💡To determine the probability of selecting a green ball, we need to consider the probability of selecting each container and then the probability of selecting a green ball from that container.First, calculate the probability of selecting each container:- Probability of selecting any container = (frac{1}{4}) (since there are four containers).Next, compute the probability of picking a green ball from each container:- **Container I**: [ frac{6}{16} = frac{3}{8} quad text{(since there are 16 balls in total in Container I)} ]- **Container II**: [ frac{5}{8} quad text{(since there are 8 balls in total in Container II)} ]- **Container III**: [ frac{6}{8} = frac{3}{4} quad text{(since there are 8 balls in total in Container III)} ]- **Container IV**: [ frac{4}{8} = frac{1}{2} quad text{(since there are 8 balls in total in Container IV)} ]Finally, calculate the total probability of picking a green ball by summing the products of the probabilities of selecting each container and the probability of picking a green ball from that container:[frac{1}{4} cdot frac{3}{8} + frac{1}{4} cdot frac{5}{8} + frac{1}{4} cdot frac{3}{4} + frac{1}{4} cdot frac{1}{2} = frac{3}{32} + frac{5}{32} + frac{6}{32} + frac{4}{32} = frac{18}{32} = frac{9}{16}]

💡First, I need to determine the total weight of the original group of 8 people. Since the average weight is ( W ) kg, the total weight is ( 8W ) kg.When two people weighing 68 kg each leave the group, the total weight decreases by ( 68 times 2 = 136 ) kg. Therefore, the new total weight of the remaining 6 people is ( 8W - 136 ) kg.Two new people join the group, and the average weight increases by 5.5 kg. This means the new average weight is ( W + 5.5 ) kg. The total weight of the group with the two new people is ( 8(W + 5.5) ) kg.Setting up the equation for the total weight after the new people join:[8W - 136 + N = 8(W + 5.5)]Solving for ( N ), the combined weight of the two new people:[N = 8(W + 5.5) - 8W + 136 = 180 text{ kg}]Given that the sum of the new people's weights is no more than 180 kg, we have:[x + y leq 180]Additionally, since the average weight increased, each new person must weigh more than the original average ( W ):[x > W quad text{and} quad y > W]Without the specific value of ( W ), the possible weights of the two new people are any pair of weights greater than ( W ) that add up to 180 kg or less.

💡First, I recognize that the dot product of two vectors is related to the cosine of the angle between them. The formula for the dot product is:[overrightarrow{a} cdot overrightarrow{b} = |overrightarrow{a}| |overrightarrow{b}| cos(theta)]Given the values:- (overrightarrow{a} cdot overrightarrow{b} = -1)- (|overrightarrow{a}| = 2)- (|overrightarrow{b}| = 1)I substitute these into the formula to solve for (cos(theta)):[-1 = 2 times 1 times cos(theta)][cos(theta) = -frac{1}{2}]Next, I determine the angle (theta) whose cosine is (-frac{1}{2}). From trigonometric knowledge, I know that:[cosleft(frac{2pi}{3}right) = -frac{1}{2}]Therefore, the angle between vectors (overrightarrow{a}) and (overrightarrow{b}) is (frac{2pi}{3}).

💡Okay, so I have this problem about a rectangle ABCD with all its vertices on a sphere centered at O. The sides AB and BC are given as √3 and 3 respectively, and the volume of the pyramid O-ABCD is 4√3. I need to find the surface area of the sphere O.First, I should visualize the problem. There's a sphere with center O, and a rectangle ABCD lying on its surface. Since all four vertices of the rectangle are on the sphere, the rectangle is inscribed in the sphere. The pyramid formed by connecting the center O to the rectangle ABCD has a volume of 4√3.I remember that the volume of a pyramid is given by (1/3) * base area * height. In this case, the base is the rectangle ABCD, and the height is the distance from the center O to the plane containing ABCD.So, maybe I can start by finding the area of the rectangle ABCD. Since it's a rectangle, the area is just the product of its length and width. AB is √3 and BC is 3, so the area should be √3 * 3, which is 3√3.Next, I know the volume of the pyramid is 4√3. Using the formula for the volume of a pyramid:Volume = (1/3) * base area * heightPlugging in the values I have:4√3 = (1/3) * 3√3 * heightSimplifying the right side:(1/3) * 3√3 = √3, so:4√3 = √3 * heightDividing both sides by √3:height = 4So, the distance from the center O to the plane ABCD is 4 units.Now, I need to relate this distance to the radius of the sphere. Since all the vertices of the rectangle are on the sphere, the distance from O to each vertex is equal to the radius R of the sphere.Let me consider the rectangle ABCD. Since it's a rectangle, its diagonals are equal and bisect each other. The length of the diagonal can be found using the Pythagorean theorem:Diagonal AC = √(AB² + BC²) = √((√3)² + 3²) = √(3 + 9) = √12 = 2√3So, the diagonal of the rectangle is 2√3. The midpoint of the diagonal is equidistant from all four vertices, and this midpoint is also the point where the perpendicular from O to the plane ABCD meets the plane.Let me denote the midpoint as M. So, OM is the distance from O to the plane ABCD, which we found to be 4. The distance from M to any vertex, say A, is half the diagonal, which is √3.Now, considering the right triangle OMA, where OA is the radius R of the sphere, OM is 4, and MA is √3. Using the Pythagorean theorem:OA² = OM² + MA²So,R² = 4² + (√3)² = 16 + 3 = 19Therefore, R = √19Now, the surface area of the sphere is given by 4πR². Plugging in R² = 19:Surface Area = 4π * 19 = 76πSo, the surface area of the sphere is 76π.Wait, let me double-check my steps to make sure I didn't make any mistakes.1. Calculated the area of the rectangle: √3 * 3 = 3√3. That seems correct.2. Used the volume formula for the pyramid: (1/3)*base area*height = 4√3. Plugged in 3√3 for the base area and solved for height, which gave me 4. That seems right.3. Found the diagonal of the rectangle using Pythagoras: √( (√3)^2 + 3^2 ) = √(3 + 9) = √12 = 2√3. Correct.4. The midpoint M is halfway along the diagonal, so MA = √3. Then, considering triangle OMA with sides OM=4, MA=√3, and OA=R. Applied Pythagoras: R² = 16 + 3 = 19. So, R=√19. That seems correct.5. Calculated surface area as 4πR² = 4π*19 = 76π. Yes, that's correct.I think all the steps are logically sound and the calculations are accurate. So, the surface area of the sphere should indeed be 76π.

💡Okay, so I have this problem where there are 30 sixth graders, each with a pen, a pencil, and a ruler. After an olympiad, 26 students lost their pen, 23 lost their ruler, and 21 lost their pencil. I need to find the smallest possible number of students who lost all three items. Hmm, that sounds like a problem involving sets and maybe some overlapping.Let me think. I remember something about the principle of inclusion-exclusion, which is used to calculate the number of elements in the union of multiple sets. Maybe that can help here. So, if I denote the sets of students who lost their pen, ruler, and pencil as A, B, and C respectively, then:- |A| = 26- |B| = 23- |C| = 21And the total number of students is 30, so |A ∪ B ∪ C| ≤ 30.The inclusion-exclusion principle for three sets says:|A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |B ∩ C| - |A ∩ C| + |A ∩ B ∩ C|I need to find the minimum value of |A ∩ B ∩ C|, which is the number of students who lost all three items.Let me plug in the numbers:|A ∪ B ∪ C| = 26 + 23 + 21 - |A ∩ B| - |B ∩ C| - |A ∩ C| + |A ∩ B ∩ C|Simplifying that:|A ∪ B ∪ C| = 70 - (|A ∩ B| + |B ∩ C| + |A ∩ C|) + |A ∩ B ∩ C|But we know that |A ∪ B ∪ C| can't be more than 30 because there are only 30 students. So:70 - (|A ∩ B| + |B ∩ C| + |A ∩ C|) + |A ∩ B ∩ C| ≤ 30Let me rearrange that:(|A ∩ B| + |B ∩ C| + |A ∩ C|) - |A ∩ B ∩ C| ≥ 40Hmm, so the sum of the pairwise intersections minus the triple intersection is at least 40. I need to find the minimum value of |A ∩ B ∩ C|.I think to minimize |A ∩ B ∩ C|, I need to maximize the sum of the pairwise intersections. But wait, the pairwise intersections can't be more than the sizes of the individual sets. For example, |A ∩ B| can't be more than |A| or |B|, which are 26 and 23 respectively. So the maximum possible |A ∩ B| is 23.Similarly, |B ∩ C| can't be more than 21, and |A ∩ C| can't be more than 21. So the maximum sum of the pairwise intersections would be 23 + 21 + 21 = 65.But plugging that back into the equation:65 - |A ∩ B ∩ C| ≥ 40So:|A ∩ B ∩ C| ≥ 65 - 40 = 25Wait, that can't be right because 25 is more than the total number of students who lost any item. Maybe I made a mistake.Let me think again. Maybe I should consider that the maximum possible sum of pairwise intersections is constrained by the total number of students. Since there are only 30 students, the maximum number of students in any intersection can't exceed 30.But I'm getting confused. Maybe I should approach it differently. Let's consider the maximum number of students who could have lost only two items. If I want to minimize the number who lost all three, I should maximize the number who lost exactly two.The total number of losses is 26 + 23 + 21 = 70. But since there are only 30 students, each student can account for multiple losses. If x students lost all three, then the total number of losses can be expressed as:Total losses = (Number of students who lost exactly one item) + 2*(Number of students who lost exactly two items) + 3*(Number of students who lost all three items)But I don't know how many lost exactly one or exactly two. Maybe I can set up an equation.Let me denote:- a = number of students who lost only a pen- b = number of students who lost only a ruler- c = number of students who lost only a pencil- ab = number of students who lost a pen and ruler but not pencil- ac = number of students who lost a pen and pencil but not ruler- bc = number of students who lost a ruler and pencil but not pen- abc = number of students who lost all threeThen:a + ab + ac + abc = 26 (total pen losses)b + ab + bc + abc = 23 (total ruler losses)c + ac + bc + abc = 21 (total pencil losses)Also, the total number of students is:a + b + c + ab + ac + bc + abc = 30I need to find the minimum value of abc.Let me try to express everything in terms of abc.From the first equation:a = 26 - ab - ac - abcFrom the second equation:b = 23 - ab - bc - abcFrom the third equation:c = 21 - ac - bc - abcSubstituting these into the total number of students:(26 - ab - ac - abc) + (23 - ab - bc - abc) + (21 - ac - bc - abc) + ab + ac + bc + abc = 30Simplify:26 + 23 + 21 - ab - ac - abc - ab - bc - abc - ac - bc - abc + ab + ac + bc + abc = 30Combine like terms:70 - 2ab - 2ac - 2bc - 2abc + ab + ac + bc + abc = 30Simplify further:70 - ab - ac - bc - abc = 30So:ab + ac + bc + abc = 40But ab, ac, bc are all non-negative integers. To minimize abc, I need to maximize ab + ac + bc.What's the maximum possible value of ab + ac + bc?Each of ab, ac, bc can't exceed the sizes of the respective pairs.For example, ab ≤ min(|A|, |B|) = min(26,23) =23Similarly, ac ≤ min(26,21)=21bc ≤ min(23,21)=21So maximum ab + ac + bc =23+21+21=65But ab + ac + bc + abc=40, so if ab + ac + bc=65, then abc=40-65=-25, which is impossible.So that approach isn't working. Maybe I need to think differently.Wait, perhaps I should consider that the sum ab + ac + bc can't exceed the total number of students times 2, but I'm not sure.Alternatively, maybe I can use the formula for the minimum overlap.I recall that for three sets, the minimum number of elements common to all three is given by:|A ∩ B ∩ C| ≥ |A| + |B| + |C| - 2NWhere N is the total number of elements.So plugging in the numbers:|A ∩ B ∩ C| ≥ 26 + 23 + 21 - 2*30 = 70 - 60 =10So the minimum number of students who lost all three items is 10.Wait, that seems to make sense. Let me check.If 10 students lost all three, then the number of students who lost only pen and ruler would be 26 -10=16, but wait, no, that's not quite right.Actually, if 10 students lost all three, then the number of students who lost pen and ruler but not pencil would be |A ∩ B| - |A ∩ B ∩ C|.Similarly for the others.But I think the formula I used gives the lower bound, so the minimum is 10.Yeah, I think that's the answer.

💡Okay, let's try to solve this problem step by step. So, we have two propositions, p and q. First, proposition p is: |m + 1| ≤ 2. That means the absolute value of (m + 1) is less than or equal to 2. I remember that absolute value inequalities can be rewritten without the absolute value by considering two cases. So, |m + 1| ≤ 2 translates to -2 ≤ m + 1 ≤ 2. If I subtract 1 from all parts of the inequality, that should give me the range for m. Let me do that: subtracting 1 from -2 gives -3, subtracting 1 from m + 1 gives m, and subtracting 1 from 2 gives 1. So, p is true when m is between -3 and 1, inclusive. So, p: -3 ≤ m ≤ 1.Next, proposition q is: The equation x² - 2mx + 1 = 0 has real roots. I remember that for a quadratic equation ax² + bx + c = 0 to have real roots, the discriminant must be greater than or equal to zero. The discriminant D is b² - 4ac. In this equation, a is 1, b is -2m, and c is 1. So, D = (-2m)² - 4*1*1 = 4m² - 4. For the equation to have real roots, D ≥ 0. So, 4m² - 4 ≥ 0. Let me solve this inequality.First, factor out the 4: 4(m² - 1) ≥ 0. Then, m² - 1 ≥ 0. That's a difference of squares, so it factors to (m - 1)(m + 1) ≥ 0. To solve this inequality, I can consider the critical points where the expression equals zero, which are m = 1 and m = -1. These points divide the number line into three intervals: m < -1, -1 < m < 1, and m > 1. I'll test each interval.For m < -1, say m = -2: (-2 - 1)(-2 + 1) = (-3)(-1) = 3, which is positive. So, the inequality holds here.For -1 < m < 1, say m = 0: (0 - 1)(0 + 1) = (-1)(1) = -1, which is negative. So, the inequality doesn't hold here.For m > 1, say m = 2: (2 - 1)(2 + 1) = (1)(3) = 3, which is positive. So, the inequality holds here.Therefore, the solution to m² - 1 ≥ 0 is m ≤ -1 or m ≥ 1. So, q is true when m is less than or equal to -1 or greater than or equal to 1.Now, the problem states that ¬p is false. Since ¬p is false, that means p is true. So, p is true, which we already determined means -3 ≤ m ≤ 1.Additionally, it says that p ∧ q is false. The logical "and" (p ∧ q) is false if at least one of p or q is false. But since we already know p is true, the only way for p ∧ q to be false is if q is false. So, q must be false.We found earlier that q is true when m ≤ -1 or m ≥ 1. Therefore, q is false when -1 < m < 1.So, we have p is true: -3 ≤ m ≤ 1, and q is false: -1 < m < 1. To find the range of m that satisfies both conditions, we need the intersection of these two intervals.The intersection of -3 ≤ m ≤ 1 and -1 < m < 1 is -1 < m < 1. Because -1 < m < 1 is entirely within -3 ≤ m ≤ 1.Wait, let me double-check that. If p is true, m is between -3 and 1, inclusive. But q is false only when m is between -1 and 1, not including -1 and 1. So, the overlap is indeed -1 < m < 1.Therefore, the range of real numbers m is -1 < m < 1.I think that makes sense. Let me recap: p is true, so m is between -3 and 1. q is false, so m is between -1 and 1. The overlap is m between -1 and 1. So, the final answer should be -1 < m < 1.**Final Answer**The range of real numbers ( m ) is boxed{(-1, 1)}.