🧠:<think>Okay, so I have two problems here about Medications A and B. Let me try to figure them out step by step. Starting with Medication A. The effectiveness is modeled by a logistic growth function: ( E_A(t) = frac{L}{1 + e^{-k(t - t_0)}} ). The parameters given are ( L = 100 ), ( k = 0.2 ), and ( t_0 = 5 ). I need to find the time ( t ) when the effectiveness reaches 75% of its maximum. First, 75% of the maximum effectiveness ( L ) would be 0.75 * 100 = 75. So, I need to solve for ( t ) when ( E_A(t) = 75 ).Plugging into the equation: ( 75 = frac{100}{1 + e^{-0.2(t - 5)}} )Let me write that down:( 75 = frac{100}{1 + e^{-0.2(t - 5)}} )I need to solve for ( t ). Let me rearrange the equation. First, divide both sides by 100:( frac{75}{100} = frac{1}{1 + e^{-0.2(t - 5)}} )Simplify the left side:( 0.75 = frac{1}{1 + e^{-0.2(t - 5)}} )Take reciprocals on both sides:( frac{1}{0.75} = 1 + e^{-0.2(t - 5)} )Calculate ( frac{1}{0.75} ). Hmm, 0.75 is 3/4, so reciprocal is 4/3 ≈ 1.3333.So,( 1.3333 = 1 + e^{-0.2(t - 5)} )Subtract 1 from both sides:( 1.3333 - 1 = e^{-0.2(t - 5)} )Which is:( 0.3333 = e^{-0.2(t - 5)} )Now, take the natural logarithm of both sides to solve for the exponent:( ln(0.3333) = -0.2(t - 5) )Calculate ( ln(0.3333) ). I remember that ( ln(1/3) ) is approximately -1.0986.So,( -1.0986 = -0.2(t - 5) )Multiply both sides by -1 to make it positive:( 1.0986 = 0.2(t - 5) )Now, divide both sides by 0.2:( frac{1.0986}{0.2} = t - 5 )Calculate that division: 1.0986 / 0.2 is 5.493.So,( 5.493 = t - 5 )Add 5 to both sides:( t = 5 + 5.493 )Which is approximately:( t ≈ 10.493 ) days.Hmm, so about 10.49 days. Let me check my steps to make sure I didn't make a mistake.1. Set ( E_A(t) = 75 ).2. Plugged into the logistic equation.3. Divided both sides by 100, got 0.75.4. Took reciprocal, got 4/3.5. Subtracted 1, got 1/3.6. Took natural log, got approximately -1.0986.7. Divided both sides by -0.2, which gave me positive 5.493.8. Added 5, got t ≈ 10.493.Seems correct. Maybe I can write it as approximately 10.49 days or round it to two decimal places, 10.49.Now, moving on to Medication B. The side effect severity is modeled by an exponential decay function: ( S_B(t) = S_0 e^{-lambda t} ). The parameters are ( S_0 = 50 ) and ( lambda = 0.1 ). I need to find the time ( t ) when the side effect severity ( S_B(t) ) reaches 10.So, set ( S_B(t) = 10 ):( 10 = 50 e^{-0.1 t} )Let me write that down:( 10 = 50 e^{-0.1 t} )I need to solve for ( t ). Let me divide both sides by 50:( frac{10}{50} = e^{-0.1 t} )Simplify:( 0.2 = e^{-0.1 t} )Take the natural logarithm of both sides:( ln(0.2) = -0.1 t )Calculate ( ln(0.2) ). I remember that ( ln(1/5) ) is approximately -1.6094.So,( -1.6094 = -0.1 t )Multiply both sides by -1:( 1.6094 = 0.1 t )Divide both sides by 0.1:( t = frac{1.6094}{0.1} )Which is:( t = 16.094 ) days.Let me verify my steps:1. Set ( S_B(t) = 10 ).2. Plugged into the exponential decay equation.3. Divided both sides by 50, got 0.2.4. Took natural log, got approximately -1.6094.5. Divided both sides by -0.1, got positive 16.094.Looks good. So, approximately 16.094 days. Maybe round it to two decimal places, 16.09 days.Wait, let me check the calculations again because sometimes with exponentials, it's easy to make a mistake.For Medication A, I had:( 75 = frac{100}{1 + e^{-0.2(t - 5)}} )Which led to:( 0.75 = frac{1}{1 + e^{-0.2(t - 5)}} )Then reciprocal is 4/3, subtract 1 is 1/3, ln(1/3) is about -1.0986, divide by -0.2 is 5.493, add 5 is 10.493. That seems correct.For Medication B:( 10 = 50 e^{-0.1 t} )Divide by 50: 0.2 = e^{-0.1 t}ln(0.2) is -1.6094, so -1.6094 = -0.1 tMultiply by -1: 1.6094 = 0.1 tDivide by 0.1: t = 16.094. Correct.So, I think both answers are correct.Final Answer1. The time when Medication A reaches 75% effectiveness is boxed{10.49} days.2. The time when Medication B's side effect severity reaches 10 is boxed{16.09} days.</think>

🧠:<think>Okay, so I have these two problems to solve. Let me start with the first one about the blog's visitor growth. Hmm, the problem says the number of unique visitors V(t) follows the differential equation dV/dt = kV(1 - V/N). That looks like the logistic growth model, right? I remember that the logistic equation models population growth where there's a carrying capacity N. Given that the blog started with 1000 visitors, so V(0) = 1000. It reached half of its carrying capacity in 6 months, so V(6) = N/2. The growth rate constant k is given as 0.05 per month. I need to find N.First, I should recall the solution to the logistic differential equation. The general solution is V(t) = N / (1 + (N/V0 - 1) e^{-kt}), where V0 is the initial population. So plugging in V0 = 1000, we get V(t) = N / (1 + (N/1000 - 1) e^{-0.05t}).Now, at t = 6 months, V(6) = N/2. Let me plug that into the equation:N/2 = N / (1 + (N/1000 - 1) e^{-0.05*6})Let me simplify this equation. First, divide both sides by N:1/2 = 1 / (1 + (N/1000 - 1) e^{-0.3})Then, take reciprocals on both sides:2 = 1 + (N/1000 - 1) e^{-0.3}Subtract 1 from both sides:1 = (N/1000 - 1) e^{-0.3}Now, solve for (N/1000 - 1):(N/1000 - 1) = e^{0.3} / 1Because if I divide both sides by e^{-0.3}, it becomes multiplying by e^{0.3}.So, N/1000 - 1 = e^{0.3}Calculate e^{0.3}. Let me approximate that. e^{0.3} is approximately 1.349858.So, N/1000 - 1 ≈ 1.349858Therefore, N/1000 ≈ 2.349858Multiply both sides by 1000:N ≈ 2.349858 * 1000 ≈ 2349.858Since the number of visitors should be a whole number, I can round this to approximately 2350. So, the carrying capacity N is about 2350 visitors.Wait, let me double-check my steps. Starting from V(t) = N / (1 + (N/1000 - 1) e^{-0.05t}), plugging t=6:V(6) = N / (1 + (N/1000 - 1) e^{-0.3}) = N/2So, 1 / (1 + (N/1000 - 1) e^{-0.3}) = 1/2Which leads to 1 + (N/1000 - 1) e^{-0.3} = 2Subtract 1: (N/1000 - 1) e^{-0.3} = 1Then, (N/1000 - 1) = e^{0.3} ≈ 1.349858So, N/1000 ≈ 2.349858, so N ≈ 2349.858, which is approximately 2350. That seems correct.Alright, moving on to the second problem. It's about the engagement rate E as a function of the number of illustrations I. The function is given as E(I) = (aI)/(b + I). We have two data points: when I=4, E=0.8; and when I=8, E=0.9. We need to find constants a and b.So, we can set up two equations:1) 0.8 = (a*4)/(b + 4)2) 0.9 = (a*8)/(b + 8)Let me write these as:(4a)/(b + 4) = 0.8 --> equation 1(8a)/(b + 8) = 0.9 --> equation 2I can solve these two equations for a and b. Let me solve equation 1 for a:4a = 0.8*(b + 4)So, 4a = 0.8b + 3.2Divide both sides by 4:a = (0.8b + 3.2)/4Simplify:a = 0.2b + 0.8 --> equation 1aSimilarly, equation 2:8a = 0.9*(b + 8)8a = 0.9b + 7.2Divide both sides by 8:a = (0.9b + 7.2)/8Simplify:a = 0.1125b + 0.9 --> equation 2aNow, set equation 1a equal to equation 2a:0.2b + 0.8 = 0.1125b + 0.9Subtract 0.1125b from both sides:0.0875b + 0.8 = 0.9Subtract 0.8 from both sides:0.0875b = 0.1Solve for b:b = 0.1 / 0.0875Calculate that: 0.1 divided by 0.0875. Let me compute:0.0875 is 7/80, so 0.1 / (7/80) = 0.1 * (80/7) = 8/7 ≈ 1.142857So, b ≈ 1.142857Now, plug this back into equation 1a:a = 0.2*(8/7) + 0.8Calculate 0.2*(8/7) = (1.6)/7 ≈ 0.228571So, a ≈ 0.228571 + 0.8 ≈ 1.028571Alternatively, let's do it with fractions to be precise.0.2 is 1/5, so 1/5 * 8/7 = 8/350.8 is 4/5, so 8/35 + 4/5 = 8/35 + 28/35 = 36/35So, a = 36/35 ≈ 1.028571Therefore, a = 36/35 and b = 8/7.Let me check if these satisfy the original equations.First, equation 1: (4a)/(b + 4) = (4*(36/35))/(8/7 + 4)Compute numerator: 4*(36/35) = 144/35Denominator: 8/7 + 4 = 8/7 + 28/7 = 36/7So, (144/35)/(36/7) = (144/35)*(7/36) = (144*7)/(35*36) = (1008)/(1260) = 0.8. Correct.Equation 2: (8a)/(b + 8) = (8*(36/35))/(8/7 + 8)Numerator: 8*(36/35) = 288/35Denominator: 8/7 + 8 = 8/7 + 56/7 = 64/7So, (288/35)/(64/7) = (288/35)*(7/64) = (2016)/(2240) = 0.9. Correct.So, the constants are a = 36/35 and b = 8/7.Alternatively, in decimal form, a ≈ 1.0286 and b ≈ 1.1429.But since the problem doesn't specify the form, fractions are probably better.So, summarizing:1. The carrying capacity N is approximately 2350.2. The constants are a = 36/35 and b = 8/7.Final Answer1. The carrying capacity ( N ) is boxed{2350}.2. The constants are ( a = boxed{dfrac{36}{35}} ) and ( b = boxed{dfrac{8}{7}} ).</think>

🧠:<think>Okay, so I have two problems to solve here. The first one is about calculating the average response time for a city grid, and the second one is about determining the time it takes for a nurse's efficiency to reach a certain level. Let me tackle them one by one.Starting with the first problem: The city grid is a 10x10 matrix, so that's 100 blocks in total. Each block has a different time to reach from the central hospital, given by the formula ( T_{ij} = 1 + sqrt{(i-5)^2 + (j-5)^2} ). I need to calculate the average response time to all blocks.Hmm, okay. So each cell (i, j) in the grid has a response time based on its distance from the center, which is at (5,5). The formula is 1 plus the square root of the sum of the squares of (i-5) and (j-5). That sounds like the Euclidean distance from the center, scaled and shifted by 1.First, I need to figure out how to compute this for all 100 blocks and then find the average. Since it's a 10x10 grid, i and j both range from 1 to 10. So for each i from 1 to 10 and each j from 1 to 10, I calculate ( T_{ij} ) and then sum all those up and divide by 100.But wait, calculating this manually for 100 blocks would be tedious. Maybe there's a pattern or symmetry I can exploit. The grid is symmetric around the center, so the distances from the center will be the same for blocks that are symmetrically opposite. That might help reduce the number of calculations.Let me think about the possible distances. The maximum distance from the center would be at the corners, which are (1,1), (1,10), (10,1), and (10,10). Let me compute that distance first. For (1,1), it's ( sqrt{(1-5)^2 + (1-5)^2} = sqrt{16 + 16} = sqrt{32} approx 5.656 ). So ( T_{11} = 1 + 5.656 approx 6.656 ) minutes.Similarly, the minimum distance is at the center itself, which is (5,5). Plugging in, we get ( sqrt{(5-5)^2 + (5-5)^2} = 0 ), so ( T_{55} = 1 + 0 = 1 ) minute.So the response times range from 1 minute at the center to approximately 6.656 minutes at the corners.Since the grid is symmetric, I can compute the response times for one quadrant and then multiply by 4, but I have to be careful with the center blocks which don't have symmetric counterparts.Wait, actually, the grid is 10x10, so the center isn't a single block but a point between blocks. Hmm, actually, in a 10x10 grid, the center would be between blocks (5,5) and (6,6), but since the formula uses (i-5) and (j-5), it's treating (5,5) as the center. So the center block is (5,5), and the grid is symmetric around that point.Therefore, each block (i,j) has a symmetric counterpart (10 - i + 1, 10 - j + 1). So for each block not on the central row or column, there are four symmetric blocks. But for blocks on the central row or column, they have two symmetric counterparts, and the center block is unique.Wait, actually, in a 10x10 grid, the central point is between blocks, but since the formula uses (i-5) and (j-5), the distances are calculated from the center point (5,5), which is actually the center of the grid. So each block's distance is relative to that point.Therefore, for blocks not on the central rows or columns, their distances will be mirrored across both axes. So for example, block (1,1) is symmetric to (10,10), (1,10), and (10,1). Similarly, block (2,3) is symmetric to (9,8), (2,8), and (9,3). So each unique distance in the first quadrant (excluding the axes) corresponds to four blocks.However, blocks on the central rows (i=5 or i=6) or central columns (j=5 or j=6) will have fewer symmetric counterparts. For example, a block on the central row (i=5) will have a mirror image on the same row but opposite side, so only two blocks. Similarly for the central column.Wait, actually, since the grid is 10x10, the central rows are i=5 and i=6, and central columns are j=5 and j=6. So blocks on these rows or columns will have symmetric counterparts on the same row or column.But maybe instead of trying to figure out the symmetry, it's easier to realize that the grid is symmetric, so the sum of all T_ij can be calculated by computing the sum for one quadrant and then multiplying appropriately, but considering the overlaps.Alternatively, perhaps it's simpler to compute the sum by iterating over all i and j, compute T_ij, and then sum them up. Since it's a 10x10 grid, it's manageable, though a bit time-consuming.Alternatively, maybe we can find a mathematical expression for the sum.Let me consider the formula: ( T_{ij} = 1 + sqrt{(i-5)^2 + (j-5)^2} ).So the total sum S is the sum over i=1 to 10 and j=1 to 10 of [1 + sqrt((i-5)^2 + (j-5)^2)].This can be split into two sums: sum_{i,j} 1 + sum_{i,j} sqrt((i-5)^2 + (j-5)^2).The first sum is simply 100, since there are 100 blocks, each contributing 1.The second sum is the sum of sqrt((i-5)^2 + (j-5)^2) for all i and j from 1 to 10.So S = 100 + sum_{i=1 to 10} sum_{j=1 to 10} sqrt((i-5)^2 + (j-5)^2).Therefore, the average response time is (100 + sum) / 100.So I need to compute the sum of sqrt((i-5)^2 + (j-5)^2) for all i and j from 1 to 10.This seems like a lot, but maybe I can find a pattern or use symmetry.Let me consider the distances from the center (5,5). For each possible distance d, how many blocks have that distance?Wait, the distance is sqrt((i-5)^2 + (j-5)^2). So for each possible pair (a,b) where a = |i-5| and b = |j-5|, the distance is sqrt(a^2 + b^2). Since i and j range from 1 to 10, a and b can range from 0 to 5.So for each a from 0 to 5 and b from 0 to 5, we can compute the distance sqrt(a^2 + b^2), and then count how many blocks correspond to each (a,b).But since the grid is symmetric, each (a,b) corresponds to multiple blocks. Specifically, for a=0 and b=0, it's just the center block (5,5). For a=0 and b>0, each (0,b) corresponds to two blocks: (5,5+b) and (5,5-b). Similarly, for a>0 and b=0, each (a,0) corresponds to two blocks: (5+a,5) and (5-a,5). For a>0 and b>0, each (a,b) corresponds to four blocks: (5+a,5+b), (5+a,5-b), (5-a,5+b), (5-a,5-b).Therefore, for each (a,b):- If a=0 and b=0: 1 block- If a=0 and b>0: 2 blocks- If a>0 and b=0: 2 blocks- If a>0 and b>0: 4 blocksSo, we can compute the sum by iterating over a from 0 to 5 and b from 0 to 5, compute the distance for each (a,b), multiply by the number of blocks for that (a,b), and sum them all up.Let me make a table for a and b from 0 to 5.First, a=0:- b=0: distance=0, count=1- b=1: distance=1, count=2- b=2: distance=2, count=2- b=3: distance=3, count=2- b=4: distance=4, count=2- b=5: distance=5, count=2Wait, no. Wait, the distance is sqrt(a^2 + b^2). So for a=0, b=1: sqrt(0 + 1) = 1. For a=0, b=2: sqrt(0 + 4) = 2, etc.Similarly, for a=1:- b=0: sqrt(1 + 0) = 1, count=2- b=1: sqrt(1 + 1) = sqrt(2) ≈1.414, count=4- b=2: sqrt(1 + 4) = sqrt(5)≈2.236, count=4- b=3: sqrt(1 + 9)=sqrt(10)≈3.162, count=4- b=4: sqrt(1 + 16)=sqrt(17)≈4.123, count=4- b=5: sqrt(1 + 25)=sqrt(26)≈5.099, count=4Similarly, for a=2:- b=0: sqrt(4 + 0)=2, count=2- b=1: sqrt(4 + 1)=sqrt(5)≈2.236, count=4- b=2: sqrt(4 + 4)=sqrt(8)≈2.828, count=4- b=3: sqrt(4 + 9)=sqrt(13)≈3.606, count=4- b=4: sqrt(4 + 16)=sqrt(20)≈4.472, count=4- b=5: sqrt(4 + 25)=sqrt(29)≈5.385, count=4a=3:- b=0: sqrt(9 + 0)=3, count=2- b=1: sqrt(9 + 1)=sqrt(10)≈3.162, count=4- b=2: sqrt(9 + 4)=sqrt(13)≈3.606, count=4- b=3: sqrt(9 + 9)=sqrt(18)≈4.243, count=4- b=4: sqrt(9 + 16)=sqrt(25)=5, count=4- b=5: sqrt(9 + 25)=sqrt(34)≈5.831, count=4a=4:- b=0: sqrt(16 + 0)=4, count=2- b=1: sqrt(16 + 1)=sqrt(17)≈4.123, count=4- b=2: sqrt(16 + 4)=sqrt(20)≈4.472, count=4- b=3: sqrt(16 + 9)=sqrt(25)=5, count=4- b=4: sqrt(16 + 16)=sqrt(32)≈5.657, count=4- b=5: sqrt(16 + 25)=sqrt(41)≈6.403, count=4a=5:- b=0: sqrt(25 + 0)=5, count=2- b=1: sqrt(25 + 1)=sqrt(26)≈5.099, count=4- b=2: sqrt(25 + 4)=sqrt(29)≈5.385, count=4- b=3: sqrt(25 + 9)=sqrt(34)≈5.831, count=4- b=4: sqrt(25 + 16)=sqrt(41)≈6.403, count=4- b=5: sqrt(25 + 25)=sqrt(50)≈7.071, count=4Okay, so now I can compute the sum by going through each a and b, compute the distance, multiply by the count, and add them all up.Let me start with a=0:- b=0: 0 * 1 = 0- b=1: 1 * 2 = 2- b=2: 2 * 2 = 4- b=3: 3 * 2 = 6- b=4: 4 * 2 = 8- b=5: 5 * 2 = 10Total for a=0: 0 + 2 + 4 + 6 + 8 + 10 = 30a=1:- b=0: 1 * 2 = 2- b=1: sqrt(2) * 4 ≈1.414 * 4 ≈5.656- b=2: sqrt(5) * 4 ≈2.236 * 4 ≈8.944- b=3: sqrt(10) * 4 ≈3.162 * 4 ≈12.648- b=4: sqrt(17) * 4 ≈4.123 * 4 ≈16.492- b=5: sqrt(26) * 4 ≈5.099 * 4 ≈20.396Total for a=1: 2 + 5.656 + 8.944 + 12.648 + 16.492 + 20.396 ≈63.136a=2:- b=0: 2 * 2 = 4- b=1: sqrt(5) * 4 ≈2.236 * 4 ≈8.944- b=2: sqrt(8) * 4 ≈2.828 * 4 ≈11.312- b=3: sqrt(13) * 4 ≈3.606 * 4 ≈14.424- b=4: sqrt(20) * 4 ≈4.472 * 4 ≈17.888- b=5: sqrt(29) * 4 ≈5.385 * 4 ≈21.54Total for a=2: 4 + 8.944 + 11.312 + 14.424 + 17.888 + 21.54 ≈78.008a=3:- b=0: 3 * 2 = 6- b=1: sqrt(10) * 4 ≈3.162 * 4 ≈12.648- b=2: sqrt(13) * 4 ≈3.606 * 4 ≈14.424- b=3: sqrt(18) * 4 ≈4.243 * 4 ≈16.972- b=4: 5 * 4 = 20- b=5: sqrt(34) * 4 ≈5.831 * 4 ≈23.324Total for a=3: 6 + 12.648 + 14.424 + 16.972 + 20 + 23.324 ≈93.368a=4:- b=0: 4 * 2 = 8- b=1: sqrt(17) * 4 ≈4.123 * 4 ≈16.492- b=2: sqrt(20) * 4 ≈4.472 * 4 ≈17.888- b=3: 5 * 4 = 20- b=4: sqrt(32) * 4 ≈5.657 * 4 ≈22.628- b=5: sqrt(41) * 4 ≈6.403 * 4 ≈25.612Total for a=4: 8 + 16.492 + 17.888 + 20 + 22.628 + 25.612 ≈109.62a=5:- b=0: 5 * 2 = 10- b=1: sqrt(26) * 4 ≈5.099 * 4 ≈20.396- b=2: sqrt(29) * 4 ≈5.385 * 4 ≈21.54- b=3: sqrt(34) * 4 ≈5.831 * 4 ≈23.324- b=4: sqrt(41) * 4 ≈6.403 * 4 ≈25.612- b=5: sqrt(50) * 4 ≈7.071 * 4 ≈28.284Total for a=5: 10 + 20.396 + 21.54 + 23.324 + 25.612 + 28.284 ≈129.156Now, let's sum up all these totals:a=0: 30a=1: ≈63.136a=2: ≈78.008a=3: ≈93.368a=4: ≈109.62a=5: ≈129.156Adding them up:30 + 63.136 = 93.13693.136 + 78.008 = 171.144171.144 + 93.368 = 264.512264.512 + 109.62 = 374.132374.132 + 129.156 = 503.288So the total sum of sqrt((i-5)^2 + (j-5)^2) over all blocks is approximately 503.288.Therefore, the total response time S is 100 + 503.288 = 603.288 minutes.The average response time is S / 100 = 603.288 / 100 ≈6.03288 minutes.So approximately 6.03 minutes.Wait, let me double-check my calculations because I might have made an error in adding up the totals.Let me list the totals again:a=0: 30a=1: ≈63.136a=2: ≈78.008a=3: ≈93.368a=4: ≈109.62a=5: ≈129.156Adding them step by step:30 + 63.136 = 93.13693.136 + 78.008 = 171.144171.144 + 93.368 = 264.512264.512 + 109.62 = 374.132374.132 + 129.156 = 503.288Yes, that seems correct.So total sum is 503.288, plus 100 gives 603.288, average is 6.03288.Rounded to, say, three decimal places, 6.033 minutes.Alternatively, maybe we can compute it more accurately by using exact values instead of approximations.Wait, I approximated sqrt(2) as 1.414, sqrt(5) as 2.236, etc. Maybe using more precise values would give a better total.But given that the problem doesn't specify the need for extreme precision, and since the distances are in minutes, probably two decimal places are sufficient.So, the average response time is approximately 6.03 minutes.Now, moving on to the second problem: The officer wants to establish a new training module for nurses that improves their efficiency. The improvement in efficiency E over time t (in months) follows the function E(t) = 100(1 - e^{-0.1t}). We need to determine the time t it will take for a nurse's efficiency to reach at least 90% of its maximum potential.First, let's understand the function. The maximum efficiency is when t approaches infinity, so E(t) approaches 100(1 - 0) = 100. So the maximum potential efficiency is 100. Therefore, 90% of that is 90.So we need to find t such that E(t) >= 90.So set up the equation:100(1 - e^{-0.1t}) >= 90Divide both sides by 100:1 - e^{-0.1t} >= 0.9Subtract 1:-e^{-0.1t} >= -0.1Multiply both sides by -1 (which reverses the inequality):e^{-0.1t} <= 0.1Take natural logarithm on both sides:ln(e^{-0.1t}) <= ln(0.1)Simplify left side:-0.1t <= ln(0.1)Multiply both sides by -1 (which reverses the inequality again):0.1t >= -ln(0.1)Compute ln(0.1):ln(0.1) = ln(1/10) = -ln(10) ≈ -2.302585So:0.1t >= 2.302585Divide both sides by 0.1:t >= 23.02585So t must be at least approximately 23.02585 months.Since the question asks for the time t in months, we can round this to the nearest month, which would be 23 months. But sometimes, depending on the context, they might want it in decimal form. But since 0.02585 is about 0.03, so 23.03 months. Depending on the requirement, but likely 23 months is sufficient.Let me verify the calculation:E(t) = 100(1 - e^{-0.1t})We set E(t) = 90:90 = 100(1 - e^{-0.1t})Divide both sides by 100:0.9 = 1 - e^{-0.1t}Subtract 1:-0.1 = -e^{-0.1t}Multiply by -1:0.1 = e^{-0.1t}Take natural log:ln(0.1) = -0.1tSo t = -ln(0.1)/0.1 ≈ -(-2.302585)/0.1 ≈23.02585Yes, that's correct.So t ≈23.03 months.Therefore, the nurse's efficiency will reach at least 90% after approximately 23.03 months. Depending on whether partial months are considered, but since the question asks for time in months, 23.03 months is acceptable, or we can round up to 24 months if partial months aren't considered. But since 0.03 of a month is about a day, it's negligible, so 23 months is sufficient.So, summarizing:1. The average response time is approximately 6.03 minutes.2. The time required for efficiency to reach 90% is approximately 23.03 months.Final Answer1. The average response time is boxed{6.03} minutes.2. The time required for the nurse's efficiency to reach at least 90% is boxed{23.03} months.</think>

🧠:<think>Alright, so I have this problem about analyzing fight sequences in Hong Kong action movies using math. It's split into two parts. Let me tackle them one by one.Starting with the first part: The function given is ( f(t) = a sin(bt) + c cos(dt) ). We need to find the constants ( a, b, c, ) and ( d ) based on the given conditions.The conditions are:1. The maximum height of a punch is 5 units.2. The frequency of the kicks is twice that of the punches.3. The amplitude of the acrobatics is 3 units.4. The period of the entire sequence is 10 seconds.Hmm, okay. So, let's break this down.First, I need to understand what each term in the function represents. It seems like ( a sin(bt) ) could be modeling the punches, and ( c cos(dt) ) could be modeling the kicks or acrobatics? Wait, the problem mentions punches, kicks, and acrobatics. So, maybe each term corresponds to a different action.But the function is a combination of sine and cosine. Maybe each term represents a different type of movement. Let's see.Condition 1: The maximum height of a punch is 5 units. So, if ( a sin(bt) ) is the punch, then the amplitude ( a ) would be 5. Because the maximum value of sine is 1, so ( a times 1 = 5 ). So, ( a = 5 ).Condition 3: The amplitude of the acrobatics is 3 units. Hmm, so maybe the other term, ( c cos(dt) ), is the acrobatics. So, the amplitude ( c ) is 3. So, ( c = 3 ).Wait, but the problem mentions punches, kicks, and acrobatics. But the function only has two terms. Maybe kicks are part of the sine or cosine term? Or perhaps kicks are modeled differently? Hmm, maybe the frequency relates to kicks.Condition 2: The frequency of the kicks is twice that of the punches. So, if punches are modeled by ( a sin(bt) ), then their frequency is ( b/(2pi) ). So, the frequency of kicks would be ( 2 times (b/(2pi)) = b/pi ). But where is the kicks term in the function? The function only has two terms, so maybe the kicks are part of the sine or cosine term? Or perhaps the kicks are modeled by another function, but the problem only gives us a two-term function.Wait, maybe the kicks are part of the sine term, and the acrobatics are the cosine term. So, punches have a maximum height of 5, so ( a = 5 ). The amplitude of acrobatics is 3, so ( c = 3 ). Then, the frequency of kicks is twice that of punches. So, if the punches have a frequency ( f_p ), then kicks have ( f_k = 2f_p ).But in the function, the sine term is ( sin(bt) ), so its frequency is ( b/(2pi) ). The cosine term is ( cos(dt) ), so its frequency is ( d/(2pi) ). If kicks have twice the frequency of punches, then if punches are the sine term, then ( d/(2pi) = 2 times (b/(2pi)) ), so ( d = 2b ).Alternatively, if kicks are the cosine term, then ( d/(2pi) = 2 times (b/(2pi)) ), so same result, ( d = 2b ).Condition 4: The period of the entire sequence is 10 seconds. The period of a function like ( f(t) = sin(bt) + cos(dt) ) is the least common multiple (LCM) of the periods of the individual sine and cosine functions. The period of ( sin(bt) ) is ( 2pi/b ), and the period of ( cos(dt) ) is ( 2pi/d ). So, the overall period is LCM of ( 2pi/b ) and ( 2pi/d ).Given that the overall period is 10 seconds, so LCM( (2pi/b, 2pi/d) ) = 10 ).But we also have that ( d = 2b ). So, let's substitute that in.So, the periods are ( 2pi/b ) and ( 2pi/(2b) = pi/b ). So, LCM of ( 2pi/b ) and ( pi/b ).The LCM of two numbers is the smallest number that is a multiple of both. So, ( 2pi/b ) is a multiple of ( pi/b ), because ( 2pi/b = 2 times (pi/b) ). So, the LCM is ( 2pi/b ).Therefore, ( 2pi/b = 10 ).Solving for ( b ):( 2pi/b = 10 )Multiply both sides by ( b ):( 2pi = 10b )Divide both sides by 10:( b = 2pi/10 = pi/5 )So, ( b = pi/5 ). Then, since ( d = 2b ), ( d = 2pi/5 ).So, summarizing:- ( a = 5 )- ( b = pi/5 )- ( c = 3 )- ( d = 2pi/5 )Let me double-check.The function is ( f(t) = 5 sin(pi t /5) + 3 cos(2pi t /5) ).The period of the sine term is ( 2pi / (pi/5) ) = 10 ) seconds.The period of the cosine term is ( 2pi / (2pi/5) ) = 5 ) seconds.So, the LCM of 10 and 5 is 10, which matches the given period. Good.The frequency of the sine term (punches) is ( 1/10 ) Hz, and the frequency of the cosine term (kicks) is ( 1/5 ) Hz, which is twice that of the punches. Perfect.Amplitude of punches is 5, amplitude of acrobatics is 3. That fits.Alright, so that should be the answer for part 1.Moving on to part 2: The jumps are modeled by parametric equations ( x(t) = A cos(omega t) ) and ( y(t) = B sin(omega t) ). We need to find ( A ) and ( B ) given that the total horizontal distance is 20 units and vertical distance is 15 units, with ( omega = pi/5 ).Wait, parametric equations for jumps. So, ( x(t) ) is horizontal position, ( y(t) ) is vertical position.Total horizontal distance covered: 20 units. Total vertical distance: 15 units.But wait, in parametric equations, the total distance isn't just the maximum x and y, but the total path length. Hmm, but the problem says "total horizontal distance covered" and "vertical distance". Maybe it's referring to the maximum horizontal and vertical distances? Or perhaps the total displacement?Wait, but in parametric equations, ( x(t) ) and ( y(t) ) can represent the position over time. So, if the jumps are periodic, the total horizontal and vertical distances might refer to the amplitude or the range.Wait, let's think. If ( x(t) = A cos(omega t) ), then the maximum horizontal position is ( A ), and the minimum is ( -A ). So, the total horizontal distance covered in one period would be from ( -A ) to ( A ), which is ( 2A ). Similarly, for ( y(t) = B sin(omega t) ), the maximum vertical position is ( B ), and the minimum is ( -B ), so total vertical distance is ( 2B ).But the problem says "total horizontal distance covered by the jumps is 20 units" and "vertical distance is 15 units". So, if it's referring to the total range, then ( 2A = 20 ) and ( 2B = 15 ). So, ( A = 10 ) and ( B = 7.5 ).Alternatively, if it's referring to the amplitude, then ( A = 20 ) and ( B = 15 ). But the term "total distance covered" usually refers to the range, i.e., from one extreme to the other, which is ( 2A ) and ( 2B ).But let me think again. If the parametric equations are ( x(t) = A cos(omega t) ) and ( y(t) = B sin(omega t) ), then as ( t ) varies, the point ( (x(t), y(t)) ) traces an ellipse. The horizontal extent is from ( -A ) to ( A ), so the total horizontal distance is ( 2A ). Similarly, the vertical extent is ( 2B ).Given that, if the total horizontal distance is 20, then ( 2A = 20 implies A = 10 ). Similarly, total vertical distance is 15, so ( 2B = 15 implies B = 7.5 ).Alternatively, if the problem is referring to the total distance traveled during one period, that would be different. The total distance traveled would be the circumference of the ellipse, but that's more complicated. However, the problem says "total horizontal distance covered" and "vertical distance", which sounds more like the range rather than the total path length.Therefore, I think ( A = 10 ) and ( B = 7.5 ).But let me verify.If ( x(t) = 10 cos(pi t /5) ) and ( y(t) = 7.5 sin(pi t /5) ), then the horizontal distance from leftmost to rightmost is ( 20 ) units, and vertical from bottom to top is ( 15 ) units. That makes sense.Alternatively, if we consider the total distance traveled, it would be the integral of the speed over one period. But that seems more involved, and the problem doesn't specify that. It just says "total horizontal distance covered" and "vertical distance", which are likely referring to the maximum extents.So, I think ( A = 10 ) and ( B = 7.5 ).Wait, but let me check the units. The problem says "total horizontal distance covered by the jumps is 20 units" and "vertical distance is 15 units". So, if it's distance, not displacement, then it's the total path length. Hmm, that complicates things.Wait, displacement would be the straight-line distance from start to end, but distance is the total length traveled. For a periodic function, over one period, the displacement would be zero because it returns to the starting point, but the total distance would be the perimeter of the ellipse.But the problem says "total horizontal distance covered" and "vertical distance". Hmm, maybe it's referring to the maximum horizontal and vertical distances achieved, i.e., the amplitudes. So, if the horizontal distance is 20, then ( A = 20 ), and vertical distance 15, so ( B = 15 ).Wait, but in that case, the total horizontal distance covered would be the amplitude, not the range. So, if the maximum horizontal position is 20, then ( A = 20 ). Similarly, maximum vertical is 15, so ( B = 15 ).But now I'm confused because the wording is ambiguous. It could be referring to the amplitude or the total range.Wait, let's think about the parametric equations. If ( x(t) = A cos(omega t) ), then the maximum value of ( x(t) ) is ( A ), and the minimum is ( -A ). So, the total horizontal distance from one end to the other is ( 2A ). Similarly, for ( y(t) ), it's ( 2B ).But the problem says "total horizontal distance covered" and "vertical distance". If it's the total distance, meaning the sum of all movements, that would be different. But in the case of parametric equations, the total distance covered in one period is the circumference of the ellipse, which is more complex.But the problem gives specific numbers: 20 units for horizontal and 15 units for vertical. If it's referring to the range, then ( 2A = 20 ) and ( 2B = 15 ), so ( A = 10 ), ( B = 7.5 ). If it's referring to the amplitudes, then ( A = 20 ), ( B = 15 ).Given the wording, "total horizontal distance covered" sounds like the maximum extent, which would be ( 2A ). Similarly, "vertical distance" would be ( 2B ). So, I think it's safer to go with ( A = 10 ) and ( B = 7.5 ).But to be thorough, let's consider both interpretations.1. If "total horizontal distance" is the range, then ( 2A = 20 implies A = 10 ), and ( 2B = 15 implies B = 7.5 ).2. If "total horizontal distance" is the amplitude, then ( A = 20 ), ( B = 15 ).But in physics, when we talk about the distance covered by a wave, it's usually the amplitude. However, in this context, since it's a jump, the total horizontal distance covered might refer to how far the actor moves left and right, which would be the range, i.e., ( 2A ). Similarly, the vertical distance would be how high and low they go, which is ( 2B ).Therefore, I think the first interpretation is correct: ( A = 10 ), ( B = 7.5 ).So, to sum up:Part 1:- ( a = 5 )- ( b = pi/5 )- ( c = 3 )- ( d = 2pi/5 )Part 2:- ( A = 10 )- ( B = 7.5 )I think that's it.</think>