🧠:<think>Alright, so I've got these two problems to solve about a father and his son who are both track athletes. Let me try to figure them out step by step.Starting with the first problem: The son is running a 400-meter dash. He accelerates from rest until he reaches his top speed in 8 seconds, then maintains that speed for the rest of the race. His total time is 48 seconds. I need to find his top speed and the distance he covers during the acceleration phase.Hmm, okay. So, the race is divided into two parts: acceleration and constant speed. The total distance is 400 meters, and the total time is 48 seconds. He accelerates for the first 8 seconds, then runs at top speed for the remaining time.Let me denote his top speed as ( v ) (in m/s). During the acceleration phase, he starts from rest, so his initial velocity is 0. He accelerates uniformly for 8 seconds until he reaches ( v ).I remember that the distance covered during uniform acceleration can be calculated using the formula:[ d = frac{1}{2} a t^2 ]But wait, I don't know the acceleration ( a ). Alternatively, since he starts from rest, the average velocity during acceleration is ( frac{v}{2} ). So, the distance covered during acceleration is:[ d_1 = frac{v}{2} times t_1 ]Where ( t_1 = 8 ) seconds.After reaching top speed, he maintains ( v ) for the remaining time ( t_2 = 48 - 8 = 40 ) seconds. The distance covered during this phase is:[ d_2 = v times t_2 ]The total distance is ( d_1 + d_2 = 400 ) meters.So, substituting:[ frac{v}{2} times 8 + v times 40 = 400 ]Simplify:[ 4v + 40v = 400 ][ 44v = 400 ][ v = frac{400}{44} ][ v = frac{100}{11} ][ v approx 9.09 text{ m/s} ]So, the top speed is approximately 9.09 m/s.Now, the distance covered during acceleration is:[ d_1 = frac{v}{2} times 8 = frac{9.09}{2} times 8 ][ d_1 = 4.545 times 8 ][ d_1 approx 36.36 text{ meters} ]Let me double-check that. If he accelerates for 8 seconds to reach 9.09 m/s, then the distance is indeed 36.36 meters. Then, he runs 40 seconds at 9.09 m/s, which is 363.64 meters. Adding them together: 36.36 + 363.64 = 400 meters. Perfect, that checks out.Moving on to the second problem: The father runs to the school, which is 3 kilometers away. He starts at 2 m/s and accelerates uniformly at 0.1 m/s². I need to find the time it takes him to reach the school and the distance he covers by the time he reaches 5 m/s.First, let's convert 3 kilometers to meters: 3 km = 3000 meters.He starts at 2 m/s and accelerates at 0.1 m/s². I need to find the time it takes him to cover 3000 meters.But before that, the second part asks for the distance covered when he reaches 5 m/s. So, let's first find the time it takes him to reach 5 m/s.Using the formula:[ v = u + at ]Where ( v = 5 ) m/s, ( u = 2 ) m/s, ( a = 0.1 ) m/s².So,[ 5 = 2 + 0.1 t ][ 0.1 t = 3 ][ t = 30 text{ seconds} ]So, it takes him 30 seconds to reach 5 m/s.Now, the distance covered in those 30 seconds can be found using:[ s = ut + frac{1}{2} a t^2 ][ s = 2 times 30 + 0.5 times 0.1 times 30^2 ][ s = 60 + 0.05 times 900 ][ s = 60 + 45 ][ s = 105 text{ meters} ]So, he covers 105 meters by the time he reaches 5 m/s.Now, for the total time to reach the school, which is 3000 meters. Since he is accelerating, we need to check if he reaches his maximum speed before covering the distance or not. Wait, the problem doesn't specify if he stops accelerating or continues accelerating until he reaches the school. Hmm.Wait, the problem says he accelerates uniformly at 0.1 m/s². So, he keeps accelerating until he reaches the school. So, we need to find the time ( t ) when the distance covered is 3000 meters.Using the equation:[ s = ut + frac{1}{2} a t^2 ]Plugging in:[ 3000 = 2 t + 0.5 times 0.1 times t^2 ][ 3000 = 2t + 0.05 t^2 ][ 0.05 t^2 + 2 t - 3000 = 0 ]Multiply both sides by 20 to eliminate the decimal:[ t^2 + 40 t - 60000 = 0 ]Now, solving this quadratic equation:[ t = frac{-40 pm sqrt{40^2 + 4 times 1 times 60000}}{2} ][ t = frac{-40 pm sqrt{1600 + 240000}}{2} ][ t = frac{-40 pm sqrt{241600}}{2} ][ t = frac{-40 pm 491.53}{2} ]We discard the negative time:[ t = frac{-40 + 491.53}{2} ][ t = frac{451.53}{2} ][ t approx 225.765 text{ seconds} ]So, approximately 225.77 seconds to reach the school.Let me verify that. Plugging back into the distance equation:[ s = 2 times 225.765 + 0.05 times (225.765)^2 ][ s = 451.53 + 0.05 times 51000 ]Wait, 225.765 squared is approximately 51000? Let me calculate:225.765^2 = (225)^2 + 2*225*0.765 + (0.765)^2 ≈ 50625 + 348.75 + 0.585 ≈ 50974.335So,[ s = 451.53 + 0.05 times 50974.335 ][ s = 451.53 + 2548.71675 ][ s ≈ 2999.24675 text{ meters} ]Which is approximately 3000 meters. So, that checks out.So, summarizing:1. Son's top speed is approximately 9.09 m/s, and he covers about 36.36 meters during acceleration.2. Father takes approximately 225.77 seconds to reach the school and covers 105 meters when he reaches 5 m/s.Final Answer1. The son's top speed is boxed{9.09 text{ m/s}} and the distance covered during acceleration is boxed{36.36 text{ meters}}.2. The father takes boxed{225.77 text{ seconds}} to reach the school and covers boxed{105 text{ meters}} by the time he reaches 5 m/s.</think>

🧠:<think>Okay, so I have these two questions to solve. Let me take them one by one.Starting with the first question: It involves a Poisson process. I remember that a Poisson process is used to model the number of events happening in a fixed interval of time or space. In this case, the number of patients a physician sees in an hour. The average rate λ is given as 3 patients per hour. We need to find the probability that a physician will be available to see 5 or more patients in a single hour. So, essentially, we need P(X ≥ 5), where X follows a Poisson distribution with λ = 3.I recall that the Poisson probability mass function is given by:P(X = k) = (e^(-λ) * λ^k) / k!So, to find P(X ≥ 5), it's easier to calculate 1 - P(X ≤ 4). That is, 1 minus the sum of probabilities from X=0 to X=4.Let me compute each term:For k=0:P(X=0) = (e^(-3) * 3^0) / 0! = e^(-3) * 1 / 1 = e^(-3) ≈ 0.0498For k=1:P(X=1) = (e^(-3) * 3^1) / 1! = 3e^(-3) ≈ 0.1494For k=2:P(X=2) = (e^(-3) * 3^2) / 2! = (9e^(-3)) / 2 ≈ 0.2240For k=3:P(X=3) = (e^(-3) * 3^3) / 3! = (27e^(-3)) / 6 ≈ 0.2240For k=4:P(X=4) = (e^(-3) * 3^4) / 4! = (81e^(-3)) / 24 ≈ 0.1680Now, summing these up:0.0498 + 0.1494 = 0.19920.1992 + 0.2240 = 0.42320.4232 + 0.2240 = 0.64720.6472 + 0.1680 = 0.8152So, P(X ≤ 4) ≈ 0.8152Therefore, P(X ≥ 5) = 1 - 0.8152 = 0.1848Hmm, so approximately 18.48% chance.Wait, let me double-check my calculations. Maybe I made a mistake in the exponents or factorials.Calculating each term again:k=0: e^(-3) ≈ 0.0498k=1: 3 * e^(-3) ≈ 0.1494k=2: (9 / 2) * e^(-3) ≈ 4.5 * 0.0498 ≈ 0.2241k=3: (27 / 6) * e^(-3) ≈ 4.5 * 0.0498 ≈ 0.2241k=4: (81 / 24) * e^(-3) ≈ 3.375 * 0.0498 ≈ 0.1680Adding them: 0.0498 + 0.1494 = 0.19920.1992 + 0.2241 = 0.42330.4233 + 0.2241 = 0.64740.6474 + 0.1680 = 0.8154So, 1 - 0.8154 = 0.1846, which is approximately 18.46%. So, my initial calculation was correct.Therefore, the probability is approximately 18.46%.Moving on to the second question: It's about a network graph representing departments and communication channels. The graph is a connected acyclic graph with 10 departments. We need to determine the maximum number of edges while maintaining its properties.I remember that a connected acyclic graph is a tree. In a tree with n nodes, the number of edges is n - 1. So, for 10 departments, the number of edges would be 10 - 1 = 9.But wait, the question says "maximum number of edges while maintaining its properties." So, since it's a tree, which is minimally connected, it has the minimum number of edges to keep it connected. If we want the maximum number of edges while still being acyclic, that's a bit confusing because adding more edges would create cycles.Wait, no. If it's a connected acyclic graph, it's a tree, and trees have exactly n - 1 edges. So, the maximum number of edges in a connected acyclic graph is n - 1. Because if you add one more edge, it becomes cyclic.So, for 10 nodes, maximum edges without cycles is 9.But wait, hold on. The question says "the network graph is guaranteed to be a connected acyclic graph." So, it's a tree. So, the number of edges is fixed at n - 1, which is 9.But the question is asking for the maximum number of edges while maintaining its properties as a connected acyclic graph. So, since it's a tree, the number of edges is fixed. So, the maximum is 9.Alternatively, if the graph were allowed to have cycles, the maximum number of edges would be n(n - 1)/2, which is 45 for 10 nodes. But since it's acyclic, it's 9.Therefore, the maximum number of edges is 9.Wait, but let me think again. If it's a connected acyclic graph, it's a tree, which has n - 1 edges. So, 10 - 1 = 9. So, 9 is the maximum number of edges because adding any more would create a cycle, which would violate the acyclic property.Yes, that makes sense.So, summarizing:1. The probability is approximately 18.46%.2. The maximum number of edges is 9.Final Answer1. The probability is boxed{0.1848}.2. The maximum number of edges is boxed{9}.</think>

🧠:<think>Okay, so I've got this problem about gamers organizing a tournament with both classic and new games. It's divided into two parts: one about classic arcade games and another about new releases. Let me try to tackle each part step by step.Starting with the first part: Classic Arcade Games Selection.1. Total number of unique combinations of 5 games out of 10.Hmm, this sounds like a combination problem. Since the order doesn't matter when selecting games, it's just "10 choose 5." The formula for combinations is C(n, k) = n! / (k!(n - k)!).So, plugging in the numbers: C(10, 5) = 10! / (5! * 5!). Let me compute that.10! is 10 × 9 × 8 × 7 × 6 × 5! So, we can cancel out the 5! in numerator and denominator.So, C(10, 5) = (10 × 9 × 8 × 7 × 6) / (5 × 4 × 3 × 2 × 1). Let me calculate that:10 × 9 = 9090 × 8 = 720720 × 7 = 50405040 × 6 = 30240Denominator: 5 × 4 = 2020 × 3 = 6060 × 2 = 120120 × 1 = 120So, 30240 / 120. Let me divide 30240 by 120.30240 ÷ 120: 120 × 252 = 30240. So, 252.Therefore, the total number of unique combinations is 252.2. Expected total score for one randomly selected combination of 5 games.Alright, so each game's score is the sum of the digits of its release year. The release years are from 1980 to 1989. Let me list them out:1980, 1981, 1982, 1983, 1984, 1985, 1986, 1987, 1988, 1989.First, I need to compute the score for each game. Let's do that.- 1980: 1 + 9 + 8 + 0 = 18- 1981: 1 + 9 + 8 + 1 = 19- 1982: 1 + 9 + 8 + 2 = 20- 1983: 1 + 9 + 8 + 3 = 21- 1984: 1 + 9 + 8 + 4 = 22- 1985: 1 + 9 + 8 + 5 = 23- 1986: 1 + 9 + 8 + 6 = 24- 1987: 1 + 9 + 8 + 7 = 25- 1988: 1 + 9 + 8 + 8 = 26- 1989: 1 + 9 + 8 + 9 = 27So, the scores are: 18, 19, 20, 21, 22, 23, 24, 25, 26, 27.Now, to find the expected total score for a randomly selected combination of 5 games. Since expectation is linear, the expected total score is 5 times the expected score of a single randomly selected game.Wait, is that correct? Because each game is equally likely to be chosen, and the selection is without replacement. So, the expected value of the sum is the sum of the expected values. Since each game has an equal chance of being included in the combination, the expected score per game is the average of all 10 game scores, and then multiplied by 5.Yes, that makes sense. So, first, let's compute the average score of all 10 games.Scores: 18, 19, 20, 21, 22, 23, 24, 25, 26, 27.Let me add them up:18 + 19 = 3737 + 20 = 5757 + 21 = 7878 + 22 = 100100 + 23 = 123123 + 24 = 147147 + 25 = 172172 + 26 = 198198 + 27 = 225Total sum = 225.Average score = 225 / 10 = 22.5.Therefore, the expected total score for 5 games is 5 * 22.5 = 112.5.Wait, that seems straightforward. Let me think again. Since each game is equally likely to be selected, the expected value is just the average per game times the number of games selected. So, yes, that should be correct.But just to be thorough, another way to compute it is to consider that each game has a probability of 5/10 = 1/2 of being selected in a random combination. Therefore, the expected total score is the sum of each game's score multiplied by the probability of being selected.So, sum over all games: score_i * (1/2). Then, since there are 10 games, the total expected score is (sum of all scores) * (1/2). Sum of all scores is 225, so 225 * (1/2) = 112.5. Yep, same result.So, that's solid.Moving on to the second part: New Release Probability.1. Total number of possible distinct ranking orders for 4 new games.This is a permutation problem. Since each game is ranked from 1 to 4, with no ties, the number of possible rankings is 4 factorial.4! = 4 × 3 × 2 × 1 = 24.So, there are 24 possible distinct ranking orders.2. Probability that two specific games will be ranked consecutively in a random ranking.Hmm, okay. So, we have 4 games, say A, B, C, D. We want the probability that two specific games, say A and B, are ranked consecutively.First, total number of possible rankings is 24, as above.Now, how many rankings have A and B consecutive?To compute this, we can treat A and B as a single entity or "block." So, instead of 4 items, we have 3 items: [AB], C, D. These can be arranged in 3! ways. But within the [AB] block, A and B can be in two orders: AB or BA.So, total number of favorable arrangements is 3! * 2 = 6 * 2 = 12.Therefore, the probability is 12 / 24 = 1/2.Wait, that seems high. Let me think again.Alternatively, another approach: fix the position of A. For each position A can be in, count the number of positions B can be in such that they are adjacent.In 4 positions, A can be in positions 1, 2, 3, or 4.If A is in position 1, then B can only be in position 2.If A is in position 2, B can be in 1 or 3.If A is in position 3, B can be in 2 or 4.If A is in position 4, B can only be in position 3.So, total number of favorable arrangements:For each position of A:- Position 1: 1 possibility for B.- Position 2: 2 possibilities.- Position 3: 2 possibilities.- Position 4: 1 possibility.Total: 1 + 2 + 2 + 1 = 6.But wait, for each of these, the remaining two games can be arranged in 2! ways. So, total number of favorable arrangements is 6 * 2! = 6 * 2 = 12, same as before.Therefore, probability is 12 / 24 = 1/2.Wait, but intuitively, is it 1/2? Let me think of it another way.The probability that two specific games are next to each other in a permutation of 4.In general, for n items, the number of permutations where two specific items are adjacent is 2*(n-1)!.So, for n=4, it's 2*6=12, as above. So, probability is 12/24=1/2.Yes, that seems correct.Alternatively, think of the total number of possible pairs of positions for A and B. There are C(4,2)=6 possible pairs.Out of these 6, how many are consecutive?In 4 positions, the consecutive pairs are (1,2), (2,3), (3,4). So, 3 pairs.Each pair can be arranged in 2 ways (A before B or B before A). So, total favorable pairs: 3*2=6.But wait, how does that relate to the total number of permutations?Wait, actually, for each permutation, the positions of A and B are determined. So, the number of favorable permutations is 6, as above, but considering the rest of the games, it's 6 * 2! = 12.Wait, maybe I confused something.Alternatively, perhaps it's better to think in terms of probability.Once A is placed in a position, the probability that B is adjacent to A.For example, if A is in position 1, then B has to be in position 2. The probability of that is 1/3 (since B can be in any of the remaining 3 positions).Similarly, if A is in position 2, B can be in 1 or 3, so probability 2/3.If A is in position 3, similar to position 2.If A is in position 4, similar to position 1.So, the average probability is:(1/4)*(1/3) + (2/4)*(2/3) + (1/4)*(1/3) = ?Wait, let me compute this.Wait, actually, the probability that A is in position 1 is 1/4, and then B is in position 2 is 1/3 (since 3 remaining spots). Similarly for position 4.For positions 2 and 3, the probability that A is in position 2 is 1/4, and then B is in position 1 or 3, which is 2/3.So, total probability:(2 positions where A has 1 adjacent spot: positions 1 and 4) each contributing (1/4)*(1/3) = 1/12.And (2 positions where A has 2 adjacent spots: positions 2 and 3) each contributing (1/4)*(2/3) = 2/12.So, total probability:2*(1/12) + 2*(2/12) = (2 + 4)/12 = 6/12 = 1/2.Yes, same result.So, the probability is indeed 1/2.Wait, but I initially thought 1/2 seemed high, but after verifying with different methods, it seems correct.Alternatively, another way: in a permutation of 4 elements, the number of adjacent pairs is 3 (positions 1-2, 2-3, 3-4). For each adjacent pair, there are 2 possible orders (AB or BA), and the remaining two elements can be arranged in 2! ways. So, total favorable permutations: 3*2*2=12. Total permutations:24. So, 12/24=1/2.Yes, that's consistent.So, all methods point to the probability being 1/2.So, summarizing:1. Classic Arcade Games: - Number of combinations: 252 - Expected total score: 112.52. New Releases: - Number of ranking orders: 24 - Probability two specific games are consecutive: 1/2I think that's all. Let me just double-check my calculations.For the expected score, the average per game is 22.5, times 5 is 112.5. That seems right.For the probability, 1/2 is correct because treating the two games as a block gives half the total permutations.Yep, I think I'm confident with these answers.Final Answer1. The total number of unique combinations is boxed{252}, and the expected total score is boxed{112.5}.2. The total number of possible distinct ranking orders is boxed{24}, and the probability is boxed{dfrac{1}{2}}.</think>

🧠:<think>To determine the 10th term ( a_{10} ) in the sequence defined by the recurrence relation ( a_n = a_{n-1} + 2a_{n-2} ) with initial terms ( a_0 = 3 ) and ( a_1 = 5 ), I will start by calculating each term step by step.First, I'll list the known terms:- ( a_0 = 3 )- ( a_1 = 5 )Next, I'll use the recurrence relation to find the subsequent terms:- ( a_2 = a_1 + 2a_0 = 5 + 2 times 3 = 11 )- ( a_3 = a_2 + 2a_1 = 11 + 2 times 5 = 21 )- ( a_4 = a_3 + 2a_2 = 21 + 2 times 11 = 43 )- ( a_5 = a_4 + 2a_3 = 43 + 2 times 21 = 85 )- ( a_6 = a_5 + 2a_4 = 85 + 2 times 43 = 171 )- ( a_7 = a_6 + 2a_5 = 171 + 2 times 85 = 341 )- ( a_8 = a_7 + 2a_6 = 341 + 2 times 171 = 683 )- ( a_9 = a_8 + 2a_7 = 683 + 2 times 341 = 1365 )- ( a_{10} = a_9 + 2a_8 = 1365 + 2 times 683 = 2731 )Thus, the 10th term ( a_{10} ) is 2731.For the combinatorics problem, I need to find the number of ways to choose exactly 4 activities out of 10. This is a combination problem where the order of selection does not matter.The formula for combinations is:[C(n, k) = frac{n!}{k!(n - k)!}]Plugging in the values:[C(10, 4) = frac{10!}{4! times 6!} = frac{10 times 9 times 8 times 7}{4 times 3 times 2 times 1} = 210]Therefore, there are 210 different ways to choose 4 activities from the 10 available.</think>