💡Alright, let's tackle this problem step by step. I have four statements about frequency and probability, and I need to determine which ones are correct. I'll go through each statement one by one, think about what they mean, and see if they make sense based on what I know about probability and frequency.Starting with statement (1): "Frequency reflects the frequency of an event occurring, while probability reflects the magnitude of the possibility of an event occurring." Okay, so frequency is about how often something happens, right? Like, if I flip a coin 10 times and it lands on heads 5 times, the frequency of heads is 5 out of 10. Probability, on the other hand, is about the likelihood or chance of an event happening. For a fair coin, the probability of getting heads is 0.5. So, statement (1) seems to be distinguishing between frequency, which is observed over trials, and probability, which is a theoretical measure. That makes sense to me. So, I think statement (1) is correct.Moving on to statement (2): "If an event A occurs m times in n random experiments, then the frequency m/n of event A occurring is the probability of event A." Hmm, this is interesting. It's saying that if an event happens m times out of n trials, then m/n is the probability of that event. But wait, isn't probability more of a theoretical value? Like, for a fair die, the probability of rolling a 3 is 1/6, regardless of how many times you roll it. However, if you roll the die 6 times and get a 3 once, the frequency is 1/6, which matches the probability. But what if you roll it 10 times and only get a 3 once? Then the frequency is 1/10, which doesn't match the probability. So, does that mean frequency is always equal to probability? No, I don't think so. Frequency can approximate probability, especially with a large number of trials, but it's not necessarily equal. So, statement (2) seems incorrect because it equates frequency directly to probability without considering the number of trials.Now, statement (3): "Frequency cannot be separated from the specific number of experiments, it is an experimental value, while probability is a deterministic, theoretical value that does not depend on the number of experiments." This is talking about the relationship between frequency and probability in terms of their dependency on the number of experiments. Frequency is indeed tied to the number of trials because it's based on actual observations. For example, flipping a coin 10 times gives a different frequency than flipping it 1000 times. Probability, however, is a fixed value that doesn't change with the number of trials. It's a theoretical measure that predicts the likelihood of an event. So, statement (3) seems to accurately describe the difference between frequency and probability. Therefore, I think statement (3) is correct.Lastly, statement (4): "In a large number of experiments, frequency is an approximation of probability, and probability is a stable value of frequency." This is about the law of large numbers, which states that as the number of trials increases, the frequency of an event will approach its theoretical probability. So, with a large number of experiments, frequency becomes a good approximation of probability. Additionally, probability is stable in the sense that it doesn't change with more trials; it's a fixed value. So, statement (4) is also correct.To summarize my thoughts:- Statement (1) is correct because it distinguishes between frequency and probability accurately.- Statement (2) is incorrect because frequency is not always equal to probability; it's an approximation that depends on the number of trials.- Statement (3) is correct because it highlights the dependency of frequency on the number of experiments and the theoretical nature of probability.- Statement (4) is correct because it refers to the law of large numbers and the stability of probability.So, the correct statements are (1), (3), and (4).

💡First, I need to understand the problem. The man has a certain number of small boxes, and when he packs them in parcels of 3, 4, 5, or 6, he always has one box left over. However, when he packs them in some other number of boxes per parcel, there are no leftovers. The number of boxes he might have is 301, and I need to find the number of boxes that leaves no leftovers when packed in a parcel.To approach this, I'll start by checking if 301 leaves a remainder of 1 when divided by 3, 4, 5, and 6. 1. **Dividing 301 by 3:** - 301 ÷ 3 = 100 with a remainder of 1. 2. **Dividing 301 by 4:** - 301 ÷ 4 = 75 with a remainder of 1. 3. **Dividing 301 by 5:** - 301 ÷ 5 = 60 with a remainder of 1. 4. **Dividing 301 by 6:** - 301 ÷ 6 = 50 with a remainder of 1.Since 301 leaves a remainder of 1 when divided by 3, 4, 5, and 6, it fits the first condition.Next, I need to find a number that divides 301 without leaving any remainder. To do this, I'll check if 301 is a prime number or if it has any divisors other than 1 and itself.- **Checking if 301 is a prime number:** - 301 is not divisible by 2, 3, 5, 7, 11, 13, 17, or 19. - Therefore, 301 is a prime number.Since 301 is a prime number, its only divisors are 1 and 301 itself. This means the only way to pack the boxes without any leftovers is to pack all 301 boxes in one parcel.Thus, the number of boxes that leaves no boxes left over when packed in a parcel is 301.

💡First, I need to determine the distance traveled at each speed. Let's denote this equal distance as ( d ) kilometers.Next, I'll calculate the time taken for each segment of the journey using the formula:[text{Time} = frac{text{Distance}}{text{Speed}}]For each speed:1. At 5 km/hr: ( frac{d}{5} ) hours2. At 10 km/hr: ( frac{d}{10} ) hours3. At 15 km/hr: ( frac{d}{15} ) hours4. At 20 km/hr: ( frac{d}{20} ) hours5. At 25 km/hr: ( frac{d}{25} ) hoursAdding up all these times gives the total time for the journey:[frac{d}{5} + frac{d}{10} + frac{d}{15} + frac{d}{20} + frac{d}{25} = frac{15}{60} text{ hours}]To simplify the equation, I'll find a common denominator for the fractions, which is 300:[frac{60d}{300} + frac{30d}{300} + frac{20d}{300} + frac{15d}{300} + frac{12d}{300} = frac{15}{60}]Combining the terms:[frac{137d}{300} = frac{15}{60}]Solving for ( d ):[137d = 300 times frac{15}{60}][137d = 75][d = frac{75}{137}]Finally, the total distance covered is:[5d = 5 times frac{75}{137} approx 2.737 text{ km}]

💡To determine the probability that Megan dials Fatima's correct phone number, we'll follow these steps:1. **Identify the number of choices for the first three digits:** - Megan has 3 possible choices for the first three digits: 295, 296, or 299.2. **Determine the permutations for the last four digits:** - The last four digits consist of 1, 6, 7, and an additional 6 or 7. - **Case 1:** If the additional digit is 6, the digits are 1, 6, 6, 7. The number of unique permutations is calculated by: [ frac{4!}{2!} = frac{24}{2} = 12 ] - **Case 2:** If the additional digit is 7, the digits are 1, 6, 7, 7. The number of unique permutations is: [ frac{4!}{2!} = frac{24}{2} = 12 ] - **Total permutations for the last four digits:** (12 + 12 = 24)3. **Calculate the total number of possible phone numbers:** - Multiply the number of choices for the first three digits by the number of permutations for the last four digits: [ 3 times 24 = 72 ]4. **Determine the probability of dialing the correct number:** - Since there is only one correct phone number, the probability is: [ frac{1}{72} ]

💡Alright, so I've got this problem here: there's a circle with 2n equally spaced points on its circumference, and I need to figure out how many right-angled triangles can be formed by choosing three of these points as vertices. The problem specifies that n is greater than 1, so we're dealing with at least four points on the circle.First, I remember that for a triangle to be right-angled, one of its angles has to be 90 degrees. Now, in the context of a circle, I recall something about the inscribed angle theorem. If I'm not mistaken, an inscribed angle subtended by a diameter is a right angle. That means if I have a triangle where one side is the diameter of the circle, then the angle opposite that side will be 90 degrees. So, that gives me a clue: right-angled triangles on a circle must have the hypotenuse as a diameter.Okay, so if I can find all the diameters in this set of points, then for each diameter, I can pair it with any other point to form a right-angled triangle. That makes sense. Now, how many diameters are there in 2n equally spaced points?Well, since the points are equally spaced, each diameter connects two points that are directly opposite each other. So, for 2n points, there should be n diameters because each diameter connects two points, and with 2n points, you can pair them up into n diameters. Let me double-check that. If n is 2, then 2n is 4, and indeed, there are 2 diameters. If n is 3, 2n is 6, and there are 3 diameters. Yeah, that seems right.So, there are n diameters. For each diameter, I need to count how many right-angled triangles can be formed by choosing a third point. Since the triangle is determined by three points, and two of them are already fixed as the endpoints of the diameter, the third point can be any of the remaining points. How many points are left after choosing the two endpoints of the diameter?Well, there are 2n points in total, and we've already used two of them for the diameter, so that leaves 2n - 2 points. Each of these points will form a unique right-angled triangle with the diameter. So, for each diameter, there are 2n - 2 possible triangles.Therefore, if there are n diameters, and each diameter can form 2n - 2 triangles, the total number of right-angled triangles should be n multiplied by (2n - 2). Let me write that down:Total triangles = n * (2n - 2)Simplifying that, it becomes:Total triangles = 2n^2 - 2nHmm, that seems straightforward, but let me make sure I'm not missing anything. Is there any overlap or double-counting here? Each triangle is uniquely determined by its diameter and the third point, so as long as each diameter is unique and each third point is unique, there shouldn't be any overlap. Since the points are equally spaced, each diameter is distinct, and each third point is also distinct for each diameter.Wait a second, but what if n is 1? The problem says n > 1, so n starts at 2. For n = 2, 2n = 4 points, which makes sense. There are 2 diameters, each can form 2 triangles, so 2*2 = 4 triangles. Let me visualize that: four points on a circle, forming a square. Each diameter can pair with the other two points to form two right-angled triangles. Yeah, that works.For n = 3, 2n = 6 points, forming a hexagon. There are 3 diameters, each can pair with 4 other points, so 3*4 = 12 triangles. Let me see: in a hexagon, each diameter connects two opposite points, and the other four points can each form a right angle with that diameter. That seems correct.Is there another way to approach this problem? Maybe by counting all possible triangles and then determining how many are right-angled. But that might be more complicated because I'd have to consider all combinations and then check which ones are right-angled. The method I used seems more straightforward.Another thought: since the points are equally spaced, the angle subtended by each adjacent point at the center is equal. For 2n points, the angle between two adjacent points is 360/(2n) degrees, which simplifies to 180/n degrees. So, each point is 180/n degrees apart from its neighbors.But how does that relate to right-angled triangles? Well, for a triangle to be right-angled, one of its angles must be 90 degrees. In terms of the central angles, the arc corresponding to the right angle must be 180 degrees because the inscribed angle is half the central angle. Wait, no, the inscribed angle is half the measure of the central angle subtending the same arc. So, if the inscribed angle is 90 degrees, the central angle must be 180 degrees, which is a semicircle. That confirms that the hypotenuse must be a diameter.So, this reinforces the initial approach: each right-angled triangle corresponds to a diameter and a third point. Therefore, the number of right-angled triangles is equal to the number of diameters multiplied by the number of possible third points for each diameter.Just to recap:1. Number of diameters = n2. For each diameter, number of possible third points = 2n - 23. Total right-angled triangles = n * (2n - 2) = 2n^2 - 2nThis seems consistent across different values of n, and the logic holds up when I test it with small values like n = 2 and n = 3.I don't think I'm missing any cases here because every right-angled triangle must have its hypotenuse as a diameter, and every diameter can be combined with any other point to form such a triangle. There's no possibility of forming a right-angled triangle without using a diameter as the hypotenuse in this setup.Therefore, the formula 2n(n - 1) gives the total number of right-angled triangles that can be formed with three of these points as vertices.

💡To solve the equation ( sqrt{10 + 3z} = 12 ), I need to eliminate the square root. I'll start by squaring both sides of the equation to get rid of the square root.After squaring both sides, I'll have ( 10 + 3z = 144 ). Next, I'll isolate the term with ( z ) by subtracting 10 from both sides, resulting in ( 3z = 134 ). Finally, I'll solve for ( z ) by dividing both sides by 3, which gives me ( z = frac{134}{3} ).To ensure my solution is correct, I'll substitute ( z = frac{134}{3} ) back into the original equation and verify that both sides are equal.