passes the exam. 2. **Calculations for P(A) and P(B)**: - **P(A)**: Jia passes if he answers at least 2 out of 3 correctly. [ P(A) = frac{C_6^2 cdot C_4^1 + C_6^3}{C_{10}^3} = frac{binom{6}{2} cdot binom{4}{1} + binom{6}{3}}{binom{10}{3}} = frac{15 cdot 4 + 20}{120} = frac{80}{120} = frac{2}{3} ] - **P(B)**: Using a similar approach for Yi who can answer 8 questions correctly, [ P(B) = frac{C_8^2 cdot C_2^1 + C_8^3}{C_{10}^3} = frac{binom{8}{2} cdot binom{2}{1} + binom{8}{3}}{binom{10}{3}} = frac{28 cdot 2 + 56}{120} = frac{112}{120} = frac{14}{15} ] 3. **Calculations for P(bar{A}) and P(bar{B})**: Using the complement rule, [ P(bar{A}) = 1 - P(A) = 1 - frac{2}{3} = frac{1}{3} ] [ P(bar{B}) = 1 - P(B) = 1 - frac{14}{15} = frac{1}{15} ] 4. **Probabilty that both do not pass**: Since (A) and (B) are independent events, [ P(bar{A} cap bar{B}) = P(bar{A}) cdot P(bar{B}) = frac{1}{3} cdot frac{1}{15} = frac{1}{45} ] 5. **Final Probability Calculation**: The probability that at least one person passes is given by [ P(text{at least one passes}) = 1 - P(bar{A} cap bar{B}) = 1 - frac{1}{45} = frac{44}{45} ] 6. **Conclusion for Part (II)**: The probability that at least one person passes the exam is boxed{frac{44}{45}}.