🧠:<think>Okay, so I have this problem about analyzing a cybersecurity threat on the nation's communication network. It's divided into two sub-problems. Let me try to tackle them one by one.Starting with Sub-problem 1: The threat can compromise communication nodes with a probability ( p ). The network is modeled as a complete graph with ( n ) nodes. I need to find the expected number of compromised nodes. Hmm, okay.First, I remember that in probability, the expected value is like the average outcome we'd expect if we could repeat an experiment many times. So, if each node has a probability ( p ) of being compromised, then for each node, the expected number of compromised nodes would just be ( p ) times the number of nodes, right?Wait, but let me think more carefully. Each node can be considered as a Bernoulli trial where success is the node being compromised. So, for each node, the expected value is ( p ). Since expectation is linear, the total expectation is just the sum of expectations for each node. That is, ( E = n times p ). So, the expected number of compromised nodes is ( np ).But the problem mentions using advanced combinatorial techniques. Hmm, maybe I'm oversimplifying. Let me consider if there's another way to approach this.In a complete graph, every node is connected to every other node. But does that affect the expectation? I don't think so because expectation is linear regardless of dependencies. So even if the nodes are dependent in some way, the expectation of the sum is the sum of expectations. So, regardless of the graph structure, the expected number of compromised nodes is just ( np ).Wait, but maybe the problem is trying to get me to think about something else, like the number of edges or something? No, the question specifically asks for the expected number of compromised nodes, not edges. So, I think my initial thought was correct. It's just ( np ).Moving on to Sub-problem 2: Given that the disruption of ( k ) nodes would lead to systemic failure, I need to calculate the probability that the network will face systemic failure. They mention using the binomial distribution, and assuming each node's compromise is independent.So, if each node is compromised independently with probability ( p ), the number of compromised nodes follows a binomial distribution with parameters ( n ) and ( p ). The probability mass function is ( P(X = m) = binom{n}{m} p^m (1 - p)^{n - m} ).But we're interested in the probability that at least ( k ) nodes are compromised, right? Because disrupting ( k ) nodes leads to failure. So, the probability of systemic failure is the sum of probabilities from ( m = k ) to ( m = n ). That is, ( P(X geq k) = sum_{m=k}^{n} binom{n}{m} p^m (1 - p)^{n - m} ).Alternatively, sometimes people express this using the complement, but since the problem specifically mentions using the binomial distribution, I think the sum is the right approach.Wait, but is there a more concise way to write this? I know that sometimes cumulative distribution functions are used, but in terms of expressing the answer, the sum is probably acceptable.Let me just recap:Sub-problem 1: The expected number of compromised nodes is ( np ).Sub-problem 2: The probability of systemic failure is the sum from ( m = k ) to ( n ) of ( binom{n}{m} p^m (1 - p)^{n - m} ).I think that's it. I don't see any complications here because the graph is complete, but since the expectation and the binomial distribution don't depend on the graph structure, just the number of nodes and their individual probabilities, I think my answers are correct.Final AnswerSub-problem 1: The expected number of compromised nodes is boxed{np}.Sub-problem 2: The probability of systemic failure is boxed{sum_{m=k}^{n} binom{n}{m} p^m (1 - p)^{n - m}}.</think>

🧠:<think>Alright, so I've got this problem about a café owner who wants to model the number of customers over time using a sinusoidal function. She's trying to find the quietest hours to write her novel. Let me try to figure this out step by step.First, the problem gives me the function ( C(t) = A sin(Bt + phi) + D ). I need to find the values of A, B, D, and φ. They've given me some information: the maximum number of customers is 50, the minimum is 10, the period is 6 hours, and the quietest time (minimum) occurs at t = 14, which is 2 PM.Starting with part 1: finding A and D.I remember that in a sinusoidal function of the form ( A sin(Bt + phi) + D ), the amplitude A is half the difference between the maximum and minimum values. So, the amplitude is the peak deviation from the average value D. Let me write that down:Amplitude ( A = frac{text{Max} - text{Min}}{2} )Plugging in the given max and min:( A = frac{50 - 10}{2} = frac{40}{2} = 20 )So, A is 20.Next, D is the vertical shift, which is the average of the maximum and minimum values. So,( D = frac{text{Max} + text{Min}}{2} )Calculating that:( D = frac{50 + 10}{2} = frac{60}{2} = 30 )So, D is 30.Alright, that takes care of part 1. Now, moving on to part 2.They mentioned the period is 6 hours. I recall that the period of a sine function ( sin(Bt + phi) ) is given by ( frac{2pi}{B} ). So, if the period is 6, we can set up the equation:( frac{2pi}{B} = 6 )Solving for B:Multiply both sides by B:( 2pi = 6B )Then divide both sides by 6:( B = frac{2pi}{6} = frac{pi}{3} )So, B is π/3.Now, the function is ( C(t) = 20 sinleft(frac{pi}{3} t + phiright) + 30 ).Next, we need to find φ. They told us that the minimum occurs at t = 14. Since the sine function reaches its minimum at ( frac{3pi}{2} ) radians, we can set up the equation:( frac{pi}{3} times 14 + phi = frac{3pi}{2} )Let me solve for φ.First, compute ( frac{pi}{3} times 14 ):( frac{14pi}{3} )So,( frac{14pi}{3} + phi = frac{3pi}{2} )Subtract ( frac{14pi}{3} ) from both sides:( phi = frac{3pi}{2} - frac{14pi}{3} )To subtract these, I need a common denominator. The denominators are 2 and 3, so the common denominator is 6.Convert ( frac{3pi}{2} ) to sixths:( frac{3pi}{2} = frac{9pi}{6} )Convert ( frac{14pi}{3} ) to sixths:( frac{14pi}{3} = frac{28pi}{6} )Now subtract:( phi = frac{9pi}{6} - frac{28pi}{6} = frac{-19pi}{6} )Hmm, that's a negative angle. I wonder if I can express this as a positive angle by adding multiples of 2π. Let's see:( -19pi/6 + 2π = -19π/6 + 12π/6 = (-19 + 12)π/6 = -7π/6 )Still negative. Let's add another 2π:( -7π/6 + 12π/6 = 5π/6 )So, φ can be expressed as 5π/6. Alternatively, since sine is periodic, adding 2π to the angle doesn't change the value. So, φ is equivalent to 5π/6.Wait, let me double-check my calculations because I might have messed up somewhere.Starting from:( frac{pi}{3} times 14 + phi = frac{3pi}{2} )Which is:( frac{14pi}{3} + phi = frac{3pi}{2} )So, φ = ( frac{3pi}{2} - frac{14pi}{3} )Convert to sixths:( frac{9pi}{6} - frac{28pi}{6} = -frac{19pi}{6} )Yes, that's correct. So, φ is -19π/6. But since sine has a period of 2π, we can add 2π to get an equivalent angle.-19π/6 + 2π = -19π/6 + 12π/6 = -7π/6Still negative. Add another 2π:-7π/6 + 12π/6 = 5π/6So, φ is 5π/6.Alternatively, since sine is an odd function, sin(θ + φ) = sin(θ - φ + π), but I think the way I did it is correct.Wait, another thought: maybe I can express φ as a positive angle by adding 2π until it's within 0 to 2π.So, starting with φ = -19π/6.Add 2π (which is 12π/6):-19π/6 + 12π/6 = -7π/6Still negative. Add another 12π/6:-7π/6 + 12π/6 = 5π/6Yes, so φ is 5π/6.Alternatively, if we consider that the phase shift is φ, but in the equation, it's ( Bt + phi ), so the phase shift is -φ/B. But maybe that's complicating things.Wait, perhaps I should think about it differently. The general sine function is ( A sin(B(t - C)) + D ), where C is the phase shift. So, in our case, it's ( A sin(Bt + phi) + D = A sin(B(t + phi/B)) + D ). So, the phase shift is -φ/B.But maybe that's overcomplicating. Since we know that the minimum occurs at t = 14, and the sine function reaches its minimum at 3π/2, so setting ( Bt + phi = 3π/2 ) when t =14.So, yes, that's how I approached it earlier.So, φ = 3π/2 - Bt = 3π/2 - (π/3)(14) = 3π/2 - 14π/3.Calculating that, as before, gives φ = -19π/6, which is equivalent to 5π/6 when adding 2π.So, φ is 5π/6.Let me verify this.If φ is 5π/6, then the function is:( C(t) = 20 sinleft(frac{pi}{3} t + frac{5pi}{6}right) + 30 )Let me check at t =14:( C(14) = 20 sinleft(frac{pi}{3} times 14 + frac{5pi}{6}right) + 30 )Calculate the argument:( frac{14pi}{3} + frac{5pi}{6} = frac{28pi}{6} + frac{5pi}{6} = frac{33pi}{6} = frac{11pi}{2} )But 11π/2 is equivalent to 11π/2 - 4π = 11π/2 - 8π/2 = 3π/2.So, sin(3π/2) = -1, which gives:C(14) = 20*(-1) + 30 = -20 + 30 = 10, which is the minimum. Perfect, that checks out.So, φ is indeed 5π/6.Therefore, the complete function is:( C(t) = 20 sinleft(frac{pi}{3} t + frac{5pi}{6}right) + 30 )Let me just recap:1. A = 20, D = 30.2. B = π/3, φ = 5π/6.So, the function is as above.I think that's all. Let me just make sure I didn't make any calculation errors.Calculating A: (50-10)/2 = 20. Correct.Calculating D: (50+10)/2 = 30. Correct.Period is 6, so 2π/B =6 => B= π/3. Correct.Then, for φ, using t=14, which is 2 PM, the minimum occurs. So, setting the argument to 3π/2.So, (π/3)*14 + φ = 3π/2.14π/3 + φ = 3π/2.φ = 3π/2 -14π/3.Convert to sixths:3π/2 = 9π/6, 14π/3 =28π/6.So, 9π/6 -28π/6= -19π/6.Adding 2π (12π/6) gives -7π/6, still negative. Add another 12π/6: 5π/6. Correct.So, φ=5π/6.Therefore, the function is correctly determined.Final Answer1. ( A = boxed{20} ) and ( D = boxed{30} ).2. ( B = boxed{dfrac{pi}{3}} ) and the complete function is ( C(t) = 20 sinleft(dfrac{pi}{3} t + dfrac{5pi}{6}right) + 30 ). The value of ( phi ) is ( boxed{dfrac{5pi}{6}} ).</think>

🧠:<think>Okay, so I have this problem where a botanical hobbyist wants to plant two types of plants in their garden. The garden is 20 meters by 15 meters. Type A plants need 1 meter in the row direction and 2 meters in the column direction. Type B plants need 2 meters in the row direction and 1 meter in the column direction. The goal is to maximize the number of each type without overlapping.First, let me try to understand the problem. The garden is a rectangle, so I can imagine it as a grid where each plant takes up a certain amount of space in both the row and column directions. For Type A, each plant needs 1 meter along the length (which I'll consider as the row direction) and 2 meters along the width (column direction). For Type B, it's the opposite: 2 meters along the length and 1 meter along the width.I think the first step is to figure out how many Type A plants can fit in the garden. Then, with the remaining space, figure out how many Type B plants can be planted.Starting with Sub-problem 1: Maximum number of Type A plants.Type A requires 1m in the row direction and 2m in the column direction. The garden is 20m long and 15m wide. So, in the row direction (20m), each Type A plant takes 1m, so we can fit 20 / 1 = 20 plants along the length. In the column direction (15m), each Type A plant takes 2m, so we can fit 15 / 2 = 7.5, but since we can't have half a plant, we take the floor, which is 7 plants along the width.Wait, but actually, when planting in a grid, the number of plants is determined by how many fit in each dimension. So, if the garden is 20m by 15m, and each Type A plant needs 1m in row and 2m in column, then the number of plants along the row is 20 / 1 = 20, and along the column is 15 / 2 = 7.5, which we round down to 7. So the total number of Type A plants is 20 * 7 = 140.But wait, is that the maximum? Because sometimes, depending on the arrangement, you might be able to fit more by rotating or adjusting the grid, but in this case, since the spacing is fixed, I think 140 is correct.Now, moving on to Sub-problem 2: After planting Type A, how many Type B can we fit in the remaining space.First, let's visualize the garden after planting Type A. Each Type A occupies 1m in row and 2m in column. So, if we have 20 plants along the row and 7 along the column, the total area used by Type A is 20 * 7 * (1*2) = 280 square meters. The total area of the garden is 20*15=300 square meters. So, remaining area is 300 - 280 = 20 square meters.But Type B plants require 2m in row and 1m in column, so each takes 2*1=2 square meters. So, theoretically, we could fit 20 / 2 = 10 Type B plants. But it's not just about area; we have to consider the spacing and arrangement.Wait, but the remaining space isn't necessarily a continuous block. After planting Type A, the remaining space might be fragmented. So, I need to figure out how the Type A plants are arranged and then see where Type B can fit.Type A plants are arranged in rows and columns. Each row is 20m long, and each column is 15m wide. But since each Type A takes 2m in the column direction, the columns are spaced 2m apart. So, the first Type A plant is at 0m, the next at 2m, and so on, up to 14m (since 14 + 2 = 16, which is beyond 15m). So, there are 7 columns of Type A plants, each 2m apart, starting at 0m, 2m, 4m, ..., 14m.Similarly, in the row direction, each Type A is 1m apart, so they start at 0m, 1m, 2m, ..., 19m.So, the garden is divided into a grid where each cell is 1m by 2m, occupied by Type A plants.Now, to plant Type B, which requires 2m in the row direction and 1m in the column direction. So, each Type B plant needs a space that is 2m long and 1m wide.Looking at the remaining space, which is the area not occupied by Type A. Since Type A is planted in a grid of 1m (row) x 2m (column), the remaining spaces would be the gaps between these grids.Wait, actually, since Type A is planted in every 1m along the row, there are no gaps in the row direction. Similarly, in the column direction, Type A is planted every 2m, so between each Type A column, there is a 0m gap because 2m spacing is exact. Wait, no, the spacing is the distance between plants, so if each Type A is spaced 2m apart in the column direction, then the total width used is 7*2m = 14m, leaving 1m unused in the column direction.Similarly, in the row direction, Type A is spaced 1m apart, so 20 plants take up 20m, which is the full length, so no space is left in the row direction.Therefore, the remaining space is 1m in the column direction (since 15m total - 14m used = 1m). So, the remaining area is a strip of 20m (row) by 1m (column).Now, Type B plants require 2m in the row direction and 1m in the column direction. So, in the remaining 20m x 1m strip, how many Type B can we fit?In the row direction, each Type B needs 2m, so 20 / 2 = 10. In the column direction, each Type B needs 1m, and we have exactly 1m available. So, we can fit 10 Type B plants in this strip.But wait, is that the only remaining space? Because Type A is planted in columns every 2m, starting at 0m, 2m, 4m, ..., 14m. So, the remaining space is the strip from 14m to 15m in the column direction, which is 1m wide, and spans the entire 20m in the row direction.Therefore, in this 20m x 1m strip, we can plant Type B plants. Each Type B needs 2m in the row and 1m in the column. So, along the row, we can fit 20 / 2 = 10 plants. Along the column, we have exactly 1m, so we can fit 1 plant. Therefore, total Type B plants in this strip is 10 * 1 = 10.But wait, is there any other space left? Because Type A is planted in a grid, but maybe there are other areas where Type B can be planted. For example, between the columns of Type A, but since Type A is spaced 2m apart, and Type B requires 1m in the column direction, maybe we can interleave them.Wait, no, because Type A is already occupying every 2m in the column direction, leaving only 1m at the end. So, the only remaining space is that 1m strip at the end. Therefore, the maximum number of Type B plants is 10.But let me double-check. The total area used by Type A is 140 plants * 1m * 2m = 280 square meters. The remaining area is 20 square meters. Each Type B plant uses 2m * 1m = 2 square meters. So, 20 / 2 = 10 plants. That matches.Alternatively, maybe there's a different arrangement where we can fit more Type B plants by not planting all Type A first. But the problem specifies that we first plant Type A to maximum, then plant Type B in the remaining space. So, we have to follow that order.Wait, but maybe the initial arrangement of Type A isn't optimal in terms of leaving space for Type B. For example, if we arrange Type A in a different pattern, perhaps we can leave more space for Type B, but the problem says to maximize Type A first, then Type B in the remaining space. So, we have to go with the maximum Type A first.Therefore, the answers are:Sub-problem 1: 140 Type A plants.Sub-problem 2: 10 Type B plants.But let me think again. Maybe I made a mistake in the arrangement. If Type A is planted in rows and columns, each Type A is 1m apart in the row and 2m apart in the column. So, the grid is 1m x 2m. Therefore, the garden is divided into 20 columns (1m each) and 7 rows (2m each), with 1m remaining in the column direction.So, the remaining space is 20m x 1m. In this space, Type B needs 2m in the row and 1m in the column. So, along the row, we can fit 10 Type B plants (20 / 2), and along the column, we have exactly 1m, so 1 plant. So, total 10 Type B.Yes, that seems correct.Alternatively, could we rotate the garden? Like, consider the row as 15m and column as 20m? But the problem specifies row direction as 20m and column as 15m, so I think we have to stick with that.Another thought: Maybe the Type B plants can be arranged in the remaining space in a different orientation. For example, if Type B requires 2m in row and 1m in column, but maybe we can rotate them to take 1m in row and 2m in column. But the problem states the spacing requirements, so I think the direction is fixed. Type A is 1m row, 2m column; Type B is 2m row, 1m column.Therefore, I think my initial conclusion is correct.So, final answers:Sub-problem 1: 140 Type A plants.Sub-problem 2: 10 Type B plants.</think>

🧠:<think>Alright, so I've got this problem about a chef creating fruit drinks with specific percentages of apples, oranges, and berries. The goal is to figure out how much of each fruit to use and then calculate the total nutrient content. Then, there's a second part where the chef wants to maximize the nutrient content while keeping each fruit at least 10% of the drink. Hmm, okay, let's break this down step by step.Starting with part 1. The total volume is 1000 ml, and the percentages are given: 30% apples, 50% oranges, and 20% berries. So, I think I need to calculate the volume of each fruit based on these percentages. For apples, 30% of 1000 ml should be straightforward. 30% is the same as 0.3, so 0.3 * 1000 = 300 ml. That seems right. Next, oranges are 50%, so that's 0.5 * 1000 = 500 ml. Makes sense, since 50% is a common percentage and half of 1000 is 500.Then, berries are 20%, so 0.2 * 1000 = 200 ml. Okay, so the volumes are 300 ml apples, 500 ml oranges, and 200 ml berries. Let me just check that these add up to 1000 ml: 300 + 500 is 800, plus 200 is 1000. Perfect, that checks out.Now, calculating the total nutrient content. Each fruit contributes a different amount per ml. Apples are 3 mg/ml, oranges are 2 mg/ml, and berries are 5 mg/ml. So, I need to multiply each volume by their respective nutrient concentration and then sum them up.For apples: 300 ml * 3 mg/ml. Let me compute that. 300 * 3 is 900 mg.Oranges: 500 ml * 2 mg/ml. That's 500 * 2 = 1000 mg.Berries: 200 ml * 5 mg/ml. That's 200 * 5 = 1000 mg.Now, adding them all together: 900 + 1000 + 1000. Let's see, 900 + 1000 is 1900, plus another 1000 is 2900 mg. So, the total nutrient content is 2900 mg. That seems like a lot, but considering the high concentration in berries, which are 20%, it might add up.Moving on to part 2. The chef wants to create a second drink with the same total volume, 1000 ml, but now wants to maximize the nutrient content. However, each fruit has to be at least 10% of the total volume. So, we need to set up an optimization problem.First, let's define variables. Let me denote the percentage of apples as A, oranges as O, and berries as B. Since percentages are involved, A, O, B are all between 10% and 90%, but their sum must be 100%.Wait, actually, the problem says each type of fruit is at least 10%, so A >= 10%, O >= 10%, B >= 10%. But since the total is 100%, each can be at most 80%, because the other two have to be at least 10% each. So, A, O, B are each in [10, 80] and A + O + B = 100.But actually, the percentages can be more than 10%, but each must be at least 10%. So, the constraints are A >= 10, O >= 10, B >= 10, and A + O + B = 100.We need to maximize the total nutrient content. The nutrient content is calculated as (A/100)*1000*3 + (O/100)*1000*2 + (B/100)*1000*5. Wait, that's the same as (A*3 + O*2 + B*5) * 10 mg. Because (A/100)*1000 is 10A, so 10A*3 is 30A, similarly 20O and 50B. Wait, actually, let me compute it properly.Wait, no, perhaps I should think in terms of the volumes. Since the total volume is 1000 ml, the volume of apples is (A/100)*1000 = 10A ml. Similarly, oranges are 10O ml, berries are 10B ml. Then, the nutrient content is 10A*3 + 10O*2 + 10B*5. Which simplifies to 30A + 20O + 50B mg.So, the objective function to maximize is 30A + 20O + 50B, subject to A + O + B = 100, and A >=10, O >=10, B >=10.Alternatively, since A + O + B = 100, we can express one variable in terms of the others. For example, B = 100 - A - O. Then, substitute into the objective function.So, substituting B, the objective becomes 30A + 20O + 50*(100 - A - O). Let's compute that:30A + 20O + 5000 - 50A - 50O = (30A - 50A) + (20O - 50O) + 5000 = (-20A) + (-30O) + 5000.So, the objective function simplifies to -20A -30O + 5000. We need to maximize this. Since the coefficients of A and O are negative, to maximize the expression, we need to minimize A and O as much as possible.But we have constraints: A >=10, O >=10, and since B = 100 - A - O >=10, which implies A + O <=90.So, to minimize A and O, set A =10, O=10, then B=80. Let's check if that satisfies all constraints: A=10, O=10, B=80. All are >=10, and A + O + B =100. Perfect.So, substituting A=10, O=10, B=80 into the objective function: -20*10 -30*10 +5000 = -200 -300 +5000 = -500 +5000 = 4500 mg.Is that the maximum? Let me think. Since the coefficients of A and O are negative, any increase in A or O would decrease the total nutrient content. So, yes, minimizing A and O (to their lower bounds) would maximize the nutrient content.Alternatively, if I didn't substitute, the original objective was 30A + 20O + 50B. Since berries have the highest nutrient concentration, we want to maximize the percentage of berries, which is achieved by minimizing the percentages of apples and oranges, which have lower nutrient concentrations.So, the optimal solution is A=10%, O=10%, B=80%, resulting in a total nutrient content of 4500 mg.Wait, let me verify that. If A=10, O=10, B=80, then:Volume of apples: 100 ml, nutrient: 100*3=300 mg.Oranges: 100 ml, nutrient: 100*2=200 mg.Berries: 800 ml, nutrient: 800*5=4000 mg.Total nutrient: 300 + 200 + 4000 = 4500 mg. Yep, that's correct.Alternatively, if I tried other combinations, say A=20, O=10, B=70:Apples: 200 ml, 600 mg.Oranges: 100 ml, 200 mg.Berries: 700 ml, 3500 mg.Total: 600 + 200 + 3500 = 4300 mg, which is less than 4500.Similarly, A=10, O=20, B=70:Apples: 100 ml, 300 mg.Oranges: 200 ml, 400 mg.Berries: 700 ml, 3500 mg.Total: 300 + 400 + 3500 = 4200 mg, still less.So, indeed, the maximum is achieved when A and O are at their minimum, and B is maximized.Therefore, the new percentages are 10% apples, 10% oranges, 80% berries, resulting in 4500 mg of the nutrient.I think that's it. Let me just recap:1. For the first drink, volumes are 300 ml apples, 500 ml oranges, 200 ml berries, total nutrient 2900 mg.2. For the second drink, to maximize nutrient, set apples and oranges to 10% each, berries to 80%, resulting in 4500 mg.Yeah, that makes sense. I don't see any mistakes in the calculations.</think>