🧠:<think>Alright, so I have two problems to solve here. Let me tackle them one by one.Problem 1: Finding when Alex's and Jamie's recovery percentages are equalOkay, so Alex's recovery is modeled by ( A(t) = 100 cdot e^{-0.05t} + 20 sin(0.1t) ) and Jamie's by ( J(t) = 90 cdot e^{-0.08t} + 30 cos(0.1t) ). I need to find the first time ( t ) when ( A(t) = J(t) ).Hmm, setting them equal:( 100 cdot e^{-0.05t} + 20 sin(0.1t) = 90 cdot e^{-0.08t} + 30 cos(0.1t) )This looks like a transcendental equation, which probably can't be solved algebraically. I think I'll need to use numerical methods or graphing to find the solution.Maybe I can rearrange terms:( 100 cdot e^{-0.05t} - 90 cdot e^{-0.08t} + 20 sin(0.1t) - 30 cos(0.1t) = 0 )Let me denote this as ( f(t) = 100e^{-0.05t} - 90e^{-0.08t} + 20sin(0.1t) - 30cos(0.1t) ). I need to find the smallest positive ( t ) where ( f(t) = 0 ).I can try plugging in some values to approximate where this might happen.Let me compute ( f(t) ) at several points:First, at ( t = 0 ):( f(0) = 100 - 90 + 0 - 30 = 10 - 30 = -20 )Negative.At ( t = 10 ):Compute each term:( 100e^{-0.5} ≈ 100 * 0.6065 ≈ 60.65 )( 90e^{-0.8} ≈ 90 * 0.4493 ≈ 40.44 )( 20sin(1) ≈ 20 * 0.8415 ≈ 16.83 )( 30cos(1) ≈ 30 * 0.5403 ≈ 16.21 )So ( f(10) ≈ 60.65 - 40.44 + 16.83 - 16.21 ≈ (60.65 - 40.44) + (16.83 - 16.21) ≈ 20.21 + 0.62 ≈ 20.83 )Positive.So between t=0 and t=10, f(t) crosses from negative to positive. So the first root is between 0 and 10.Let me try t=5:( 100e^{-0.25} ≈ 100 * 0.7788 ≈ 77.88 )( 90e^{-0.4} ≈ 90 * 0.6703 ≈ 60.33 )( 20sin(0.5) ≈ 20 * 0.4794 ≈ 9.59 )( 30cos(0.5) ≈ 30 * 0.8776 ≈ 26.33 )So f(5) ≈ 77.88 - 60.33 + 9.59 - 26.33 ≈ (77.88 - 60.33) + (9.59 - 26.33) ≈ 17.55 - 16.74 ≈ 0.81Almost zero, but still positive.Wait, at t=5, f(t) ≈ 0.81, which is positive. But at t=0, it was -20. So the root is between 0 and 5.Wait, but at t=5, it's positive, so maybe the root is between 4 and 5?Wait, let me check t=4:Compute each term:( 100e^{-0.2} ≈ 100 * 0.8187 ≈ 81.87 )( 90e^{-0.32} ≈ 90 * 0.7261 ≈ 65.35 )( 20sin(0.4) ≈ 20 * 0.3894 ≈ 7.79 )( 30cos(0.4) ≈ 30 * 0.9211 ≈ 27.63 )So f(4) ≈ 81.87 - 65.35 + 7.79 - 27.63 ≈ (81.87 - 65.35) + (7.79 - 27.63) ≈ 16.52 - 19.84 ≈ -3.32Negative.So between t=4 and t=5, f(t) goes from -3.32 to +0.81. So the root is between 4 and 5.Let me try t=4.5:Compute each term:( 100e^{-0.225} ≈ 100 * e^{-0.225} ≈ 100 * 0.7985 ≈ 79.85 )( 90e^{-0.36} ≈ 90 * e^{-0.36} ≈ 90 * 0.6983 ≈ 62.85 )( 20sin(0.45) ≈ 20 * 0.4335 ≈ 8.67 )( 30cos(0.45) ≈ 30 * 0.9025 ≈ 27.08 )So f(4.5) ≈ 79.85 - 62.85 + 8.67 - 27.08 ≈ (79.85 - 62.85) + (8.67 - 27.08) ≈ 17 - 18.41 ≈ -1.41Still negative.t=4.75:( 100e^{-0.2375} ≈ 100 * e^{-0.2375} ≈ 100 * 0.789 ≈ 78.9 )( 90e^{-0.38} ≈ 90 * e^{-0.38} ≈ 90 * 0.684 ≈ 61.56 )( 20sin(0.475) ≈ 20 * 0.455 ≈ 9.1 )( 30cos(0.475) ≈ 30 * 0.891 ≈ 26.73 )f(4.75) ≈ 78.9 - 61.56 + 9.1 - 26.73 ≈ (78.9 - 61.56) + (9.1 - 26.73) ≈ 17.34 - 17.63 ≈ -0.29Still negative, but closer.t=4.9:( 100e^{-0.245} ≈ 100 * e^{-0.245} ≈ 100 * 0.782 ≈ 78.2 )( 90e^{-0.392} ≈ 90 * e^{-0.392} ≈ 90 * 0.674 ≈ 60.66 )( 20sin(0.49) ≈ 20 * 0.469 ≈ 9.38 )( 30cos(0.49) ≈ 30 * 0.888 ≈ 26.64 )f(4.9) ≈ 78.2 - 60.66 + 9.38 - 26.64 ≈ (78.2 - 60.66) + (9.38 - 26.64) ≈ 17.54 - 17.26 ≈ 0.28Positive.So between t=4.75 and t=4.9, f(t) crosses from negative to positive. Let's try t=4.85:Compute each term:( 100e^{-0.2425} ≈ 100 * e^{-0.2425} ≈ 100 * 0.784 ≈ 78.4 )( 90e^{-0.388} ≈ 90 * e^{-0.388} ≈ 90 * 0.678 ≈ 61.02 )( 20sin(0.485) ≈ 20 * 0.463 ≈ 9.26 )( 30cos(0.485) ≈ 30 * 0.895 ≈ 26.85 )f(4.85) ≈ 78.4 - 61.02 + 9.26 - 26.85 ≈ (78.4 - 61.02) + (9.26 - 26.85) ≈ 17.38 - 17.59 ≈ -0.21Still negative.t=4.875:( 100e^{-0.24375} ≈ 100 * e^{-0.24375} ≈ 100 * 0.783 ≈ 78.3 )( 90e^{-0.39} ≈ 90 * e^{-0.39} ≈ 90 * 0.676 ≈ 60.84 )( 20sin(0.4875) ≈ 20 * 0.465 ≈ 9.3 )( 30cos(0.4875) ≈ 30 * 0.894 ≈ 26.82 )f(4.875) ≈ 78.3 - 60.84 + 9.3 - 26.82 ≈ (78.3 - 60.84) + (9.3 - 26.82) ≈ 17.46 - 17.52 ≈ -0.06Almost zero, slightly negative.t=4.8875:( 100e^{-0.244375} ≈ 100 * e^{-0.244375} ≈ 100 * 0.782 ≈ 78.2 )( 90e^{-0.391} ≈ 90 * e^{-0.391} ≈ 90 * 0.675 ≈ 60.75 )( 20sin(0.48875) ≈ 20 * 0.466 ≈ 9.32 )( 30cos(0.48875) ≈ 30 * 0.893 ≈ 26.79 )f(4.8875) ≈ 78.2 - 60.75 + 9.32 - 26.79 ≈ (78.2 - 60.75) + (9.32 - 26.79) ≈ 17.45 - 17.47 ≈ -0.02Still negative.t=4.89:( 100e^{-0.2445} ≈ 100 * e^{-0.2445} ≈ 100 * 0.782 ≈ 78.2 )( 90e^{-0.3912} ≈ 90 * e^{-0.3912} ≈ 90 * 0.675 ≈ 60.75 )( 20sin(0.489) ≈ 20 * 0.466 ≈ 9.32 )( 30cos(0.489) ≈ 30 * 0.893 ≈ 26.79 )f(4.89) ≈ 78.2 - 60.75 + 9.32 - 26.79 ≈ same as above, ≈ -0.02Wait, maybe I need more precise calculations.Alternatively, perhaps using linear approximation between t=4.875 (-0.06) and t=4.9 (+0.28). The change is 0.34 over 0.025 weeks.We need to find t where f(t)=0.From t=4.875 (-0.06) to t=4.9 (+0.28). The difference in f(t) is 0.34 over 0.025 weeks.We need to cover 0.06 to reach zero from t=4.875.So fraction = 0.06 / 0.34 ≈ 0.176Thus, t ≈ 4.875 + 0.176*0.025 ≈ 4.875 + 0.0044 ≈ 4.8794 weeks.So approximately 4.88 weeks.Let me check t=4.88:Compute each term:( 100e^{-0.244} ≈ 100 * e^{-0.244} ≈ 100 * 0.782 ≈ 78.2 )( 90e^{-0.3904} ≈ 90 * e^{-0.3904} ≈ 90 * 0.675 ≈ 60.75 )( 20sin(0.488) ≈ 20 * 0.466 ≈ 9.32 )( 30cos(0.488) ≈ 30 * 0.893 ≈ 26.79 )So f(4.88) ≈ 78.2 - 60.75 + 9.32 - 26.79 ≈ (78.2 - 60.75) + (9.32 - 26.79) ≈ 17.45 - 17.47 ≈ -0.02Still negative. Maybe t=4.885:Compute each term:( 100e^{-0.24425} ≈ 100 * e^{-0.24425} ≈ 100 * 0.782 ≈ 78.2 )( 90e^{-0.3908} ≈ 90 * e^{-0.3908} ≈ 90 * 0.675 ≈ 60.75 )( 20sin(0.4885) ≈ 20 * 0.466 ≈ 9.32 )( 30cos(0.4885) ≈ 30 * 0.893 ≈ 26.79 )Same as above, so f(t) ≈ -0.02.Hmm, maybe my approximation is not precise enough. Alternatively, perhaps using a calculator or more precise computation.Alternatively, maybe I can use the Newton-Raphson method.Let me denote t as x.f(x) = 100e^{-0.05x} - 90e^{-0.08x} + 20sin(0.1x) - 30cos(0.1x)f'(x) = -5e^{-0.05x} + 7.2e^{-0.08x} + 2cos(0.1x) + 3sin(0.1x)Starting with x0 = 4.9, where f(x0) ≈ 0.28Compute f(4.9) ≈ 0.28Compute f'(4.9):First term: -5e^{-0.245} ≈ -5 * 0.782 ≈ -3.91Second term: 7.2e^{-0.392} ≈ 7.2 * 0.674 ≈ 4.85Third term: 2cos(0.49) ≈ 2 * 0.888 ≈ 1.776Fourth term: 3sin(0.49) ≈ 3 * 0.469 ≈ 1.407So f'(4.9) ≈ -3.91 + 4.85 + 1.776 + 1.407 ≈ (-3.91 + 4.85) + (1.776 + 1.407) ≈ 0.94 + 3.183 ≈ 4.123So Newton-Raphson update: x1 = x0 - f(x0)/f'(x0) ≈ 4.9 - 0.28 / 4.123 ≈ 4.9 - 0.068 ≈ 4.832Wait, but f(4.832) was negative. Hmm, maybe my derivative was miscalculated.Wait, let me recalculate f'(4.9):-5e^{-0.05*4.9} = -5e^{-0.245} ≈ -5*0.782 ≈ -3.917.2e^{-0.08*4.9} = 7.2e^{-0.392} ≈ 7.2*0.674 ≈ 4.852cos(0.1*4.9) = 2cos(0.49) ≈ 2*0.888 ≈ 1.7763sin(0.1*4.9) = 3sin(0.49) ≈ 3*0.469 ≈ 1.407So total f'(4.9) ≈ -3.91 + 4.85 + 1.776 + 1.407 ≈ (-3.91 + 4.85) + (1.776 + 1.407) ≈ 0.94 + 3.183 ≈ 4.123So x1 ≈ 4.9 - 0.28 / 4.123 ≈ 4.9 - 0.068 ≈ 4.832But f(4.832) was negative, so perhaps I need to try another iteration.Compute f(4.832):100e^{-0.05*4.832} ≈ 100e^{-0.2416} ≈ 100*0.784 ≈ 78.490e^{-0.08*4.832} ≈ 90e^{-0.38656} ≈ 90*0.678 ≈ 61.0220sin(0.1*4.832) ≈ 20sin(0.4832) ≈ 20*0.463 ≈ 9.2630cos(0.1*4.832) ≈ 30cos(0.4832) ≈ 30*0.895 ≈ 26.85f(4.832) ≈ 78.4 - 61.02 + 9.26 - 26.85 ≈ (78.4 - 61.02) + (9.26 - 26.85) ≈ 17.38 - 17.59 ≈ -0.21So f(4.832) ≈ -0.21Compute f'(4.832):-5e^{-0.05*4.832} ≈ -5e^{-0.2416} ≈ -5*0.784 ≈ -3.927.2e^{-0.08*4.832} ≈ 7.2e^{-0.38656} ≈ 7.2*0.678 ≈ 4.882cos(0.4832) ≈ 2*0.895 ≈ 1.793sin(0.4832) ≈ 3*0.463 ≈ 1.389f'(4.832) ≈ -3.92 + 4.88 + 1.79 + 1.389 ≈ (-3.92 + 4.88) + (1.79 + 1.389) ≈ 0.96 + 3.179 ≈ 4.139So x2 = x1 - f(x1)/f'(x1) ≈ 4.832 - (-0.21)/4.139 ≈ 4.832 + 0.0508 ≈ 4.8828Compute f(4.8828):100e^{-0.05*4.8828} ≈ 100e^{-0.24414} ≈ 100*0.782 ≈ 78.290e^{-0.08*4.8828} ≈ 90e^{-0.390624} ≈ 90*0.675 ≈ 60.7520sin(0.48828) ≈ 20*0.466 ≈ 9.3230cos(0.48828) ≈ 30*0.893 ≈ 26.79f(4.8828) ≈ 78.2 - 60.75 + 9.32 - 26.79 ≈ (78.2 - 60.75) + (9.32 - 26.79) ≈ 17.45 - 17.47 ≈ -0.02Still negative. Compute f'(4.8828):-5e^{-0.24414} ≈ -5*0.782 ≈ -3.917.2e^{-0.390624} ≈ 7.2*0.675 ≈ 4.862cos(0.48828) ≈ 2*0.893 ≈ 1.7863sin(0.48828) ≈ 3*0.466 ≈ 1.398f'(4.8828) ≈ -3.91 + 4.86 + 1.786 + 1.398 ≈ (-3.91 + 4.86) + (1.786 + 1.398) ≈ 0.95 + 3.184 ≈ 4.134So x3 = x2 - f(x2)/f'(x2) ≈ 4.8828 - (-0.02)/4.134 ≈ 4.8828 + 0.0048 ≈ 4.8876Compute f(4.8876):100e^{-0.05*4.8876} ≈ 100e^{-0.24438} ≈ 100*0.782 ≈ 78.290e^{-0.08*4.8876} ≈ 90e^{-0.391008} ≈ 90*0.675 ≈ 60.7520sin(0.48876) ≈ 20*0.466 ≈ 9.3230cos(0.48876) ≈ 30*0.893 ≈ 26.79f(4.8876) ≈ 78.2 - 60.75 + 9.32 - 26.79 ≈ same as before ≈ -0.02Hmm, seems like it's oscillating around -0.02. Maybe I need a better approach.Alternatively, perhaps using a calculator or computational tool would be more efficient, but since I'm doing this manually, maybe I can accept that the root is approximately 4.88 weeks.So, t ≈ 4.88 weeks.Now, to find the common recovery percentage, compute A(t) or J(t) at t=4.88.Compute A(4.88):100e^{-0.05*4.88} + 20sin(0.1*4.88)= 100e^{-0.244} + 20sin(0.488)≈ 100*0.782 + 20*0.466≈ 78.2 + 9.32 ≈ 87.52%Similarly, compute J(4.88):90e^{-0.08*4.88} + 30cos(0.1*4.88)= 90e^{-0.3904} + 30cos(0.488)≈ 90*0.675 + 30*0.893≈ 60.75 + 26.79 ≈ 87.54%Close enough, considering rounding errors. So approximately 87.5%.So, the first time when both have equal recovery is approximately 4.88 weeks, with a recovery percentage of about 87.5%.Problem 2: Total distance covered after 10 weeksAlex runs a distance that increases by 10% each week, starting at 5 km.Jamie swims a distance that increases by 15% each week, starting at 3 km.We need to calculate the total distance each covers after 10 weeks.This is a geometric series problem.For Alex:First term a = 5 km, common ratio r = 1.10, number of terms n=10.Total distance S_Alex = a*(r^n - 1)/(r - 1)Similarly for Jamie:a = 3 km, r = 1.15, n=10.Total distance S_Jamie = a*(r^n - 1)/(r - 1)Compute S_Alex:S_Alex = 5*(1.10^10 - 1)/(1.10 - 1) = 5*(1.10^10 - 1)/0.10Compute 1.10^10:1.1^10 ≈ 2.5937So S_Alex ≈ 5*(2.5937 - 1)/0.10 = 5*(1.5937)/0.10 = 5*15.937 ≈ 79.685 kmFor Jamie:S_Jamie = 3*(1.15^10 - 1)/(1.15 - 1) = 3*(1.15^10 - 1)/0.15Compute 1.15^10:1.15^10 ≈ 4.0456So S_Jamie ≈ 3*(4.0456 - 1)/0.15 = 3*(3.0456)/0.15 = 3*20.304 ≈ 60.912 kmSo Alex covers approximately 79.685 km, Jamie covers approximately 60.912 km.Therefore, Alex covers more distance overall.Final Answer1. The first time when both Alex's and Jamie's recovery percentages are equal is approximately boxed{4.88} weeks, with a common recovery percentage of approximately boxed{87.5%}.2. After 10 weeks, Alex will have covered a total distance of approximately boxed{79.69} km, while Jamie will have covered approximately boxed{60.91} km. Therefore, Alex covers more distance overall.</think>

🧠:<think>Okay, so I have this problem about Samantha and her spiritual group. It involves the Fibonacci sequence, which I remember is a sequence where each number is the sum of the two preceding ones. The problem has two parts. Let me try to tackle them one by one.Starting with part 1: They say that the number of participants each week follows a pattern based on the Fibonacci sequence. Specifically, they give me that in the first week, there were 8 participants, and in the second week, 13 participants. They also mention that the number of participants can be expressed as P(n) = a·F(n) + b·F(n-1), where F(n) is the nth Fibonacci number, and a and b are constants we need to find.Hmm, okay. So, first, I should probably recall the Fibonacci sequence. The standard Fibonacci sequence starts with F(1) = 1, F(2) = 1, F(3) = 2, F(4) = 3, F(5) = 5, and so on. But wait, sometimes people index it starting from 0, so F(0) = 0, F(1) = 1, F(2) = 1, etc. I need to make sure which one they're using here.Looking back at the problem, they mention the first week as week 1, so probably they are using F(1) as the first Fibonacci number. Let me confirm that. If P(1) = 8, then according to the formula, P(1) = a·F(1) + b·F(0). Wait, but if they're starting from F(1), then F(0) might be 0 or undefined. Hmm, maybe I should check.Wait, maybe it's better to define F(1) = 1, F(2) = 1, F(3) = 2, etc. So, for n=1, P(1)=8, which would be a·F(1) + b·F(0). But if F(0) is 0, then P(1) = a·1 + b·0 = a. So, a = 8.Similarly, for n=2, P(2)=13, which is a·F(2) + b·F(1). Since F(2)=1 and F(1)=1, this becomes a + b = 13. But we already found a=8, so 8 + b =13, which means b=5.Wait, that seems straightforward. Let me write that down step by step.Given:- P(1) = 8 = a·F(1) + b·F(0)- P(2) = 13 = a·F(2) + b·F(1)Assuming F(1)=1, F(2)=1, and F(0)=0 (if needed). So,For n=1:P(1) = a·F(1) + b·F(0) = a·1 + b·0 = aSo, a = 8.For n=2:P(2) = a·F(2) + b·F(1) = a·1 + b·1 = a + bWe know P(2)=13, and a=8, so 8 + b =13 => b=5.Therefore, a=8 and b=5.Let me verify this with n=3 to see if it makes sense. The Fibonacci sequence for n=3 is F(3)=2. So, P(3)=8·2 +5·1=16 +5=21. If the pattern is Fibonacci, then participants should be 8,13,21,34,... which is indeed a Fibonacci sequence starting from 8 and 13. So, that seems correct.Okay, so part 1 is done. a=8, b=5.Moving on to part 2: Samantha wants to plan a special event on the 10th week and needs to estimate the total number of participants from week 1 to week 10. So, I need to calculate the sum S = P(1) + P(2) + ... + P(10).Given that P(n) =8·F(n) +5·F(n-1), the sum S would be the sum from n=1 to n=10 of [8·F(n) +5·F(n-1)].I can split this sum into two separate sums:S = 8·ΣF(n) from n=1 to10 +5·ΣF(n-1) from n=1 to10.Let me adjust the indices for the second sum. When n=1, F(n-1)=F(0). If F(0)=0, then the term is 0. So, the second sum becomes ΣF(k) from k=0 to9, where k =n-1.Therefore, S =8·ΣF(n) from1-10 +5·ΣF(k) from0-9.But ΣF(k) from0-9 is equal to ΣF(n) from1-9 + F(0). Since F(0)=0, it's just ΣF(n) from1-9.Therefore, S=8·ΣF(1-10) +5·ΣF(1-9).Let me denote S1 = ΣF(1-10) and S2=ΣF(1-9).So, S=8·S1 +5·S2.I need to compute S1 and S2. I remember that the sum of the first n Fibonacci numbers is F(n+2) -1. Let me recall that formula.Yes, the sum from k=1 to n of F(k) is F(n+2) -1.So, for S1 = ΣF(1-10) = F(12) -1.Similarly, S2 = ΣF(1-9) = F(11) -1.So, I need to find F(11) and F(12).Let me list the Fibonacci numbers up to F(12):F(1)=1F(2)=1F(3)=2F(4)=3F(5)=5F(6)=8F(7)=13F(8)=21F(9)=34F(10)=55F(11)=89F(12)=144So, S1 = F(12) -1 =144 -1=143S2 = F(11) -1=89 -1=88Therefore, S=8·143 +5·88.Let me compute that:8·143: 143·8. Let's compute 100·8=800, 40·8=320, 3·8=24. So, 800+320=1120 +24=1144.5·88=440.So, total S=1144 +440=1584.Therefore, the total number of participants from week 1 to week 10 is 1584.Wait, let me double-check my calculations.First, verifying the sum formula: sum from k=1 to n of F(k) = F(n+2)-1.Yes, that's correct. For example, sum from 1-1: F(1)=1, which should be F(3)-1=2-1=1. Correct.Sum from 1-2:1+1=2, which is F(4)-1=3-1=2. Correct.Sum from 1-3:1+1+2=4, which is F(5)-1=5-1=4. Correct.So, the formula holds.Therefore, S1=F(12)-1=144-1=143.S2=F(11)-1=89-1=88.Thus, 8*143=1144, 5*88=440. 1144+440=1584.Yes, that seems correct.Alternatively, I could compute each P(n) from n=1 to10 and sum them up to verify.Given P(n)=8·F(n)+5·F(n-1). Let's compute each term:n=1: P(1)=8·F(1)+5·F(0)=8·1 +5·0=8n=2:8·F(2)+5·F(1)=8·1 +5·1=8+5=13n=3:8·F(3)+5·F(2)=8·2 +5·1=16+5=21n=4:8·3 +5·2=24+10=34n=5:8·5 +5·3=40+15=55n=6:8·8 +5·5=64+25=89n=7:8·13 +5·8=104+40=144n=8:8·21 +5·13=168+65=233n=9:8·34 +5·21=272+105=377n=10:8·55 +5·34=440+170=610Now, let's list these P(n):8,13,21,34,55,89,144,233,377,610.Now, let's sum them up step by step:Start with 8.8 +13=2121 +21=4242 +34=7676 +55=131131 +89=220220 +144=364364 +233=597597 +377=974974 +610=1584.Yes, same result. So, the total is indeed 1584.Therefore, the answers are:1. a=8, b=5.2. Total participants over 10 weeks:1584.Final Answer1. The values of ( a ) and ( b ) are ( boxed{8} ) and ( boxed{5} ) respectively.2. The total number of participants over the 10 weeks is ( boxed{1584} ).</think>

🧠:<think>Alright, so I have two problems here related to Sister Maria's scheduling of nurses. Let me tackle them one by one.Problem 1: Probability CalculationFirst, let's understand the setup. There are 12 nurses, each working exactly one shift on each of the three days: Christmas Eve, Christmas Day, and the day after Christmas. Sister Maria wants to maximize the total number of shifts covered, which I think means she wants as many nurses as possible to be available for Mass on one of the days. Each nurse must have equal opportunities to attend Mass, so the probability of being scheduled on any specific day is equal.Wait, each nurse has to work exactly one shift per day, so that means each nurse is assigned to one of the three days for their shift. So, for each day, some number of nurses will be working, and the rest can attend Mass. The goal is to maximize the number of shifts, which I think is a bit confusing because each nurse is already assigned to one shift per day. Maybe it's about maximizing the number of days each nurse can attend Mass? Hmm, the wording is a bit unclear.Wait, no. Each nurse must work exactly one shift on each of the three days. That means each nurse is assigned to one shift on Christmas Eve, one on Christmas Day, and one on the day after. So, each day, a nurse is either working or not. Wait, that doesn't make sense because if each nurse works exactly one shift per day, that would mean each day, all 12 nurses are working, but there are three days. Wait, no, that can't be.Wait, maybe I misread. Let me check again. It says each nurse must work exactly one shift on Christmas Eve, Christmas Day, and the day after Christmas. So, each nurse is assigned to one shift on each day, meaning each day, each nurse is assigned to a shift. But that would mean each day, all 12 nurses are working, which doesn't make sense because then there's no one to attend Mass.Wait, that can't be right. Maybe it's that each nurse works exactly one shift across the three days? So, each nurse is assigned to one of the three days, working one shift on that day, and attending Mass on the other two days. That would make more sense because then the number of nurses working on each day can vary, and the rest can attend Mass.Yes, that must be it. So, each nurse is assigned to exactly one day (either Christmas Eve, Christmas Day, or the day after) to work a shift, and on the other two days, they can attend Mass. So, Sister Maria wants to maximize the number of shifts covered, which would mean maximizing the number of nurses working each day, but also ensuring that each nurse has equal opportunity to attend Mass on one of the three days.Wait, but if each nurse is assigned to exactly one day, then the number of nurses working on each day can be variable, but the total number of shifts is 12, one per nurse. So, the total number of shifts is fixed at 12, with each day having some number of nurses working, and the rest attending Mass.But the problem says "maximize the total number of shifts covered." Wait, if each nurse is assigned to exactly one shift, then the total number of shifts is fixed at 12. So, maybe "maximize the number of shifts covered" is a misstatement, and it's actually about maximizing the number of days each nurse can attend Mass? Or perhaps it's about distributing the shifts such that the number of nurses working on each day is as balanced as possible.Wait, the problem also mentions that each nurse must have equal opportunities to attend Mass on one of the three days. So, perhaps the probability of being scheduled on any specific day is equal, meaning each nurse has a 1/3 chance of being assigned to work on Christmas Eve, 1/3 on Christmas Day, and 1/3 on the day after.So, if each nurse independently has a 1/3 chance to be assigned to each day, then the number of nurses working on Christmas Day would follow a binomial distribution with parameters n=12 and p=1/3.But the question is asking for the probability that exactly 4 nurses attend Mass on Christmas Day. Wait, if a nurse is assigned to work on Christmas Day, they can't attend Mass that day. So, the number of nurses attending Mass on Christmas Day is equal to the number of nurses not assigned to work on Christmas Day.So, if X is the number of nurses working on Christmas Day, then the number attending Mass is 12 - X. So, we need the probability that 12 - X = 4, which is equivalent to X = 8.Therefore, we need the probability that exactly 8 nurses are assigned to work on Christmas Day. Since each nurse independently has a 1/3 chance of being assigned to Christmas Day, this is a binomial probability.The formula for the binomial probability is:P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)Where n = 12, k = 8, p = 1/3.So, plugging in the numbers:C(12, 8) * (1/3)^8 * (2/3)^4C(12, 8) is the combination of 12 choose 8, which is equal to 495.So, 495 * (1/3)^8 * (2/3)^4Calculating this:First, (1/3)^8 = 1 / 6561 ≈ 0.00015241579(2/3)^4 = 16 / 81 ≈ 0.1975308642Multiplying these together: 0.00015241579 * 0.1975308642 ≈ 0.000030117Then multiply by 495: 495 * 0.000030117 ≈ 0.014899So, approximately 0.0149, or 1.49%.Wait, but let me double-check the calculations.Alternatively, using exact fractions:C(12, 8) = 495(1/3)^8 = 1 / 6561(2/3)^4 = 16 / 81So, 495 * (1/6561) * (16/81) = 495 * 16 / (6561 * 81)Calculate denominator: 6561 * 81 = 531441Numerator: 495 * 16 = 7920So, 7920 / 531441 ≈ 0.0149Yes, that's correct.So, the probability is approximately 0.0149, or 1.49%.But let me express it as a fraction:7920 / 531441Simplify numerator and denominator by dividing numerator and denominator by 3:7920 ÷ 3 = 2640531441 ÷ 3 = 177147Again, divide by 3:2640 ÷ 3 = 880177147 ÷ 3 = 59049Again, divide by 3:880 ÷ 3 ≈ 293.333, which is not an integer, so we can't divide further.So, the fraction is 880 / 59049, which is approximately 0.0149.So, the probability is 880/59049, which is approximately 1.49%.Problem 2: Markov Chain Stationary DistributionNow, moving on to the second problem. Sister Maria is using a Markov chain model to schedule shifts, minimizing disruption. Each state represents a possible assignment of shifts for a single nurse, and transitions depend on the prior day's assignments.The transition matrix P is given as:[ P = begin{bmatrix} 0.5 & 0.3 & 0.2 0.4 & 0.4 & 0.2 0.3 & 0.3 & 0.4 end{bmatrix} ]Where the rows represent shifts on Christmas Eve, Christmas Day, and the day after Christmas, respectively, and the columns represent the probability of being assigned to Christmas Eve, Christmas Day, and the day after Christmas the following year.We need to find the stationary distribution of this Markov chain.The stationary distribution π is a row vector such that π = πP, and the sum of the components of π is 1.So, let's denote the stationary distribution as π = [π₁, π₂, π₃], where π₁ is the stationary probability of being in state 1 (Christmas Eve), π₂ for state 2 (Christmas Day), and π₃ for state 3 (day after Christmas).We need to solve the system of equations:π₁ = π₁ * 0.5 + π₂ * 0.4 + π₃ * 0.3π₂ = π₁ * 0.3 + π₂ * 0.4 + π₃ * 0.3π₃ = π₁ * 0.2 + π₂ * 0.2 + π₃ * 0.4And π₁ + π₂ + π₃ = 1Let me write these equations out:1. π₁ = 0.5π₁ + 0.4π₂ + 0.3π₃2. π₂ = 0.3π₁ + 0.4π₂ + 0.3π₃3. π₃ = 0.2π₁ + 0.2π₂ + 0.4π₃And 4. π₁ + π₂ + π₃ = 1Let's rearrange equations 1, 2, 3 to bring all terms to one side.From equation 1:π₁ - 0.5π₁ - 0.4π₂ - 0.3π₃ = 00.5π₁ - 0.4π₂ - 0.3π₃ = 0 --> Equation AFrom equation 2:π₂ - 0.3π₁ - 0.4π₂ - 0.3π₃ = 0-0.3π₁ + 0.6π₂ - 0.3π₃ = 0 --> Equation BFrom equation 3:π₃ - 0.2π₁ - 0.2π₂ - 0.4π₃ = 0-0.2π₁ - 0.2π₂ + 0.6π₃ = 0 --> Equation CSo, we have three equations:A: 0.5π₁ - 0.4π₂ - 0.3π₃ = 0B: -0.3π₁ + 0.6π₂ - 0.3π₃ = 0C: -0.2π₁ - 0.2π₂ + 0.6π₃ = 0And equation D: π₁ + π₂ + π₃ = 1Let me try to solve these equations.First, let's express equations A, B, C in terms of π₁, π₂, π₃.From equation A:0.5π₁ = 0.4π₂ + 0.3π₃ --> π₁ = (0.4π₂ + 0.3π₃) / 0.5 = 0.8π₂ + 0.6π₃ --> Equation A1From equation B:-0.3π₁ + 0.6π₂ - 0.3π₃ = 0Let's substitute π₁ from A1 into equation B.-0.3*(0.8π₂ + 0.6π₃) + 0.6π₂ - 0.3π₃ = 0Calculate:-0.24π₂ - 0.18π₃ + 0.6π₂ - 0.3π₃ = 0Combine like terms:(-0.24 + 0.6)π₂ + (-0.18 - 0.3)π₃ = 00.36π₂ - 0.48π₃ = 0Divide both sides by 0.12:3π₂ - 4π₃ = 0 --> 3π₂ = 4π₃ --> π₂ = (4/3)π₃ --> Equation B1From equation C:-0.2π₁ - 0.2π₂ + 0.6π₃ = 0Again, substitute π₁ from A1 and π₂ from B1.-0.2*(0.8π₂ + 0.6π₃) - 0.2π₂ + 0.6π₃ = 0First, expand:-0.16π₂ - 0.12π₃ - 0.2π₂ + 0.6π₃ = 0Combine like terms:(-0.16 - 0.2)π₂ + (-0.12 + 0.6)π₃ = 0-0.36π₂ + 0.48π₃ = 0Divide both sides by 0.12:-3π₂ + 4π₃ = 0 --> -3π₂ + 4π₃ = 0But from equation B1, we have 3π₂ = 4π₃, so substituting into this equation:-3*(4π₃/3) + 4π₃ = -4π₃ + 4π₃ = 0Which is 0=0, so it's consistent.So, we have from B1: π₂ = (4/3)π₃And from A1: π₁ = 0.8π₂ + 0.6π₃Substitute π₂ = (4/3)π₃ into A1:π₁ = 0.8*(4/3)π₃ + 0.6π₃ = (3.2/3)π₃ + 0.6π₃Convert 0.6 to thirds: 0.6 = 1.8/3So, π₁ = (3.2/3 + 1.8/3)π₃ = (5.0/3)π₃ = (5/3)π₃So, π₁ = (5/3)π₃, π₂ = (4/3)π₃Now, using equation D: π₁ + π₂ + π₃ = 1Substitute:(5/3)π₃ + (4/3)π₃ + π₃ = 1Combine terms:(5/3 + 4/3 + 3/3)π₃ = 1(12/3)π₃ = 1 --> 4π₃ = 1 --> π₃ = 1/4Then, π₂ = (4/3)*(1/4) = 1/3And π₁ = (5/3)*(1/4) = 5/12So, the stationary distribution is π = [5/12, 1/3, 1/4]Let me verify this.Compute πP:π = [5/12, 1/3, 1/4]Multiply by P:First element: 5/12*0.5 + 1/3*0.4 + 1/4*0.3= (5/12)*(1/2) + (1/3)*(2/5) + (1/4)*(3/10)Wait, no, better to compute decimals:5/12 ≈ 0.41671/3 ≈ 0.33331/4 = 0.25So,First element: 0.4167*0.5 + 0.3333*0.4 + 0.25*0.3= 0.20835 + 0.13332 + 0.075 = 0.41667 ≈ 5/12Second element: 0.4167*0.3 + 0.3333*0.4 + 0.25*0.3= 0.125 + 0.13332 + 0.075 ≈ 0.33332 ≈ 1/3Third element: 0.4167*0.2 + 0.3333*0.2 + 0.25*0.4= 0.08334 + 0.06666 + 0.1 ≈ 0.25 ≈ 1/4So, it checks out.Therefore, the stationary distribution is [5/12, 1/3, 1/4].Final Answer1. The probability is boxed{dfrac{880}{59049}}.2. The stationary distribution is boxed{left[ dfrac{5}{12}, dfrac{1}{3}, dfrac{1}{4} right]}.</think>

🧠:<think>Okay, so I've got these two problems about tarot cards, and I need to figure them out step by step. Let me start by understanding what each question is asking.First, problem 1 is about calculating the number of possible distinct sequences when drawing 10 unique cards from a standard tarot deck of 78 cards. Each card can only be drawn once, so no repeats. Hmm, okay, so this sounds like a permutation problem because the order matters here. When you draw cards in a sequence, the order in which they appear is important, right? So, if I'm drawing 10 cards one after another without replacement, the number of different sequences would be the number of ways to arrange 10 cards out of 78.I remember that permutations are used when the order matters, and combinations are when it doesn't. Since each sequence is distinct based on the order, permutations are the way to go. The formula for permutations is P(n, k) = n! / (n - k)!, where n is the total number of items, and k is the number of items we're choosing.So, plugging in the numbers, n is 78 and k is 10. Therefore, the number of possible sequences should be 78! divided by (78 - 10)!, which is 78! / 68!. That makes sense because for the first card, there are 78 possibilities, for the second card, 77, and so on, down to 70 for the tenth card. Multiplying all these together gives the total number of sequences.Wait, let me verify that. If I have 78 cards, the first draw has 78 options, the second has 77, third 76, and so on until the tenth card, which would have 78 - 9 = 69 options? Wait, hold on, 78 minus 10 is 68, so the last term is 69? Hmm, no, wait, 78 - 10 + 1 is 69. So, actually, the number of sequences is 78 × 77 × 76 × ... × 69. Which is the same as 78! / 68!.Yes, that seems correct. So, problem 1 is solved by calculating this permutation.Moving on to problem 2. Alex wants the probability of drawing a specific 10-card sequence known as the "Celtic Cross." Since the deck is shuffled uniformly at random, each permutation is equally likely. So, the probability would be the number of favorable outcomes divided by the total number of possible outcomes.In this case, the favorable outcome is just one specific sequence—the exact "Celtic Cross" spread in the exact order. So, there's only 1 favorable outcome. The total number of possible outcomes is the number of distinct sequences we calculated in problem 1, which is 78! / 68!.Therefore, the probability should be 1 divided by (78! / 68!), which simplifies to 68! / 78!.Alternatively, since 78! / 68! is equal to 78 × 77 × ... × 69, the probability is 1 divided by that product. So, another way to write it is 1 / (78 × 77 × 76 × ... × 69).Let me double-check that. The probability of a specific permutation is always 1 over the total number of permutations, right? So yes, if there are P(78,10) possible sequences, each with equal probability, then the chance of any specific one is 1 / P(78,10). That seems correct.Wait, another thought: sometimes in probability, especially with card draws, people might confuse combinations and permutations. But in this case, since the order matters for the spread, it's definitely a permutation. If it were just the set of 10 cards regardless of order, it would be a combination, but since it's a specific spread, order matters.So, to recap, problem 1 is about permutations of 10 cards from 78, which is 78! / 68!. Problem 2 is the probability of a specific permutation, which is 1 divided by that number.Just to make sure I didn't mix up anything, let me think about a simpler case. Suppose I have a deck of 5 cards and I want to draw 2. The number of sequences would be 5 × 4 = 20. The probability of a specific sequence, say card 1 then card 2, is 1/20. That makes sense because there are 20 possible ordered pairs, each equally likely.Applying that logic here, with 78 cards and 10 draws, the number of sequences is 78 × 77 × ... × 69, which is 78! / 68!. The probability is 1 over that number.I think that solidifies my understanding. So, I can confidently say that the number of distinct sequences is 78! divided by 68!, and the probability is the reciprocal of that.Final Answer1. The number of possible distinct sequences is boxed{dfrac{78!}{68!}}.2. The probability of drawing the specific "Celtic Cross" spread is boxed{dfrac{68!}{78!}}.</think>