answer:To determine the most reasonable survey plan for understanding the vision situation of junior high school students in a city with 9 districts, let's evaluate each option: A: Testing the vision of junior high school students in a certain middle school in the city would provide data from only one school, which may not be representative of the entire city. B: Testing the vision of all junior high school students in a certain district of the city would give a broader perspective than option A, but it still focuses on a single district, which might not reflect the situation in other districts. C: Testing the vision of all junior high school students in the entire city would certainly give the most comprehensive data, but it might be impractical in terms of resources and time. D: Selecting 5 junior high schools from each district and testing the vision of students from the selected schools strikes a balance between comprehensiveness and practicality. It ensures representation from all parts of the city while keeping the survey manageable. Considering the need for a representative and practical approach, the most reasonable plan is: boxed{D}

answer:The problem requires us to determine the number of additional questions Xiaozhang should answer to maximize the probability of achieving at least 7 out of 10 points, given he already correctly answered the first 6 questions. 1. **Understand the Problem Context**: - Correct answer: +1 point - Incorrect answer: -1 point - No attempt: 0 points - Xiaozhang needs at least 7 points to pass. 2. **Initial Conditions**: - Xiaozhang has already answered 6 questions correctly, earning him 6 points. - There are 4 remaining questions, each with a probability ( frac{1}{2} ) of being answered correctly. 3. **Probability Calculations**: - Let ( X ) be the number of additional questions Xiaozhang answers. - We calculate the probability of reaching at least 7 points with 1, 2, 3, or 4 additional questions. **Case 1**: Answer 1 more question. [ text{Probability of passing} = frac{1}{2} ] - If the 1 question is correct, Xiaozhang has 7 points. - If wrong, Xiaozhang will have 5 points. **Case 2**: Answer 2 more questions. [ text{Probability of passing} = P(2 text{ correct}) + P(1 text{ correct} land 1 text{ incorrect}) ] [ P(2 text{ correct}) = left( frac{1}{2} right)^2 = frac{1}{4} ] [ P(1 text{ correct} land 1 text{ incorrect}) = 2 times left( frac{1}{2} times frac{1}{2} right) = frac{2}{4} = frac{1}{2} ] Total probability: [ P_2 = frac{1}{4} + frac{1}{2} = frac{3}{4} = 0.75 ] **Case 3**: Answer 3 more questions (At least 1 correct out of 3) [ text{Probability of passing} = sum_{k=1}^{3} binom{3}{k} left( frac{1}{2} right)^3 ] [ = binom{3}{1} cdot left(frac{1}{2} right)^3 + binom{3}{2} cdot left(frac{1}{2}right)^3 + binom{3}{3} cdot left(frac{1}{2}right)^3 ] [ = 3 cdot frac{1}{8} + 3 cdot frac{1}{8} + 1 cdot frac{1}{8} = frac{7}{8} ] **Case 4**: Answer 4 more questions [ text{Probability of passing} = sum_{k=1}^{4} binom{4}{k} left(frac{1}{2}right)^4 ] [ = binom{4}{1} cdot left(frac{1}{2}right)^4 + binom{4}{2} cdot left(frac{1}{2}right)^4 + binom{4}{3} cdot left(frac{1}{2}right)^4 + binom{4}{4} cdot left(frac{1}{2}right)^4 ] [ = 4 cdot frac{1}{16} + 6 cdot frac{1}{16} + 4 cdot frac{1}{16} + 1 cdot frac{1}{16} = frac{15}{16} ] 4. **Conclusion**: After evaluating the probabilities for each case, we find that answering 3 or 4 more questions gives the highest chances of passing the test. [ boxed{text{7 or 9 questions}} ]

answer:1. Begin with the expression: frac{sqrt{2}cdotsqrt{4}cdotsqrt{6}cdotsqrt{8}}{sqrt{3}cdotsqrt{5}cdotsqrt{7}cdotsqrt{9}}. 2. Simplify each square root and cancel when possible: - sqrt{4} = 2 and sqrt{6} = sqrt{2}cdotsqrt{3}, - sqrt{8} = 2sqrt{2} and sqrt{9} = 3. 3. Substituting the simplified values, the expression becomes: - frac{sqrt{2}cdot2cdot(sqrt{2}cdotsqrt{3})cdot(2sqrt{2})}{sqrt{3}cdotsqrt{5}cdotsqrt{7}cdot3}. 4. Further simplification and cancellation: - Numerator: sqrt{2}cdot2cdotsqrt{2}cdotsqrt{3}cdot2cdotsqrt{2} = 4cdot2cdot2 = 16. - Denominator stays as sqrt{5}cdotsqrt{7}cdot3cdotsqrt{3} = 3sqrt{15}cdotsqrt{7} = 3sqrt{105}. - The fraction is now frac{16}{3sqrt{105}}. 5. To rationalize the denominator, multiply top and bottom by sqrt{105}: - Numerator: 16cdotsqrt{105}, - Denominator: 3cdot105 = 315. - The fraction simplifies to frac{16sqrt{105}}{315}. Conclusion: Final simplified and rationalized form: boxed{frac{16sqrt{105}}{315}}.

answer:1. **Define Variables:** Let the total distance from Rita's house to her sister's house be x miles. 2. **Calculate Speeds:** Rita's usual speed is frac{x}{200} miles per minute, as under normal conditions it takes her 200 minutes to cover x miles. 3. **Distance and Speed Adjustments:** After driving frac{1}{4} of the distance, Rita covers frac{x}{4} miles before the rainstorm. The remaining distance is frac{3x}{4} miles. Since the speed reduction is 15 miles per hour, and 1 hour = 60 minutes, the reduction is equivalent to frac{15}{60} = frac{1}{4} miles per minute. Hence, her new speed during the rainstorm is frac{x}{200} - frac{1}{4} miles per minute. 4. **Set Up the Equation:** The total time taken for the trip is given as 300 minutes. We can set up the equation based on time taken before and during the rainstorm: [ frac{frac{x}{4}}{frac{x}{200}} + frac{frac{3x}{4}}{frac{x}{200} - frac{1}{4}} = 300 ] 5. **Simplify and Solve the Equation:** Simplify each term: - Time before the storm: frac{frac{x}{4}}{frac{x}{200}} = frac{x}{4} cdot frac{200}{x} = 50 minutes. - Time during the storm: frac{frac{3x}{4}}{frac{x}{200} - frac{1}{4}} = frac{frac{3x}{4}}{frac{x - 50}{200}} = frac{3x}{4} cdot frac{200}{x - 50}. Substitute these into the total time equation: [ 50 + frac{3x}{4} cdot frac{200}{x - 50} = 300 ] Simplify and solve for x: [ frac{3x cdot 200}{4(x - 50)} = 250 quad text{(since } 300 - 50 = 250text{)} ] [ frac{600x}{4x - 200} = 250 ] [ 600x = 1000x - 50000 ] [ 400x = 50000 ] [ x = frac{50000}{400} = 125 ] 6. **Conclusion:** The total distance from Rita's house to her sister's house is 125 miles. The final answer is boxed{C}.