🧠:<think>Okay, so I have this problem about a support group organizer planning events with guest speakers. The goal is to maximize attendee engagement while respecting venue constraints. There are two parts to this problem. Let me try to tackle them one by one.Starting with part 1: The organizer has a probability function ( P(t) = e^{-lambda t} ) which represents the probability that an attendee stays until time ( t ) after the event starts. The guest speaker is scheduled to start at ( t_0 ) hours into the event and speaks for ( D ) hours. We need to find the expected number of attendees who will be present for the entire duration of the guest speaker's talk. The initial number of attendees is ( N_0 ).Hmm, okay. So, first, I need to figure out the probability that an attendee is still present at time ( t_0 ) and remains until ( t_0 + D ). Since the probability of staying until time ( t ) is ( P(t) = e^{-lambda t} ), then the probability that an attendee is still there at ( t_0 ) is ( P(t_0) = e^{-lambda t_0} ). Similarly, the probability that they stay until ( t_0 + D ) is ( P(t_0 + D) = e^{-lambda (t_0 + D)} ).Wait, but the question is about the expected number of attendees who will be present for the entire duration of the guest speaker's talk. That means they need to be present from ( t_0 ) to ( t_0 + D ). So, it's the probability that they stay until ( t_0 + D ) given that they were present at ( t_0 ). Is that correct?But actually, since the probability ( P(t) ) is the probability of staying until time ( t ), regardless of when they arrived. So, if an attendee is present at ( t_0 ), the probability that they stay until ( t_0 + D ) is ( P(D) ), because from ( t_0 ) onward, the time is ( D ). So, it's like a memoryless property? Wait, no, the exponential distribution is memoryless, but here we're dealing with probabilities, not the distribution itself.Wait, let me think again. The probability that an attendee is still there at time ( t_0 + D ) is ( e^{-lambda (t_0 + D)} ). But the probability that they are present at ( t_0 ) is ( e^{-lambda t_0} ). So, the conditional probability that they stay until ( t_0 + D ) given that they were present at ( t_0 ) is ( frac{P(t_0 + D)}{P(t_0)} = e^{-lambda D} ). So, that's the probability that an attendee who is present at ( t_0 ) will stay for the entire duration ( D ).Therefore, the expected number of attendees present for the entire duration is ( N_0 times P(t_0) times e^{-lambda D} ). Wait, is that right? Let me break it down:1. The expected number of attendees at ( t_0 ) is ( N_0 times P(t_0) = N_0 e^{-lambda t_0} ).2. The probability that each of these attendees stays for the next ( D ) hours is ( e^{-lambda D} ).3. So, the expected number is ( N_0 e^{-lambda t_0} times e^{-lambda D} = N_0 e^{-lambda (t_0 + D)} ).Alternatively, since the probability of staying until ( t_0 + D ) is ( e^{-lambda (t_0 + D)} ), the expected number is ( N_0 times e^{-lambda (t_0 + D)} ).Wait, but is that the same as ( N_0 e^{-lambda t_0} e^{-lambda D} )? Yes, because ( e^{-lambda (t_0 + D)} = e^{-lambda t_0} e^{-lambda D} ). So, either way, it's the same result.So, I think that's the answer for part 1: ( N_0 e^{-lambda (t_0 + D)} ).Moving on to part 2: The venue has a seating capacity ( C ), and the organizer must ensure that the number of attendees at the start of the guest speaker's talk does not exceed this capacity. Given that the event starts with ( N_0 ) attendees, and the probability of an attendee leaving before time ( t ) is ( 1 - P(t) ), determine the maximum initial number of attendees ( N_0 ) such that the expected number of attendees at time ( t_0 ) is less than or equal to ( C ).Alright, so the expected number of attendees at time ( t_0 ) is ( N_0 e^{-lambda t_0} ). We need this to be less than or equal to ( C ). So, we can set up the inequality:( N_0 e^{-lambda t_0} leq C )To find the maximum ( N_0 ), we solve for ( N_0 ):( N_0 leq C e^{lambda t_0} )Therefore, the maximum initial number of attendees ( N_0 ) is ( C e^{lambda t_0} ).Wait, let me verify that. The expected number at ( t_0 ) is ( N_0 e^{-lambda t_0} ), so to have this less than or equal to ( C ), we rearrange:( N_0 leq C e^{lambda t_0} )Yes, that seems correct. So, the maximum ( N_0 ) is ( C e^{lambda t_0} ).But just to make sure, let's think about it. If we have more people initially, more people will leave over time, but the expected number at ( t_0 ) is ( N_0 e^{-lambda t_0} ). So, to not exceed capacity ( C ) at ( t_0 ), we need ( N_0 ) to be such that when multiplied by ( e^{-lambda t_0} ), it's less than or equal to ( C ). So, solving for ( N_0 ), we get ( N_0 leq C e^{lambda t_0} ). That makes sense because as ( t_0 ) increases, ( N_0 ) can be larger, which aligns with intuition since more people would have left by then.So, summarizing:1. The expected number of attendees for the entire guest speaker's talk is ( N_0 e^{-lambda (t_0 + D)} ).2. The maximum initial number of attendees ( N_0 ) is ( C e^{lambda t_0} ).I think that's it. Let me just write the final answers clearly.Final Answer1. The expected number of attendees is boxed{N_0 e^{-lambda (t_0 + D)}}.2. The maximum initial number of attendees is boxed{C e^{lambda t_0}}.</think>

🧠:<think>Okay, so I have this problem about a film producer planning a movie with a complex special effects scene involving a controlled explosion. The producer is concerned about both safety and feasibility, which translates into two parts: calculating the expected value of the safety probability ( P_s ) and determining the expected total cost ( mathbb{E}[C] ). Let me tackle each part step by step.Starting with the first part: Safety. The probability ( P_s ) of a safe explosion depends on three independent variables: stability of the explosives ( S ), skill of the special effects team ( T ), and quality of the safety equipment ( E ). The relationships are given by their respective probability density functions (pdfs):- ( S sim text{Uniform}(0.75, 1) )- ( T sim text{Beta}(2, 5) )- ( E sim text{Normal}(0.8, 0.1) ) truncated to [0.6, 1]And ( P_s = S cdot T cdot E ). I need to find the expected value ( mathbb{E}[P_s] ).Since ( S ), ( T ), and ( E ) are independent, the expectation of their product is the product of their expectations. That is,[mathbb{E}[P_s] = mathbb{E}[S] cdot mathbb{E}[T] cdot mathbb{E}[E]]So, I need to compute the expected values of each of these variables separately.First, for ( S sim text{Uniform}(0.75, 1) ). The expectation of a uniform distribution ( text{Uniform}(a, b) ) is ( frac{a + b}{2} ). Therefore,[mathbb{E}[S] = frac{0.75 + 1}{2} = frac{1.75}{2} = 0.875]Next, ( T sim text{Beta}(2, 5) ). The expectation of a Beta distribution ( text{Beta}(alpha, beta) ) is ( frac{alpha}{alpha + beta} ). So,[mathbb{E}[T] = frac{2}{2 + 5} = frac{2}{7} approx 0.2857]Wait, that seems low. Let me double-check. Beta distribution parameters are shape parameters, so yes, the expectation is ( alpha/(alpha + beta) ). So 2/(2+5) is indeed 2/7, which is approximately 0.2857. Okay, that seems correct.Now, ( E sim text{Normal}(0.8, 0.1) ) truncated to [0.6, 1]. Hmm, this is a bit more involved. The expectation of a truncated normal distribution isn't as straightforward as the regular normal distribution. I need to recall the formula for the expectation of a truncated normal.The expectation of a truncated normal distribution with parameters ( mu ), ( sigma^2 ), lower bound ( a ), and upper bound ( b ) is given by:[mathbb{E}[E] = mu + frac{phi(alpha) - phi(beta)}{Phi(beta) - Phi(alpha)} cdot sigma]Where:- ( alpha = frac{a - mu}{sigma} )- ( beta = frac{b - mu}{sigma} )- ( phi ) is the standard normal PDF- ( Phi ) is the standard normal CDFGiven ( E sim text{Normal}(0.8, 0.1) ), so ( mu = 0.8 ), ( sigma = 0.1 ). The truncation is to [0.6, 1], so ( a = 0.6 ), ( b = 1 ).Calculating ( alpha ) and ( beta ):[alpha = frac{0.6 - 0.8}{0.1} = frac{-0.2}{0.1} = -2][beta = frac{1 - 0.8}{0.1} = frac{0.2}{0.1} = 2]So, ( alpha = -2 ), ( beta = 2 ).Now, I need to compute ( phi(-2) ), ( phi(2) ), ( Phi(-2) ), and ( Phi(2) ).Recall that ( phi(z) = frac{1}{sqrt{2pi}} e^{-z^2/2} ), and ( Phi(z) ) is the integral of ( phi ) from ( -infty ) to ( z ).Calculating ( phi(-2) ) and ( phi(2) ):Since the standard normal distribution is symmetric, ( phi(-2) = phi(2) ).Calculating ( phi(2) ):[phi(2) = frac{1}{sqrt{2pi}} e^{-2^2/2} = frac{1}{sqrt{2pi}} e^{-2} approx frac{1}{2.5066} times 0.1353 approx 0.05399]So, ( phi(-2) = phi(2) approx 0.05399 ).Now, ( Phi(-2) ) and ( Phi(2) ):From standard normal tables or using a calculator:- ( Phi(-2) approx 0.02275 )- ( Phi(2) approx 0.97725 )Therefore, the denominator ( Phi(beta) - Phi(alpha) = 0.97725 - 0.02275 = 0.9545 ).The numerator ( phi(alpha) - phi(beta) = 0.05399 - 0.05399 = 0 ). Wait, that can't be right. Wait, no, hold on. The numerator is ( phi(alpha) - phi(beta) ). But since ( alpha = -2 ) and ( beta = 2 ), and ( phi(-2) = phi(2) ), so the numerator is zero. That would imply that the expectation is just ( mu ), but that can't be correct because truncation usually affects the expectation.Wait, maybe I made a mistake in the formula. Let me double-check the expectation formula for a truncated normal.Upon reviewing, I realize that the formula for the expectation of a truncated normal distribution is:[mathbb{E}[E] = mu + frac{phi(alpha) - phi(beta)}{Phi(beta) - Phi(alpha)} cdot sigma]But in this case, since ( alpha = -2 ) and ( beta = 2 ), both ( phi(alpha) ) and ( phi(beta) ) are equal, so their difference is zero. Therefore, the expectation simplifies to:[mathbb{E}[E] = mu + 0 cdot sigma = mu = 0.8]Wait, that seems counterintuitive. If we truncate a normal distribution symmetrically around the mean, does the expectation remain the same? Hmm, actually, yes, because the truncation is symmetric. The distribution is symmetric around 0.8, so truncating equally on both sides doesn't shift the mean. Therefore, the expectation remains 0.8.Therefore, ( mathbb{E}[E] = 0.8 ).So, putting it all together:[mathbb{E}[P_s] = mathbb{E}[S] cdot mathbb{E}[T] cdot mathbb{E}[E] = 0.875 times 0.2857 times 0.8]Let me compute that step by step.First, 0.875 multiplied by 0.2857.0.875 * 0.2857 ≈ 0.25 (since 0.875 is 7/8, and 0.2857 is approximately 2/7; 7/8 * 2/7 = 2/8 = 0.25). Let me compute it more accurately:0.875 * 0.2857:0.8 * 0.2857 = 0.228560.075 * 0.2857 ≈ 0.0214275Adding together: 0.22856 + 0.0214275 ≈ 0.2499875 ≈ 0.25.So, approximately 0.25.Then, multiplying by 0.8:0.25 * 0.8 = 0.2So, the expected value of ( P_s ) is 0.2, or 20%.Wait, that seems low. Let me verify the calculations again.First, ( mathbb{E}[S] = 0.875 ), correct.( mathbb{E}[T] = 2/7 ≈ 0.2857 ), correct.( mathbb{E}[E] = 0.8 ), correct.Multiplying them together: 0.875 * 0.2857 ≈ 0.25, then 0.25 * 0.8 = 0.2.Yes, that seems correct. So, the expected probability of a safe explosion is 20%.Moving on to the second part: Feasibility. The total cost ( C ) is given by:[C = 5000 + 200X + 150Y + 50XY]Where ( X ) follows a Poisson distribution with parameter ( lambda = 3 ), and ( Y ) follows an exponential distribution with rate parameter ( beta = 0.5 ). I need to find ( mathbb{E}[C] ).Since expectation is linear, I can break this down:[mathbb{E}[C] = mathbb{E}[5000] + mathbb{E}[200X] + mathbb{E}[150Y] + mathbb{E}[50XY]]Simplifying each term:- ( mathbb{E}[5000] = 5000 )- ( mathbb{E}[200X] = 200 cdot mathbb{E}[X] )- ( mathbb{E}[150Y] = 150 cdot mathbb{E}[Y] )- ( mathbb{E}[50XY] = 50 cdot mathbb{E}[XY] )So, I need to compute ( mathbb{E}[X] ), ( mathbb{E}[Y] ), and ( mathbb{E}[XY] ).First, ( X sim text{Poisson}(lambda = 3) ). The expectation of a Poisson distribution is equal to its parameter ( lambda ). Therefore,[mathbb{E}[X] = 3]Next, ( Y sim text{Exponential}(beta = 0.5) ). The expectation of an exponential distribution with rate parameter ( beta ) is ( 1/beta ). So,[mathbb{E}[Y] = frac{1}{0.5} = 2]Now, the tricky part is ( mathbb{E}[XY] ). Since ( X ) and ( Y ) are independent (as they are different variables with no stated dependence), the expectation of their product is the product of their expectations. Therefore,[mathbb{E}[XY] = mathbb{E}[X] cdot mathbb{E}[Y] = 3 times 2 = 6]Therefore, putting it all together:[mathbb{E}[C] = 5000 + 200 times 3 + 150 times 2 + 50 times 6]Calculating each term:- 200 * 3 = 600- 150 * 2 = 300- 50 * 6 = 300Adding them up:5000 + 600 = 56005600 + 300 = 59005900 + 300 = 6200So, the expected total cost ( mathbb{E}[C] ) is 6200.Wait, let me double-check the calculation:5000 + (200*3) + (150*2) + (50*6) = 5000 + 600 + 300 + 300 = 5000 + 1200 = 6200. Yes, that's correct.So, summarizing:1. The expected value of ( P_s ) is 0.2, or 20%.2. The expected total cost ( mathbb{E}[C] ) is 6200.I think that's it. Let me just make sure I didn't make any calculation errors.For the first part, the key was recognizing that the expectation of the product is the product of expectations because of independence. For the truncated normal, since the truncation was symmetric around the mean, the expectation remained the same as the original mean, which was 0.8. That was a crucial point.For the second part, breaking down the expectation using linearity was straightforward, and recognizing that ( X ) and ( Y ) are independent allowed me to compute ( mathbb{E}[XY] ) as the product of their expectations. All steps seem solid, so I feel confident in the results.</think>

🧠:<think>Alright, so I've got these two statistics problems to solve, and I need to think through them step by step. Let's start with the first one.Problem 1: Data Analysis and ProbabilityDr. Smith is looking at sexual satisfaction levels before and after therapy. The before levels are normally distributed with a mean of 40 and a standard deviation of 15. After therapy, they're normally distributed with a mean of 65 and a standard deviation of 10. We need to find the probability that a randomly selected patient has an increase greater than 30 points.Hmm, okay. So, the increase is the after score minus the before score. Since both are normally distributed, the difference should also be normally distributed. Let me recall: if X and Y are independent normal variables, then X - Y is also normal with mean μ_X - μ_Y and variance σ_X² + σ_Y².Wait, actually, since we're subtracting, it's Y - X, so the mean would be μ_Y - μ_X, and the variance would be σ_Y² + σ_X². Let me write that down.Let X be the before therapy score: X ~ N(40, 15²)Let Y be the after therapy score: Y ~ N(65, 10²)We are interested in D = Y - X. Since Y and X are independent, D is normal with:Mean of D: μ_D = μ_Y - μ_X = 65 - 40 = 25Variance of D: σ_D² = σ_Y² + σ_X² = 10² + 15² = 100 + 225 = 325So, standard deviation of D: σ_D = sqrt(325) ≈ 18.03We need P(D > 30). So, we can standardize D:Z = (D - μ_D) / σ_D = (30 - 25) / 18.03 ≈ 5 / 18.03 ≈ 0.277So, we need the probability that Z > 0.277. Looking at the standard normal distribution table, the area to the left of Z=0.277 is approximately 0.608. Therefore, the area to the right is 1 - 0.608 = 0.392.Wait, let me double-check that Z value. 0.277 is roughly 0.28. The Z-table for 0.28 is about 0.6103. So, 1 - 0.6103 = 0.3897, which is approximately 0.39 or 39%.So, the probability is roughly 39%.But let me make sure I didn't make a mistake in the variance. Since we're subtracting two independent variables, the variances add up. So, 10² + 15² is indeed 100 + 225 = 325. So, standard deviation is sqrt(325) ≈ 18.03. That seems right.Alternatively, maybe I can compute it more precisely. Let's calculate sqrt(325):325 is 25*13, so sqrt(25*13) = 5*sqrt(13). sqrt(13) is approximately 3.6055, so 5*3.6055 ≈ 18.0275. So, yeah, approximately 18.03.So, Z = (30 - 25)/18.03 ≈ 5/18.03 ≈ 0.277. So, that's correct.Looking up 0.277 in the Z-table: 0.27 is 0.6066, 0.28 is 0.6103. So, 0.277 is about 0.6103 - (0.28 - 0.277)*(0.6103 - 0.6066). Wait, that might be overcomplicating. Alternatively, using linear approximation.Difference between 0.27 and 0.28 is 0.01, which corresponds to a difference of 0.6103 - 0.6066 = 0.0037. So, for 0.277, which is 0.27 + 0.007, the area would be 0.6066 + (0.007/0.01)*0.0037 ≈ 0.6066 + 0.00259 ≈ 0.6092. So, approximately 0.6092. Therefore, the area to the right is 1 - 0.6092 = 0.3908, so about 39.08%.So, approximately 39.1%.Alternatively, using a calculator, the exact value for Z=0.277 is about 0.6093, so 1 - 0.6093 = 0.3907, which is about 39.07%.So, roughly 39.1%.Therefore, the probability is approximately 39.1%.Problem 2: Hypothesis TestingDr. Smith wants to test whether the mean increase in satisfaction levels is significantly different from zero at a 5% significance level. The observed mean increase is 25 points with a sample standard deviation of 12 points.We need to formulate the null and alternative hypotheses and determine whether the therapy is effective.First, let's set up the hypotheses.Null hypothesis (H0): The mean increase is zero. So, μ = 0.Alternative hypothesis (H1): The mean increase is not zero. So, μ ≠ 0.Since the alternative is two-tailed, we'll perform a two-tailed test.Given that the sample size is 100, which is large, we can use the z-test. However, the problem mentions the sample standard deviation, which is 12. Wait, but for a z-test, we need the population standard deviation. Hmm, but since the sample size is large, sometimes people use the sample standard deviation as an estimate.Wait, actually, in practice, when the population standard deviation is unknown, even for large samples, we use the t-test. But with n=100, the t-test is very similar to the z-test.But let's see what the problem says. It says to use the appropriate statistical test. Since the sample size is 100, which is large, and the population standard deviation isn't given, so we have to use the sample standard deviation. Therefore, it's a t-test.But wait, in the first problem, the before and after were independent, but in this case, since it's the same patients, it's a paired test. Wait, actually, in the first problem, the before and after were independent variables, but in reality, for the same patients, the before and after are dependent.Wait, hold on. Wait, in the first problem, the data is before and after for the same patients, but they were modeled as independent. But in reality, they are paired. However, in the first problem, it's just about computing the probability, so maybe they are treated as independent for that purpose.But in the second problem, when doing the hypothesis test, since it's the same patients, we need to consider the paired difference. So, the increase is D = Y - X, which is a single sample of differences.Given that, the sample size is 100, the observed mean increase is 25, and the sample standard deviation is 12.So, we can perform a one-sample t-test on the differences.So, the hypotheses are:H0: μ_D = 0H1: μ_D ≠ 0The test statistic is t = (sample mean - hypothesized mean) / (sample standard deviation / sqrt(n))So, t = (25 - 0) / (12 / sqrt(100)) = 25 / (12 / 10) = 25 / 1.2 ≈ 20.8333Wait, that seems extremely large. Let me check.Sample mean increase: 25Hypothesized mean: 0Sample standard deviation: 12Sample size: 100So, standard error (SE) = 12 / sqrt(100) = 12 / 10 = 1.2So, t = 25 / 1.2 ≈ 20.8333That's a huge t-value. The critical t-value for a two-tailed test with α=0.05 and degrees of freedom = 99 is approximately ±1.984 (since for large degrees of freedom, it approaches 1.96). So, 20.83 is way beyond that.Therefore, we can reject the null hypothesis at the 5% significance level.Alternatively, the p-value would be practically zero, as such a large t-value is extremely unlikely under the null hypothesis.Therefore, the therapy is effective.Wait, but let me think again. Is this a paired test or an independent test? Since it's the same patients before and after, it's a paired test, so we're correct in using the paired differences. So, yes, the test is appropriate.Alternatively, if we had treated it as two independent samples, we would have used a different test, but in this case, since it's paired, the one-sample t-test on the differences is correct.So, to summarize:Null hypothesis: μ_D = 0Alternative hypothesis: μ_D ≠ 0Test statistic t ≈ 20.83, which is way beyond the critical value. Therefore, reject H0. Therapy is effective.Final Answer1. The probability is boxed{0.391}.2. The null hypothesis is rejected, and the therapy is effective. So, the conclusion is boxed{text{Reject } H_0}.</think>

🧠:<think>Okay, so I have this problem about a botanist creating a unique botanical garden with seven circular plots arranged in a hexagonal pattern. There are two parts to the problem, and I need to solve both. Let me start with the first one.Problem 1: Calculating the total area available for planting flowersThe garden has seven circular plots. Each plot is a circle with radius ( r ). The botanist wants each plot to have a non-overlapping area that covers 75% of the area of each circle. So, I need to find the total area available for planting flowers in all seven plots.First, let me recall the formula for the area of a circle. The area ( A ) of a circle with radius ( r ) is given by:[ A = pi r^2 ]Since each plot is a circle, each plot has an area of ( pi r^2 ). The botanist wants 75% of each plot's area to be used for planting. So, the area available for flowers in one plot is:[ 0.75 times pi r^2 ]Now, since there are seven plots, the total area available for planting flowers would be seven times the area of one plot:[ 7 times 0.75 times pi r^2 ]Let me compute that:First, 7 multiplied by 0.75. 7 times 0.75 is 5.25, right? Because 7 times 0.7 is 4.9, and 7 times 0.05 is 0.35, so 4.9 + 0.35 = 5.25.So, the total area is:[ 5.25 times pi r^2 ]But 5.25 can be expressed as a fraction. 5.25 is the same as ( frac{21}{4} ). So, another way to write this is:[ frac{21}{4} pi r^2 ]Alternatively, if I want to write it as a decimal, it's 5.25πr². But since the problem says to express it in terms of ( r ), either form should be acceptable, but maybe the fractional form is more precise.Wait, let me double-check. Each plot is 75% covered, so each plot contributes 0.75πr², and there are seven plots. So, 7 * 0.75 is indeed 5.25, so 5.25πr². Yeah, that seems right.So, I think that's the answer for the first part. The total area is ( frac{21}{4} pi r^2 ) or 5.25πr².Problem 2: Calculating the total length of pipes required for the watering systemNow, the second part is about designing a watering system. The botanist uses the points where circles intersect as nodes and connects them using pipes along the shortest paths between nodes. The circles are arranged in a hexagonal pattern, with one at the center and six surrounding it, each tangent to its neighbors. The intersections form a regular hexagon within the garden. I need to find the total length of the pipes required.Hmm, okay. So, let me visualize this. There's a central circle, and six circles around it, each tangent to the central one and to their two neighbors. The points where the surrounding circles intersect each other form a regular hexagon.Wait, but each surrounding circle is tangent to its neighbors, so the distance between centers of adjacent surrounding circles is 2r, right? Because each has radius r, so the distance between centers is 2r.But the centers of the surrounding circles form a regular hexagon around the central circle. The side length of this hexagon is 2r, since each surrounding circle is 2r apart from its neighbor.But the problem says the intersections form a regular hexagon. Hmm, so the points where the surrounding circles intersect each other are the nodes. So, each surrounding circle intersects its two neighbors at two points each, but the nodes are the intersection points that form a regular hexagon.Wait, so if each surrounding circle is tangent to its neighbor, they only touch at one point, right? Because if two circles are tangent, they intersect at exactly one point. So, if the surrounding circles are tangent to each other, then the points of tangency are the nodes.But the problem says the intersections form a regular hexagon. So, the nodes are the points where the surrounding circles are tangent to each other, which are six points forming a regular hexagon.So, the watering system connects these nodes with pipes along the shortest paths. So, the pipes would be the sides of the regular hexagon.Therefore, the total length of the pipes would be the perimeter of the regular hexagon.But wait, let me think again. The nodes are the points where the surrounding circles intersect, which in this case are the points of tangency. So, each side of the hexagon is the distance between two adjacent nodes, which is the same as the side length of the hexagon.But what is the side length of this hexagon? Since each surrounding circle has radius r, and the centers of the surrounding circles form a regular hexagon with side length 2r (distance between centers is 2r because the circles are tangent). So, the distance between centers is 2r.But the nodes (points of tangency) are located at a distance of r from each center. So, the distance between two adjacent nodes is equal to the side length of the hexagon formed by the nodes.Wait, maybe I need to compute the distance between two adjacent nodes. Let me consider two adjacent surrounding circles. Each has radius r, and their centers are 2r apart. The point where they are tangent is exactly halfway between the two centers, right? Because if two circles of radius r are tangent, the point of tangency is at a distance r from each center along the line connecting the centers.So, the distance between two adjacent nodes is the distance between two points that are each r distance from their respective centers, which are 2r apart. So, the two nodes are each on the line connecting the centers, at a distance r from each center. So, the distance between the two nodes is 2r - r - r = 0? Wait, that can't be.Wait, no. If two circles are tangent, they only touch at one point. So, the point of tangency is the only intersection point, and that point is located exactly halfway between the two centers. So, the distance between two adjacent nodes (points of tangency) is zero because they are the same point? That doesn't make sense.Wait, perhaps I'm misunderstanding the problem. It says the intersections form a regular hexagon. So, maybe the surrounding circles intersect each other at two points each, forming a hexagon. But if the surrounding circles are only tangent, they only intersect at one point. So, perhaps the circles are not just tangent, but overlapping slightly so that they intersect at two points each, forming a regular hexagon.But the problem says each circle is tangent to its neighboring circles, so they only touch at one point. So, the nodes are the points of tangency, which are six points forming a regular hexagon.So, the watering system connects these nodes with pipes along the shortest paths. So, the pipes are the sides of the hexagon.But the side length of the hexagon is the distance between two adjacent nodes, which is the distance between two points of tangency of adjacent surrounding circles.Wait, but if each surrounding circle is tangent to its neighbor, the point of tangency is a single point. So, the nodes are these single points, each shared between two surrounding circles.So, the regular hexagon is formed by connecting these points of tangency. So, each side of the hexagon is the distance between two adjacent points of tangency.But how do we compute that distance?Let me think. The centers of the surrounding circles form a regular hexagon with side length 2r. Each surrounding circle has radius r, and the central circle also has radius r.The points of tangency between two surrounding circles are located at a distance of r from each center. So, if I have two centers separated by 2r, the point of tangency is exactly halfway between them, at a distance r from each center.So, the distance between two adjacent points of tangency (nodes) would be the length of the side of the hexagon formed by these points.Wait, but in a regular hexagon, all sides are equal, and the distance between adjacent nodes is equal to the side length.But in this case, the nodes are points of tangency between surrounding circles, each located at a distance r from their respective centers.So, let me model this. Let's consider two adjacent surrounding circles, say Circle A and Circle B, each with radius r, whose centers are 2r apart. The point of tangency P is exactly halfway between the centers, so it's at a distance r from both centers.Now, the next surrounding circle, Circle C, is also 2r away from Circle B, and its point of tangency with Circle B is another point Q, also halfway between the centers of B and C.So, the distance between points P and Q is the side length of the hexagon.To find the distance PQ, let me consider the triangle formed by the centers of A, B, and C. Since the centers form a regular hexagon, the angle between AB and BC is 60 degrees.Wait, in a regular hexagon, each internal angle is 120 degrees, but the angle between two adjacent sides from the center is 60 degrees.Wait, no. In a regular hexagon, the central angle between two adjacent vertices is 60 degrees because 360/6 = 60.So, the centers of the surrounding circles form a regular hexagon with side length 2r, and each center is 2r apart from its neighbors, with each central angle being 60 degrees.So, if I consider the centers of A, B, and C, they form an equilateral triangle with side length 2r, because each side is 2r, and each angle is 60 degrees.Now, points P and Q are the midpoints of AB and BC, respectively, each at a distance r from B.So, in triangle ABC, which is equilateral with side length 2r, points P and Q are midpoints of AB and BC.So, the distance PQ can be found using the coordinates or using vector geometry.Let me assign coordinates to make it easier.Let me place center B at the origin (0,0). Then, center A is at (-2r, 0), and center C is at (2r, 0), but wait, no. If the centers form a regular hexagon, the coordinates would be different.Wait, perhaps it's better to place center B at (0,0), center A at (-r, 0), and center C at (r, 0), but that might not form a regular hexagon.Wait, no. Let me think again.In a regular hexagon, each vertex is at a distance equal to the side length from the center. So, if the centers of the surrounding circles form a regular hexagon with side length 2r, then the distance from the center of the hexagon (which is the center of the central circle) to each surrounding center is 2r.Wait, no. Wait, the distance from the central circle's center to each surrounding circle's center is equal to the radius of the central circle plus the radius of the surrounding circles if they are tangent. But in this case, the central circle has radius r, and each surrounding circle also has radius r, and they are tangent. So, the distance between the central circle's center and each surrounding circle's center is 2r.Therefore, the centers of the surrounding circles lie on a circle of radius 2r centered at the central circle's center.So, the centers of the surrounding circles form a regular hexagon with side length equal to the distance between two adjacent surrounding centers.Wait, in a regular hexagon inscribed in a circle of radius 2r, the side length is equal to the radius. So, the side length is 2r.Wait, no. In a regular hexagon, the side length is equal to the radius of the circumscribed circle. So, if the centers of the surrounding circles are on a circle of radius 2r, then the side length of the hexagon is 2r.Therefore, the distance between centers of two adjacent surrounding circles is 2r.So, going back, centers A, B, and C form an equilateral triangle with side length 2r.Points P and Q are the midpoints of AB and BC, respectively, each at a distance r from B.So, to find the distance PQ, we can use coordinates.Let me place center B at (0,0). Then, center A is at (-2r, 0), and center C is at (2r, 0). Wait, but that would make the triangle ABC a straight line, which is not an equilateral triangle.Wait, no. If centers A, B, and C form an equilateral triangle with side length 2r, then their coordinates should be arranged such that each is 2r apart.Let me place center B at (0,0). Then, center A can be at (2r, 0). Center C would be at (r, r√3), forming an equilateral triangle.Wait, yes, because in an equilateral triangle with side length 2r, the height is √3 r. So, if center B is at (0,0), center A is at (2r, 0), and center C is at (r, r√3).So, point P is the midpoint of AB, which is between (0,0) and (2r, 0), so P is at (r, 0).Point Q is the midpoint of BC, which is between (0,0) and (r, r√3). So, midpoint Q is at ((0 + r)/2, (0 + r√3)/2) = (r/2, (r√3)/2).Now, the distance between P (r, 0) and Q (r/2, (r√3)/2) can be calculated using the distance formula.Distance PQ = √[(r - r/2)² + (0 - (r√3)/2)²] = √[(r/2)² + ( - (r√3)/2)²] = √[(r²/4) + (3r²/4)] = √[(4r²/4)] = √[r²] = r.So, the distance between two adjacent nodes (points of tangency) is r.Therefore, each side of the regular hexagon formed by the nodes is of length r.Since a regular hexagon has six sides, the total length of the pipes required is 6 * r = 6r.Wait, but let me confirm this. The nodes are the points of tangency between the surrounding circles, and each side of the hexagon connecting these nodes is length r. So, the perimeter is 6r.But wait, in the problem statement, it says the botanist connects the nodes using pipes along the shortest paths between nodes. So, the shortest path between two nodes is the straight line connecting them, which is the side of the hexagon, which we found to be r.Therefore, the total length of the pipes is 6r.But let me think again. If the nodes form a regular hexagon with side length r, then yes, the perimeter is 6r. So, that should be the total length of the pipes.Wait, but earlier, I thought the centers of the surrounding circles form a regular hexagon with side length 2r, and the nodes (points of tangency) form another regular hexagon inside, with side length r. So, the pipes are the sides of this inner hexagon, each of length r, so total length 6r.Yes, that seems correct.So, the total length of the pipes required is 6r.Wait, but let me make sure I didn't make a mistake in the coordinate calculation.I placed center B at (0,0), center A at (2r, 0), and center C at (r, r√3). Then, point P is the midpoint of AB, which is (r, 0). Point Q is the midpoint of BC, which is (r/2, (r√3)/2). Then, the distance between P and Q is √[(r - r/2)² + (0 - (r√3)/2)²] = √[(r/2)² + ( - (r√3)/2)²] = √[r²/4 + 3r²/4] = √[r²] = r. So, yes, that's correct.Therefore, each side of the hexagon is r, so the total length is 6r.So, the total length of the pipes is 6r.Summary of Thoughts:1. For the first problem, each of the seven plots contributes 75% of its area to planting. Each plot's area is πr², so 0.75πr² per plot. Seven plots give 5.25πr², which is 21/4 πr².2. For the second problem, the nodes are the points where the surrounding circles are tangent to each other, forming a regular hexagon. The distance between adjacent nodes is r, so the total pipe length is 6r.Final Answer1. The total area available for planting is boxed{dfrac{21}{4} pi r^2}.2. The total length of the pipes required is boxed{6r}.</think>