🧠:<think>Alright, so I have this problem about leadership circles in Bayelsa, which is modeled as a graph. There are two parts to the problem. Let me try to tackle them one by one.Starting with part 1: We need to find the number of distinct ways to select a subset of edges that forms a spanning tree. The function is given as f(n, m), and we need to express it using the Laplacian matrix and the Matrix Tree Theorem.Hmm, okay. I remember that the Matrix Tree Theorem relates the number of spanning trees in a graph to the determinant of a certain matrix derived from the Laplacian matrix. Let me recall the exact statement.The Laplacian matrix, L, of a graph is defined as D - A, where D is the degree matrix and A is the adjacency matrix. The Matrix Tree Theorem says that the number of spanning trees is equal to any cofactor of the Laplacian matrix. That is, if you remove any row and the corresponding column, the determinant of the resulting matrix gives the number of spanning trees.So, if we denote the Laplacian matrix as L, then f(n, m) is equal to the determinant of any (n-1)x(n-1) principal minor of L. So, mathematically, f(n, m) = det(L'), where L' is L with one row and column removed.Wait, but the problem says to express f(n, m) in terms of the Laplacian matrix. So, maybe it's just the determinant of a certain minor? Yeah, that seems right.So, part 1 is about applying the Matrix Tree Theorem, which gives us f(n, m) as the determinant of any cofactor of the Laplacian matrix. I think that's the answer for part 1.Moving on to part 2: We have a policy where at least half of the leadership circles need to approve a new initiative. So, for a random subset S of nodes with size |S| = ceil(n/2), we need to find the expected value of g(S), where g(S) is the number of edges with both endpoints in S.Okay, so g(S) is essentially the number of edges in the induced subgraph of S. We need to compute E[g(S)] where S is a random subset of size ceil(n/2).Let me think about how to compute this expectation. Since expectation is linear, maybe we can compute the expected number of edges by summing over all edges and computing the probability that both endpoints are in S.Yes, that sounds right. So, for each edge e in the graph, let X_e be the indicator random variable which is 1 if both endpoints of e are in S, and 0 otherwise. Then, g(S) = sum_{e in E} X_e. Therefore, E[g(S)] = sum_{e in E} E[X_e].Since all edges are treated the same in expectation, E[X_e] is the same for each edge. So, E[g(S)] = m * E[X_e], where E[X_e] is the probability that both endpoints of a given edge are in S.So, what is E[X_e]? It's the probability that both endpoints are selected in S. Since S is a random subset of size k = ceil(n/2), the probability that both endpoints are in S is C(n-2, k-2)/C(n, k).Let me compute that. The number of ways to choose S such that both endpoints are included is C(n-2, k-2). The total number of ways to choose S is C(n, k). So, E[X_e] = C(n-2, k-2)/C(n, k).Simplify this expression. Let's compute C(n-2, k-2)/C(n, k). C(n-2, k-2) = (n-2)! / [(k-2)! (n - k)!]C(n, k) = n! / [k! (n - k)!]So, the ratio is [(n-2)! / (k-2)! (n - k)!)] / [n! / (k! (n - k)!)] = [ (n-2)! k! ] / [n! (k-2)! ] Simplify numerator and denominator:(n-2)! / n! = 1 / [n(n-1)]k! / (k-2)! = k(k - 1)So, putting it together: [1 / (n(n-1))] * [k(k - 1)] = [k(k - 1)] / [n(n - 1)]Therefore, E[X_e] = [k(k - 1)] / [n(n - 1)]Since k = ceil(n/2), let's write that as k = ⎡n/2⎤.Therefore, E[g(S)] = m * [k(k - 1)] / [n(n - 1)]So, substituting k = ceil(n/2), the expected number of edges is m multiplied by [ceil(n/2)(ceil(n/2) - 1)] divided by [n(n - 1)].Let me check if this makes sense. For example, if n is even, say n=4, then k=2. Then, E[g(S)] = m * [2*1]/[4*3] = m*(2/12) = m/6. Hmm, is that correct?Wait, for n=4, the number of possible subsets S of size 2 is 6. Each edge has two endpoints, so the probability that both are in S is C(2,2)/C(4,2) = 1/6. So, for each edge, the probability is 1/6, so the expected number of edges is m*(1/6). That seems correct.Similarly, if n=5, then k=3. Then, E[g(S)] = m * [3*2]/[5*4] = m*(6/20) = (3/10)m. Let's see, for n=5, the number of subsets S of size 3 is 10. For a given edge, the number of subsets containing both endpoints is C(3,1) = 3, since we need to choose 1 more node from the remaining 3. So, probability is 3/10, which matches. So, that seems correct.Therefore, the formula seems to hold.So, summarizing:1. The number of spanning trees is given by the determinant of any (n-1)x(n-1) principal minor of the Laplacian matrix, by the Matrix Tree Theorem.2. The expected value of g(S) is m multiplied by [k(k - 1)] / [n(n - 1)], where k = ceil(n/2).So, I think that's the solution.Final Answer1. The number of spanning trees is given by the determinant of a cofactor of the Laplacian matrix, so ( f(n, m) = boxed{det(L')} ) where ( L' ) is any ((n-1) times (n-1)) principal minor of the Laplacian matrix.2. The expected value of ( g(S) ) is ( boxed{frac{m cdot lceil frac{n}{2} rceil cdot left( lceil frac{n}{2} rceil - 1 right)}{n(n - 1)}} ).</think>

🧠:<think>First, I need to calculate the total kilocalories burned in a week based on the current sedentary lifestyle. The individual spends 12 hours sitting and 12 hours sleeping each day.For sitting, the calorie burn rate is 1.2 kilocalories per minute. Converting 12 hours to minutes gives 720 minutes. Multiplying 720 by 1.2 results in 864 kilocalories burned per day from sitting.For sleeping, the calorie burn rate is 0.9 kilocalories per minute. Converting 12 hours to minutes also gives 720 minutes. Multiplying 720 by 0.9 results in 648 kilocalories burned per day from sleeping.Adding these together, the total daily kilocalories burned is 864 plus 648, which equals 1,512 kilocalories per day. Over a week, this amounts to 1,512 multiplied by 7, totaling 10,584 kilocalories.Next, if the individual decides to exercise for 1 hour a day, burning an additional 8 kilocalories per minute, I need to calculate the new weekly kilocalories burned. One hour is 60 minutes, so the additional calories burned per day from exercise are 60 multiplied by 8, which equals 480 kilocalories.Adding this to the previous daily total of 1,512 kilocalories gives a new daily total of 1,992 kilocalories. Over a week, this amounts to 1,992 multiplied by 7, totaling 13,944 kilocalories.To find the percentage increase in total weekly kilocalories burned, I subtract the original weekly total from the new total: 13,944 minus 10,584 equals 3,360 kilocalories. Dividing this difference by the original total (10,584) and multiplying by 100 gives the percentage increase: (3,360 / 10,584) * 100 ≈ 31.75%.</think>

🧠:<think>Okay, so I have this problem where a parent is analyzing how temperature affects their child's outdoor playtime. They've given me a quadratic model, P(T) = aT² + bT + c, and some data points. I need to find the coefficients a, b, and c. Then, in part 2, I have to calculate the expected playtime when temperature follows a normal distribution with mean 25°C and standard deviation 5°C.Starting with part 1. They've given me three data points: when T is 20, P is 2; when T is 25, P is 3; and when T is 30, P is 1. So, I can set up three equations based on these points and solve for a, b, and c.Let me write down the equations:1. When T = 20, P = 2: 2 = a*(20)² + b*(20) + c Simplify that: 2 = 400a + 20b + c2. When T = 25, P = 3: 3 = a*(25)² + b*(25) + c Simplify: 3 = 625a + 25b + c3. When T = 30, P = 1: 1 = a*(30)² + b*(30) + c Simplify: 1 = 900a + 30b + cSo now I have a system of three equations:1. 400a + 20b + c = 22. 625a + 25b + c = 33. 900a + 30b + c = 1I need to solve for a, b, c. Let me write them down again:Equation 1: 400a + 20b + c = 2Equation 2: 625a + 25b + c = 3Equation 3: 900a + 30b + c = 1I can solve this system using elimination. Let me subtract Equation 1 from Equation 2 to eliminate c.Equation 2 - Equation 1:(625a - 400a) + (25b - 20b) + (c - c) = 3 - 2That gives:225a + 5b = 1Let me call this Equation 4: 225a + 5b = 1Similarly, subtract Equation 2 from Equation 3:(900a - 625a) + (30b - 25b) + (c - c) = 1 - 3Which simplifies to:275a + 5b = -2Let me call this Equation 5: 275a + 5b = -2Now, I have Equations 4 and 5:Equation 4: 225a + 5b = 1Equation 5: 275a + 5b = -2Subtract Equation 4 from Equation 5:(275a - 225a) + (5b - 5b) = -2 - 1That gives:50a = -3So, a = -3 / 50 = -0.06Now, plug a back into Equation 4 to find b.Equation 4: 225a + 5b = 1225*(-0.06) + 5b = 1Calculate 225 * (-0.06):225 * 0.06 = 13.5, so 225*(-0.06) = -13.5So:-13.5 + 5b = 1Add 13.5 to both sides:5b = 1 + 13.5 = 14.5So, b = 14.5 / 5 = 2.9Hmm, 14.5 divided by 5 is 2.9, which is 29/10 or 2.9.Now, with a and b known, plug back into Equation 1 to find c.Equation 1: 400a + 20b + c = 2400*(-0.06) + 20*(2.9) + c = 2Calculate each term:400*(-0.06) = -2420*(2.9) = 58So:-24 + 58 + c = 2Simplify:34 + c = 2So, c = 2 - 34 = -32Therefore, the coefficients are:a = -0.06, b = 2.9, c = -32Wait, let me double-check these calculations because sometimes when dealing with decimals, it's easy to make a mistake.First, a = -3/50 is indeed -0.06.Then, plugging into Equation 4:225*(-0.06) = -13.5-13.5 + 5b = 1 => 5b = 14.5 => b = 2.9. That seems correct.Then, plugging into Equation 1:400*(-0.06) = -2420*(2.9) = 58-24 + 58 = 3434 + c = 2 => c = -32. Correct.So, the quadratic model is P(T) = -0.06T² + 2.9T - 32.Wait, let me check if this model fits the given data points.For T = 20:P(20) = -0.06*(400) + 2.9*(20) -32= -24 + 58 -32= (-24 -32) +58 = (-56) +58 = 2. Correct.For T =25:P(25) = -0.06*(625) + 2.9*(25) -32= -37.5 + 72.5 -32= (-37.5 -32) +72.5 = (-69.5) +72.5 = 3. Correct.For T =30:P(30) = -0.06*(900) + 2.9*(30) -32= -54 + 87 -32= (-54 -32) +87 = (-86) +87 = 1. Correct.Okay, so the coefficients are correct.So, part 1 is done: a = -0.06, b = 2.9, c = -32.Moving on to part 2. The parent wants to find the expected value of outdoor playtime when temperature T follows a normal distribution with mean μ =25°C and standard deviation σ=5°C.So, we need to compute E[P(T)] where P(T) = -0.06T² + 2.9T -32, and T ~ N(25, 5²).Since expectation is linear, we can write:E[P(T)] = -0.06 E[T²] + 2.9 E[T] -32We know that for a normal distribution, E[T] = μ =25.E[T²] can be found using Var(T) = E[T²] - (E[T])²We know Var(T) = σ² =25, so:25 = E[T²] - (25)²Thus, E[T²] =25 + 625 =650Therefore, plug back into E[P(T)]:E[P(T)] = -0.06*(650) + 2.9*(25) -32Calculate each term:-0.06*650 = -392.9*25 =72.5So:-39 +72.5 -32 = (-39 -32) +72.5 = (-71) +72.5 =1.5So, the expected playtime is 1.5 hours.Wait, let me verify the calculations step by step.First, E[T] =25, correct.Var(T) =25, so E[T²] = Var(T) + (E[T])² =25 +625=650, correct.Then, E[P(T)] = -0.06*650 +2.9*25 -32.Compute each term:-0.06*650: 0.06*600=36, 0.06*50=3, so total 36+3=39, so -0.06*650=-39.2.9*25: 2*25=50, 0.9*25=22.5, so total 50+22.5=72.5.Then, -39 +72.5 -32.Compute step by step:-39 +72.5 =33.533.5 -32=1.5Yes, that's correct.So, the expected playtime is 1.5 hours.Therefore, the answers are:1. a = -0.06, b =2.9, c =-322. Expected playtime is 1.5 hours.But wait, let me think again about part 2. Since P(T) is a quadratic function, and T is normally distributed, is there a different approach?Wait, no, because expectation is linear, regardless of the distribution. So, even if P(T) is quadratic, E[P(T)] = a E[T²] + b E[T] + c.So, that approach is correct.Alternatively, if I had to compute E[P(T)] as an integral over the normal distribution, but that would be more complicated and unnecessary since we can use the linearity of expectation.So, I think my approach is correct.Just to recap:E[P(T)] = a E[T²] + b E[T] + cWe found E[T] =25, E[T²] =650So, plug in a=-0.06, b=2.9, c=-32:E[P(T)] = (-0.06)(650) +2.9(25) + (-32) = -39 +72.5 -32 =1.5.Yes, that's correct.So, I think I'm confident with these results.Final Answer1. The values of the constants are ( a = boxed{-0.06} ), ( b = boxed{2.9} ), and ( c = boxed{-32} ).2. The expected value of the outdoor playtime is ( boxed{1.5} ) hours.</think>

🧠:<think>Alright, so I'm trying to solve this problem about a university student planning their trip to five different countries. They need to exchange their British Pounds (GBP) into each country's local currency and manage their budget considering inflation. Let me break this down step by step.Starting with part 1. The student has a budget of B GBP and wants to exchange this into five different currencies. Each country has its own exchange rate, denoted as r1, r2, r3, r4, and r5. The function f(i) represents the amount of local currency they'll have in country i after exchanging GBP. So, if I think about it, the amount of local currency they get is just the amount of GBP they exchange multiplied by the exchange rate. But wait, the student wants to spend the same amount S in each country. That means they need to figure out how much GBP to exchange into each currency so that when multiplied by the exchange rate, it equals S. Let me write that out. For each country i, the amount of local currency they have is f(i) = x_i * r_i, where x_i is the amount of GBP exchanged into country i's currency. But since they want to spend the same S in each country, f(i) should be equal to S. So, x_i = S / r_i. But the total amount of GBP they exchange, T, is the sum of all x_i from i=1 to 5. So, T = x1 + x2 + x3 + x4 + x5. Substituting x_i, we get T = (S / r1) + (S / r2) + (S / r3) + (S / r4) + (S / r5). I can factor out the S, so T = S * (1/r1 + 1/r2 + 1/r3 + 1/r4 + 1/r5). That makes sense because the total GBP needed depends on how much they want to spend in each country and the exchange rates. If the exchange rates are higher (more local currency per GBP), they need less GBP for the same spending.Moving on to part 2. Now, the student has to consider inflation in each country. The inflation rates are I1 to I5, compounded continuously, and they stay in each country for t_i days. I remember that continuous compounding uses the formula A = P * e^(rt), where P is the principal amount, r is the rate, and t is time. So, in this case, the amount of local currency they effectively have after inflation would be A_i = f(i) * e^(-I_i * t_i). Wait, why negative? Because inflation reduces the purchasing power, so the effective amount decreases. So, it's exponential decay. Therefore, for each country, A_i = S * e^(-I_i * t_i). Since they spend S in each country, but inflation affects the value, the effective amount is less. To find the total effective amount across all five countries, I just sum up all A_i. So, total A = A1 + A2 + A3 + A4 + A5 = S * [e^(-I1*t1) + e^(-I2*t2) + e^(-I3*t3) + e^(-I4*t4) + e^(-I5*t5)]. Let me double-check if I got the signs right. If inflation is positive, then e^(-I*t) should be less than 1, which reduces the amount. That seems correct. So, the total effective amount is the sum of each country's adjusted spending.Wait, but in part 1, T was the total GBP needed, which is S times the sum of reciprocals of exchange rates. Then in part 2, the total effective local currency is S times the sum of exponential terms. I think that's correct because each country's spending is adjusted individually for inflation.I don't see any mistakes in the reasoning. The key was recognizing that for each country, the spending is S, so the exchange amount is S divided by the rate, and then summing those up for the total GBP. For inflation, applying continuous compounding correctly with the negative exponent to account for the decrease in value.Final Answer1. The total amount ( T ) of GBP needed is ( boxed{T = S left( frac{1}{r_1} + frac{1}{r_2} + frac{1}{r_3} + frac{1}{r_4} + frac{1}{r_5} right)} ).2. The total effective amount of local currency across all countries is ( boxed{A = S left( e^{-I_1 t_1} + e^{-I_2 t_2} + e^{-I_3 t_3} + e^{-I_4 t_4} + e^{-I_5 t_5} right)} ).</think>