💡Alright, so I've got this multiple-choice question here, and I need to figure out which statement is correct. Let's go through each option one by one and see if I can understand them properly.Starting with option A: "If a person shoots 10 times and hits the target 7 times, then the probability of hitting the target is 0.7." Hmm, okay. So, this person has a success rate of 7 out of 10, which is 70%, or 0.7. At first glance, it seems like the probability is just the ratio of successful hits to total attempts. But wait, is that really the case? I remember that probability is more about the likelihood of an event happening in the future, not just the ratio of past successes. So, if someone hits 7 out of 10 times, that gives us a frequency of 0.7, but does that mean the probability is 0.7? I'm not entirely sure. Maybe it's an estimate of the probability, but it's not necessarily the exact probability. So, I think A might not be entirely correct because it's conflating frequency with probability.Moving on to option B: "A student conducts a coin-tossing experiment 6 times, and there will definitely be 3 times 'heads up'." Okay, so if you flip a coin 6 times, is it guaranteed to get exactly 3 heads? That doesn't sound right. I remember that each coin flip is an independent event with a 50% chance of heads and 50% chance of tails. Just because you flip it 6 times doesn't mean you'll get exactly half heads and half tails. It's possible to get more or fewer heads. For example, you could get all tails or all heads, though those are less likely. So, saying it will definitely be 3 heads is incorrect. It's a random event, and the number of heads can vary. So, B is definitely wrong.Next, option C: "A welfare lottery issued in a certain place has a return rate of 47%, and if someone spends 100 yuan on lottery tickets, there will definitely be a return of 47 yuan." Hmm, okay, so the return rate is 47%, meaning that on average, for every yuan spent, you get back 0.47 yuan. But does that mean if you spend 100 yuan, you'll definitely get back 47 yuan? That doesn't sound right either. I think the return rate is an average over all tickets sold, not a guarantee for each individual purchase. So, someone could spend 100 yuan and maybe win more or less, but it's not a certainty. It's probabilistic, so you can't guarantee a specific return. Therefore, C is also incorrect.Finally, option D: "An event with a probability of 1 is not necessarily a certain event." Wait, that seems a bit confusing. I thought that if an event has a probability of 1, it means it's certain to happen. But the statement is saying it's not necessarily certain. How is that possible? Let me think. Maybe in some contexts, like continuous probability distributions, an event can have probability 1 but still not be certain? For example, if you pick a random number between 0 and 1, the probability of picking any specific number is 0, but the probability of picking a number in the interval is 1. So, in that case, even though the probability is 1, it's not certain that you'll pick a specific number. Wait, no, that's not exactly the same. Maybe it's more about measure theory. In continuous probability spaces, events with probability 1 are almost sure, but not certain in the sense that they exclude a set of measure zero. So, in some cases, an event can have probability 1 without being absolutely certain. Therefore, D might actually be correct.Let me recap:- A: Talks about frequency as probability, which isn't exactly accurate.- B: Claims a definite outcome in a random experiment, which is wrong.- C: Mistakenly interprets return rate as a guaranteed return, which isn't the case.- D: Correctly points out that probability 1 doesn't always mean certainty, especially in continuous probability spaces.So, after considering each option, I think D is the correct answer.

💡Alright, so I've got this problem here about the per capita disposable income in China from 2020 to 2022. It says that in 2020, the income was 32,000 yuan, and in 2022, it was 37,000 yuan. They want me to find the average annual growth rate, denoted by x, and set up an equation based on that information.Okay, first, I need to understand what an average annual growth rate means. I think it's like the percentage increase each year that, when compounded over the number of years, gives the total growth from the start to the end. So, if I have an initial amount and it grows by x each year, after two years, it should be the initial amount multiplied by (1 + x) squared, right?Let me write that down. If the initial income is 32,000 yuan, and after two years it's 37,000 yuan, then:37,000 = 32,000 * (1 + x)^2Hmm, that seems logical. But wait, the options given are in thousands, like 3.2 and 3.7. Maybe they simplified the numbers by dividing by 1,000 to make it easier. So, 32,000 becomes 3.2, and 37,000 becomes 3.7. That would make the equation:3.7 = 3.2 * (1 + x)^2Looking at the options, that's option B: 3.2(1 + x)^2 = 3.7.But let me double-check. Is it possible that the growth rate is decreasing instead of increasing? That would mean using (1 - x) instead of (1 + x). But since the income went up from 32,000 to 37,000, it's definitely increasing, so x should be positive. Therefore, (1 + x) makes sense.Wait, let me think about the formula again. The general formula for compound growth is:Final amount = Initial amount * (1 + growth rate)^number of periodsIn this case, the number of periods is 2 years, so it's squared. Yeah, that matches what I wrote earlier.Is there any chance that the equation should be set up differently? Like, maybe starting from 3.7 and going back to 3.2? That would be if we were finding a decay rate, but since it's growth, starting from the lower number and growing to the higher number makes more sense.Also, the options C and D start with 3.7, which is the higher number, and subtract or add x. But since we're going from 3.2 to 3.7, it's a growth, so starting with 3.2 and multiplying by (1 + x)^2 is correct.I think I'm confident now that option B is the right equation to represent the average annual growth rate over the two years.

💡Okay, so I have this problem about graph coloring. It says that if a graph G with n vertices is uniquely 3-colorable and n is at least 3, then G must have at least 2n - 3 edges. Hmm, interesting. I need to prove this. Let me try to break it down.First, let me recall what a k-coloring is. It's assigning one of k colors to each vertex so that no two adjacent vertices share the same color. Unique 3-colorable means that there's only one way to color the graph with three colors, right? So, no two different colorings exist where some vertices have different colors in one coloring compared to another. That seems pretty restrictive.So, G is uniquely 3-colorable, which implies that the coloring is fixed in a way. Maybe the structure of the graph is such that it doesn't allow for any flexibility in coloring. Maybe it's a tree or something similar? But trees are 2-colorable, so that's not it. Maybe it's a more connected graph.I need to show that such a graph has at least 2n - 3 edges. Let me think about the number of edges in a graph. For a connected graph, the minimum number of edges is n - 1, which is a tree. But here, we need something more, 2n - 3. So, it's more connected than a tree but not necessarily a complete graph.Maybe I can use some properties of 3-colorable graphs. If a graph is 3-colorable, it doesn't contain any odd-length cycles, right? Or wait, no, that's for bipartite graphs, which are 2-colorable. For 3-colorable graphs, they can have cycles of various lengths, but not necessarily all even.Wait, maybe I should think about the structure of uniquely 3-colorable graphs. If a graph is uniquely 3-colorable, then it must have a unique partition into three color classes. So, no matter how you try to color it, the color classes remain the same. That suggests that the graph has a lot of constraints on how vertices can be connected.Perhaps I can model this by considering the color classes. Let's say the three color classes have sizes n1, n2, and n3, such that n1 + n2 + n3 = n. Since the coloring is unique, the way these color classes are connected must be very specific.Maybe I can consider the bipartite graphs between each pair of color classes. For example, between color 1 and color 2, color 2 and color 3, and color 3 and color 1. If these bipartite graphs are connected, then the number of edges between each pair would be at least the number of vertices in each color class minus one, because a connected bipartite graph has at least that many edges.Wait, let me think about that. If I have a bipartite graph with partitions of size a and b, the minimum number of edges to make it connected is a + b - 1. So, if between color 1 and color 2, the number of edges is at least n1 + n2 - 1, similarly for the other pairs.So, adding up these minimum edges:Edges between color 1 and color 2: n1 + n2 - 1Edges between color 2 and color 3: n2 + n3 - 1Edges between color 3 and color 1: n3 + n1 - 1Total edges: (n1 + n2 - 1) + (n2 + n3 - 1) + (n3 + n1 - 1) = 2(n1 + n2 + n3) - 3 = 2n - 3Ah, so that gives us the required number of edges. But wait, does this hold? I need to make sure that each bipartite subgraph is connected. Is that necessarily true for a uniquely 3-colorable graph?Suppose, for contradiction, that one of these bipartite subgraphs is disconnected. Then, we could permute the colors within one of the disconnected components, right? That would give a different coloring, which would contradict the uniqueness of the coloring. So, the bipartite subgraphs must be connected.Therefore, each bipartite subgraph between two color classes must be connected, which means they have at least n1 + n2 - 1, n2 + n3 - 1, and n3 + n1 - 1 edges respectively. Adding these up gives us 2n - 3 edges in total.So, that seems to work. But let me double-check. If the bipartite subgraphs are connected, then the total number of edges is at least 2n - 3. And since the graph is uniquely 3-colorable, the bipartite subgraphs can't be disconnected, otherwise, we could have different colorings.Therefore, the graph must have at least 2n - 3 edges. I think that makes sense.Wait, but is there a case where the bipartite subgraphs could have more edges? Of course, but the problem is asking for the minimum number of edges, so 2n - 3 is the lower bound.Let me think about an example. If n = 3, then 2n - 3 = 3. So, a triangle graph, which is uniquely 3-colorable, has exactly 3 edges. That works.If n = 4, then 2n - 3 = 5. Let's see, a complete graph K4 is 4-colorable, but not 3-colorable. So, maybe a graph like a triangle with an additional vertex connected to all three vertices of the triangle. That would be 3 + 3 = 6 edges, which is more than 5. Hmm, but is it uniquely 3-colorable?Wait, in that case, the additional vertex can be colored with the third color, but since it's connected to all three vertices, it can only take one color. So, maybe it's uniquely 3-colorable. But the number of edges is 6, which is more than 5. So, maybe the minimal case is when the graph is a tree plus some edges.Wait, no, trees are 2-colorable. So, maybe the minimal uniquely 3-colorable graph is something like a triangle with additional edges. But I need to think more carefully.Alternatively, maybe the minimal case is when the graph is a complete bipartite graph K2,2, which is bipartite and hence 2-colorable, but not 3-colorable. So, that doesn't help.Wait, maybe I should think about the structure of uniquely 3-colorable graphs. They must have a unique partition into three independent sets, and the connections between these sets must be such that no two vertices in the same set are connected, and the connections between sets are such that the coloring is forced.So, perhaps the graph is a complete tripartite graph with each partition having at least one vertex. But complete tripartite graphs have more edges than 2n - 3.Wait, no. For example, if we have three partitions with sizes 1,1, and n-2, then the number of edges would be 1*(n-2) + 1*(n-2) + 1*1 = 2(n-2) + 1 = 2n - 3. So, that's exactly the number of edges we're supposed to prove.So, in this case, the graph is a complete tripartite graph with partitions of size 1,1,n-2. That graph is uniquely 3-colorable because the two single-vertex partitions must be colored with two distinct colors, and the remaining n-2 vertices must be colored with the third color. Since the single vertices are connected to all others, their colors are fixed, and the rest must take the third color.Moreover, the number of edges is 2n - 3, which matches the lower bound. So, this seems like the minimal case.Therefore, the minimal uniquely 3-colorable graph with n vertices has exactly 2n - 3 edges, and any uniquely 3-colorable graph must have at least that many edges.I think that solidifies the proof. So, to summarize:1. Assume G is uniquely 3-colorable with color classes of sizes n1, n2, n3.2. The bipartite subgraphs between each pair of color classes must be connected; otherwise, we could permute colors within a disconnected component, leading to a different coloring, contradicting uniqueness.3. Each connected bipartite subgraph has at least (n_i + n_j - 1) edges.4. Summing these gives a total of at least 2n - 3 edges.5. An example of a graph achieving this bound is the complete tripartite graph with partitions 1,1,n-2.So, the proof seems complete.

💡First, I need to determine the ratio of the investments made by A, B, and C. A invested Rs. 6300, B invested Rs. 4200, and C invested Rs. 10500. To simplify the ratio, I'll find the greatest common divisor of these amounts, which is 2100.Dividing each investment by 2100:- A's share ratio = 6300 / 2100 = 3- B's share ratio = 4200 / 2100 = 2- C's share ratio = 10500 / 2100 = 5This gives the simplified ratio of their investments as 3:2:5.Next, I'll calculate the total parts of the ratio by adding them together: 3 + 2 + 5 = 10.The total profit is Rs. 12600. To find A's share, I'll calculate the fraction of the total profit corresponding to A's ratio:A's share = (3 / 10) * 12600 = 0.3 * 12600 = Rs. 3780.Therefore, A's share in the profit is Rs. 3780.

💡Alright, so I have this problem here: Given positive numbers ( a ) and ( b ), I need to find the largest positive number ( c ) such that for any positive real number ( x ), the inequality ( c leqslant max left{a x+frac{1}{a x}, b x+frac{1}{b x}right} ) holds. Hmm, okay, let me try to break this down step by step.First, I notice that the expression inside the max function is of the form ( t + frac{1}{t} ) where ( t ) is either ( a x ) or ( b x ). I remember that the function ( f(t) = t + frac{1}{t} ) has some nice properties. Specifically, it's known that for any positive ( t ), ( f(t) ) is always greater than or equal to 2, and this minimum value of 2 occurs when ( t = 1 ). So, maybe this function will be useful here.Let me define ( f(t) = t + frac{1}{t} ) for ( t > 0 ). Then, the given expression becomes ( max { f(a x), f(b x) } ). So, the problem is asking for the largest ( c ) such that ( c leq max { f(a x), f(b x) } ) for all ( x > 0 ).I think I need to find the minimum value of ( max { f(a x), f(b x) } ) over all positive ( x ). Because if I can find the smallest value that this maximum can take, then that will be the largest ( c ) satisfying the inequality for all ( x ).So, let me consider two cases: when ( a = b ) and when ( a neq b ).**Case 1: ( a = b )**If ( a = b ), then both expressions inside the max function are the same: ( f(a x) ). So, the max is just ( f(a x) ). Therefore, the problem reduces to finding the minimum value of ( f(a x) ) over all ( x > 0 ).But ( f(t) ) has its minimum at ( t = 1 ), so the minimum value of ( f(a x) ) occurs when ( a x = 1 ), which gives ( x = frac{1}{a} ). Plugging this back in, we get ( f(1) = 1 + 1 = 2 ). Therefore, in this case, the largest ( c ) is 2.**Case 2: ( a neq b )**Without loss of generality, let's assume ( a < b ). So, ( a ) is smaller than ( b ). I need to analyze how ( f(a x) ) and ( f(b x) ) behave as ( x ) varies.Let me think about the behavior of ( f(t) ). As ( t ) increases, ( f(t) ) increases beyond 2. As ( t ) decreases towards 0, ( f(t) ) also increases beyond 2. So, ( f(t) ) has a minimum at ( t = 1 ), and it's symmetric in a way around that point.Now, if I consider ( f(a x) ) and ( f(b x) ), since ( a < b ), as ( x ) increases, ( b x ) will grow faster than ( a x ), so ( f(b x) ) will eventually become larger than ( f(a x) ). Conversely, as ( x ) decreases, ( a x ) will decrease faster than ( b x ), so ( f(a x) ) will become larger than ( f(b x) ).Therefore, there must be some critical value of ( x ) where ( f(a x) = f(b x) ). Let me find this point.Set ( f(a x) = f(b x) ):[a x + frac{1}{a x} = b x + frac{1}{b x}]Let me rearrange this equation:[a x - b x = frac{1}{b x} - frac{1}{a x}]Factor out ( x ) on the left and ( frac{1}{x} ) on the right:[(a - b)x = frac{1}{x} left( frac{1}{b} - frac{1}{a} right)]Multiply both sides by ( x ):[(a - b)x^2 = frac{1}{b} - frac{1}{a}]Simplify the right-hand side:[frac{1}{b} - frac{1}{a} = frac{a - b}{ab}]So, we have:[(a - b)x^2 = frac{a - b}{ab}]Since ( a neq b ), we can divide both sides by ( a - b ):[x^2 = frac{1}{ab}]Taking square roots (and considering ( x > 0 )):[x = frac{1}{sqrt{ab}}]So, at ( x = frac{1}{sqrt{ab}} ), both ( f(a x) ) and ( f(b x) ) are equal. Let me compute this value:[f(a x) = a cdot frac{1}{sqrt{ab}} + frac{1}{a cdot frac{1}{sqrt{ab}}} = frac{sqrt{a}}{sqrt{b}} + frac{sqrt{b}}{sqrt{a}} = sqrt{frac{a}{b}} + sqrt{frac{b}{a}}]Similarly, ( f(b x) ) will give the same result. So, at this critical point, both functions are equal to ( sqrt{frac{a}{b}} + sqrt{frac{b}{a}} ).Now, I need to see how ( max { f(a x), f(b x) } ) behaves around this critical point.For ( x > frac{1}{sqrt{ab}} ), since ( b > a ), ( b x ) is larger than ( a x ), so ( f(b x) ) will be larger than ( f(a x) ). As ( x ) increases beyond ( frac{1}{sqrt{ab}} ), ( f(b x) ) increases beyond ( sqrt{frac{a}{b}} + sqrt{frac{b}{a}} ).For ( x < frac{1}{sqrt{ab}} ), ( a x ) is smaller than ( b x ), so ( f(a x) ) will be larger than ( f(b x) ). As ( x ) decreases below ( frac{1}{sqrt{ab}} ), ( f(a x) ) increases beyond ( sqrt{frac{a}{b}} + sqrt{frac{b}{a}} ).Therefore, the minimum value of ( max { f(a x), f(b x) } ) occurs exactly at ( x = frac{1}{sqrt{ab}} ), and this minimum value is ( sqrt{frac{a}{b}} + sqrt{frac{b}{a}} ).So, the largest ( c ) such that ( c leq max { f(a x), f(b x) } ) for all ( x > 0 ) is ( sqrt{frac{a}{b}} + sqrt{frac{b}{a}} ).Wait, let me double-check this. If I set ( c ) to this value, then for all ( x ), the max of the two functions will be at least ( c ). And since this is the minimal value of the max function, it's indeed the largest possible ( c ).Just to make sure, let's test with specific numbers. Suppose ( a = 1 ) and ( b = 2 ). Then, ( c = sqrt{frac{1}{2}} + sqrt{frac{2}{1}} = frac{sqrt{2}}{2} + sqrt{2} = frac{3sqrt{2}}{2} approx 2.121 ).Let me check for ( x = frac{1}{sqrt{1 cdot 2}} = frac{1}{sqrt{2}} approx 0.707 ). Then, ( f(a x) = fleft( frac{1}{sqrt{2}} right) = frac{1}{sqrt{2}} + sqrt{2} approx 0.707 + 1.414 = 2.121 ), and ( f(b x) = fleft( sqrt{2} right) = sqrt{2} + frac{1}{sqrt{2}} approx 1.414 + 0.707 = 2.121 ). So, both are equal, as expected.Now, let's try ( x = 1 ). Then, ( f(a x) = 1 + 1 = 2 ), and ( f(b x) = 2 + frac{1}{2} = 2.5 ). The max is 2.5, which is greater than ( c approx 2.121 ).What about ( x = 0.5 )? Then, ( f(a x) = 0.5 + 2 = 2.5 ), and ( f(b x) = 1 + 1 = 2 ). The max is 2.5, again greater than ( c ).And if I take ( x = 2 ), ( f(a x) = 2 + 0.5 = 2.5 ), ( f(b x) = 4 + 0.25 = 4.25 ). The max is 4.25, which is way larger than ( c ).So, it seems consistent that ( c = sqrt{frac{a}{b}} + sqrt{frac{b}{a}} ) is indeed the minimal value of the max function, and thus the largest ( c ) satisfying the inequality for all ( x > 0 ).Therefore, after all this thinking and checking, I'm confident that the answer is ( sqrt{frac{a}{b}} + sqrt{frac{b}{a}} ).**Final Answer**The largest positive number ( c ) is boxed{sqrt{frac{a}{b}} + sqrt{frac{b}{a}}}.

💡Okay, so I have this problem here where I need to figure out the range of values that ( s^2 - c^2 ) can take. Let me try to understand what's given and what I need to find.First, the problem says that ( r ) is the distance from the origin to a point ( P ) with coordinates ( x ) and ( y ). So, I know that ( r ) is the radius in polar coordinates, right? That means ( r = sqrt{x^2 + y^2} ). Got that down.Next, they define ( s ) as the ratio ( frac{y}{r} ) and ( c ) as the ratio ( frac{x}{r} ). Hmm, that seems familiar. Wait, isn't that similar to sine and cosine in trigonometry? Because if you think about it, in a unit circle, ( sin(theta) = frac{y}{r} ) and ( cos(theta) = frac{x}{r} ). So, ( s ) is like sine and ( c ) is like cosine, but not necessarily on the unit circle because ( r ) isn't necessarily 1 here. It's just any point ( (x, y) ) in the plane.So, the problem is asking about ( s^2 - c^2 ). Let me write that out:[s^2 - c^2 = left( frac{y}{r} right)^2 - left( frac{x}{r} right)^2]Simplifying that, I get:[s^2 - c^2 = frac{y^2}{r^2} - frac{x^2}{r^2} = frac{y^2 - x^2}{r^2}]But since ( r^2 = x^2 + y^2 ), I can substitute that in:[s^2 - c^2 = frac{y^2 - x^2}{x^2 + y^2}]Okay, so now I have ( s^2 - c^2 ) expressed in terms of ( x ) and ( y ). I need to figure out the range of possible values this expression can take. Let's denote ( s^2 - c^2 ) as ( k ):[k = frac{y^2 - x^2}{x^2 + y^2}]I need to find the maximum and minimum values of ( k ). To do this, maybe I can analyze the expression. Let's see.First, note that both the numerator and the denominator are real numbers, and the denominator is always positive because it's the sum of squares. So, the sign of ( k ) depends on the numerator ( y^2 - x^2 ).If ( y^2 > x^2 ), then ( k ) is positive. If ( y^2 < x^2 ), then ( k ) is negative. If ( y^2 = x^2 ), then ( k = 0 ).Now, let's find the maximum value of ( k ). Since ( k = frac{y^2 - x^2}{x^2 + y^2} ), let's consider when ( y^2 ) is as large as possible relative to ( x^2 ). The maximum occurs when ( x^2 ) is as small as possible, which is zero. So, if ( x = 0 ), then:[k = frac{y^2 - 0}{0 + y^2} = frac{y^2}{y^2} = 1]Similarly, the minimum value occurs when ( y^2 ) is as small as possible relative to ( x^2 ). The minimum occurs when ( y = 0 ), so:[k = frac{0 - x^2}{x^2 + 0} = frac{-x^2}{x^2} = -1]Therefore, ( k ) can take any value between -1 and 1. But wait, can it actually reach -1 and 1?Yes, when ( x = 0 ), ( k = 1 ), and when ( y = 0 ), ( k = -1 ). So, the range of ( k ) is from -1 to 1, inclusive.Let me double-check this. Suppose ( x = y ). Then:[k = frac{x^2 - x^2}{x^2 + x^2} = frac{0}{2x^2} = 0]Which is within the range. If ( y ) is larger than ( x ), say ( y = 2x ), then:[k = frac{(2x)^2 - x^2}{x^2 + (2x)^2} = frac{4x^2 - x^2}{x^2 + 4x^2} = frac{3x^2}{5x^2} = frac{3}{5} = 0.6]Which is between -1 and 1. Similarly, if ( x ) is larger than ( y ), say ( x = 2y ), then:[k = frac{y^2 - (2y)^2}{(2y)^2 + y^2} = frac{y^2 - 4y^2}{4y^2 + y^2} = frac{-3y^2}{5y^2} = -frac{3}{5} = -0.6]Again, within the range. So, it seems consistent that ( k ) can take any value between -1 and 1, including -1 and 1.Therefore, the values of ( s^2 - c^2 ) are limited to the numbers between -1 and +1, both included.**Final Answer**boxed{D}