answer:Each club has a frac{1}{3} chance of being selected. Let n be the number of students in the chosen club. The total ways to choose four students from n members are given by binom{n}{4}. The ways to choose exactly two co-presidents among three and two other members from the remaining n-3 are binom{3}{2} times binom{n-3}{2}. For the clubs with 6, 9, and 10 students respectively: 1. For the club with 6 students: frac{binom{3}{2} binom{3}{2}}{binom{6}{4}} = frac{3 times 3}{15} = frac{9}{15} = frac{3}{5}. 2. For the club with 9 students: frac{binom{3}{2} binom{6}{2}}{binom{9}{4}} = frac{3 times 15}{126} = frac{45}{126} = frac{15}{42} = frac{5}{14}. 3. For the club with 10 students: frac{binom{3}{2} binom{7}{2}}{binom{10}{4}} = frac{3 times 21}{210} = frac{63}{210} = frac{9}{30} = frac{3}{10}. Thus, the total probability is: frac{1}{3} left( frac{3}{5} + frac{5}{14} + frac{3}{10} right) = frac{1}{3} left( frac{42}{70} + frac{25}{70} + frac{21}{70} right) = frac{1}{3} times frac{88}{70} = frac{88}{210} = frac{44}{105} . The final answer is thus boxed{frac{44}{105}}.

answer:To determine the value of ( m^2 ) such that the line ( y = mx + 1 ) intersects the ellipse ( x^2 + 4y^2 = 1 ) at exactly one point, we need to ensure that the quadratic equation resulting from substituting ( y ) from the line equation into the ellipse equation has exactly one solution. This implies its discriminant should be zero. 1. **Substitute** the line equation ( y = mx + 1 ) into the ellipse equation ( x^2 + 4y^2 = 1 ). [ x^2 + 4(mx + 1)^2 = 1 ] 2. **Expand** the squared term: [ x^2 + 4(mx + 1)^2 = x^2 + 4(m^2x^2 + 2mx + 1) ] Which simplifies to: [ x^2 + 4(m^2x^2 + 2mx + 1) = x^2 + 4m^2x^2 + 8mx + 4 ] 3. **Combine** like terms: [ (1 + 4m^2)x^2 + 8mx + 4 = 1 ] 4. **Rearrange** the equation into standard quadratic form ( ax^2 + bx + c = 0 ): [ (1 + 4m^2)x^2 + 8mx + 3 = 0 ] 5. **Identify** the coefficients (a), (b), and (c): [ a = 1 + 4m^2, quad b = 8m, quad c = 3 ] 6. **Set the discriminant** ( Delta ) to zero for the quadratic equation to have exactly one solution: [ Delta = b^2 - 4ac ] 7. **Substitute** the coefficients: [ Delta = (8m)^2 - 4(1 + 4m^2) cdot 3 ] Simplifying: [ Delta = 64m^2 - 4(1 + 4m^2) cdot 3 ] [ Delta = 64m^2 - 12(1 + 4m^2) ] [ Delta = 64m^2 - 12 - 48m^2 ] [ Delta = 16m^2 - 12 ] 8. **Set the discriminant to zero** and solve for (m^2): [ 16m^2 - 12 = 0 ] [ 16m^2 = 12 ] [ m^2 = frac{12}{16} ] [ m^2 = frac{3}{4} ] 9. **Conclusion**: Therefore, the value of ( m^2 ) is [ boxed{frac{3}{4}} ] Therefore, the correct option is (C).

answer:Given vectors overrightarrow{a}=(-2,0), overrightarrow{b}=(1,2), overrightarrow{c}=(1,sqrt{3}), we will evaluate each option step by step. **Option A: overrightarrow{a} parallel overrightarrow{b}** To check if two vectors are parallel, we can compare their direction ratios. For overrightarrow{a} and overrightarrow{b}, the direction ratios are (-2,0) and (1,2) respectively. If overrightarrow{a} parallel overrightarrow{b}, then -2/1 should equal 0/2. However, -2 neq 0, thus overrightarrow{a} and overrightarrow{b} are not parallel. Therefore, option A is incorrect. **Option B: overrightarrow{a} cdot overrightarrow{b}=2** The dot product of overrightarrow{a} and overrightarrow{b} is calculated as follows: [ overrightarrow{a} cdot overrightarrow{b} = (-2) cdot 1 + 0 cdot 2 = -2 ] Since overrightarrow{a} cdot overrightarrow{b} = -2 neq 2, option B is incorrect. **Option C: |overrightarrow{b}|=2|overrightarrow{c}|** First, calculate the magnitude of overrightarrow{b}: [ |overrightarrow{b}| = sqrt{1^2 + 2^2} = sqrt{5} ] Then, calculate the magnitude of overrightarrow{c}: [ |overrightarrow{c}| = sqrt{1^2 + (sqrt{3})^2} = sqrt{4} = 2 ] Comparing |overrightarrow{b}| and 2|overrightarrow{c}|: [ sqrt{5} neq 2 times 2 ] Thus, |overrightarrow{b}| neq 2|overrightarrow{c}|, making option C incorrect. **Option D: The angle between overrightarrow{a} and overrightarrow{c} is 120^{circ}** To find the angle between overrightarrow{a} and overrightarrow{c}, we use the dot product formula: [ costheta = frac{overrightarrow{a} cdot overrightarrow{c}}{|overrightarrow{a}||overrightarrow{c}|} ] Calculating overrightarrow{a} cdot overrightarrow{c}: [ overrightarrow{a} cdot overrightarrow{c} = (-2) cdot 1 + 0 cdot sqrt{3} = -2 ] The magnitudes are |overrightarrow{a}| = sqrt{(-2)^2 + 0^2} = 2 and |overrightarrow{c}| = 2 as calculated before. Thus, [ costheta = frac{-2}{2 times 2} = -frac{1}{2} ] Given costheta = -frac{1}{2} and knowing that theta in [0^{circ},180^{circ}], we conclude theta = 120^{circ}. Therefore, the angle between overrightarrow{a} and overrightarrow{c} is indeed 120^{circ}, making option D correct. Therefore, the correct option is boxed{D}.

answer:1. **Understanding the Problem**: - We are given point ( M ) on the median ( AD ) of triangle ( ABC ), where ( D ) is the midpoint of ( BC ). - Line ( BM ) intersects side ( AC ) at point ( N ). - Line ( AB ) is a tangent to the circumcircle of ( triangle NBC ) at point ( B ). - We are provided ( frac{BC}{BN} = lambda ), and we need to find ( frac{BM}{MN} ) in terms of ( lambda ). 2. **Using Menelaus' Theorem**: - Applying Menelaus' theorem on ( triangle BCN ) with transversal ( AMD ), we have: [ frac{BM}{MN} cdot frac{NA}{AC} cdot frac{CD}{DB} = 1. ] - Since ( D ) is the midpoint of ( BC ), we know ( CD = DB ). Thus, ( frac{CD}{DB} = 1 ), simplifying our equation to: [ frac{BM}{MN} cdot frac{NA}{AC} cdot 1 = 1. ] [ frac{BM}{MN} = frac{AC}{NA}. ] 3. **Utilizing Similar Triangles**: - Because ( AB ) is tangent to the circumcircle of ( triangle NBC ), angle ( angle ABN = angle ACB ). Consequently, triangles ( triangle ABN ) and ( triangle ACB ) are similar: [ triangle ABN sim triangle ACB. ] - From the similarity, we get the following proportionality: [ frac{AB}{AN} = frac{AC}{AB} = frac{CB}{BN}. ] Hence, we have: [ frac{AB}{AN} = frac{CB}{BN}. ] 4. **Solving for (frac{AC}{AN})**: - Squaring both sides, we get: [ frac{AC^2}{AN^2} = left(frac{CB}{BN}right)^2. ] - Thus, [ frac{AC}{AN} = left(frac{CB}{BN}right)^2. ] Given ( frac{BC}{BN} = lambda ), we have: [ frac{AC}{AN} = lambda^2. ] 5. **Conclusion**: - Substituting into our earlier result from Menelaus' theorem: [ frac{BM}{MN} = frac{AC}{AN} = lambda^2. ] [ boxed{lambda^2} ]