answer：Okay, so I have this problem about a parent reading a book to their child about crossing the street safely. The book uses a Poisson distribution to model car arrivals at a busy intersection. There are two parts to the problem. Let me try to tackle them one by one.Starting with the first question: Calculate the probability that no cars arrive from either street in a 30-second interval while the parent and child are crossing. They want the answer as a decimal rounded to four decimal places.Alright, so Poisson distribution is used here. I remember that the Poisson distribution gives the probability of a certain number of events happening in a fixed interval of time or space. The formula is:P(k) = (λ^k * e^(-λ)) / k!Where λ is the average rate (the expected number of occurrences), k is the number of occurrences, and e is the base of the natural logarithm.In this case, cars arrive from each of two perpendicular streets with an average rate of 10 cars per minute. So, first, I need to figure out the rate for a 30-second interval because the question is about a 30-second window.Since 30 seconds is half a minute, I can calculate the average number of cars arriving in 30 seconds from one street. If it's 10 cars per minute, then in 30 seconds, it should be 10 * (30/60) = 5 cars.But wait, the parent and child are crossing, so we need to consider cars arriving from both streets. Since the streets are perpendicular, I assume the arrivals are independent. So, the total rate for both streets combined would be 5 + 5 = 10 cars in 30 seconds.So, λ for the combined rate is 10 cars in 30 seconds. Now, we need the probability that no cars arrive from either street in that interval. That is, k = 0.Plugging into the Poisson formula:P(0) = (10^0 * e^(-10)) / 0! = (1 * e^(-10)) / 1 = e^(-10)Calculating e^(-10). I remember that e is approximately 2.71828. So, e^(-10) is 1 / e^10.Calculating e^10: Let me approximate it. e^1 is about 2.718, e^2 is about 7.389, e^3 is about 20.085, e^4 is about 54.598, e^5 is about 148.413, e^6 is about 403.429, e^7 is about 1096.633, e^8 is about 2980.911, e^9 is about 8103.083, and e^10 is about 22026.465.So, e^(-10) is approximately 1 / 22026.465 ≈ 0.000045399.Wait, is that right? Because 1 / 22026 is roughly 0.0000454. So, about 0.0000454.But let me double-check. Maybe I can use a calculator for a more precise value. Alternatively, I know that ln(10) is about 2.302585, so e^(-10) is 10^(-10 / 2.302585) ≈ 10^(-4.34294) ≈ 10^(-4) * 10^(-0.34294). 10^(-4) is 0.0001, and 10^(-0.34294) is approximately 0.454. So, 0.0001 * 0.454 ≈ 0.0000454. So, same result.Therefore, the probability is approximately 0.0000454, which is 0.0000 when rounded to four decimal places. Wait, that can't be right because 0.0000454 is 0.0000 when rounded to four decimal places, but that seems too low. Is there a mistake here?Wait, hold on. Maybe I misapplied the rates. Let me go back.The problem says cars arrive according to a Poisson distribution with an average rate of 10 cars per minute from each of two perpendicular streets. So, each street has 10 cars per minute. So, for a 30-second interval, each street would have λ = 10 * (30/60) = 5 cars.But since the parent and child are crossing, they have to consider cars from both streets. So, the combined rate is 5 + 5 = 10 cars per 30 seconds.Wait, but is that correct? Because if the arrivals are independent, the total number of cars from both streets is the sum of two independent Poisson variables, which is also Poisson with λ = λ1 + λ2.So, yes, the combined rate is 10 cars per 30 seconds.Therefore, the probability that no cars arrive from either street is P(0) = e^(-10) ≈ 0.0000454.But 0.0000454 is 0.0000 when rounded to four decimal places. Hmm, that seems really low. Is that correct?Alternatively, maybe I should model each street separately and then find the joint probability.Since the arrivals from each street are independent, the probability that no cars arrive from the first street is e^(-5), and similarly for the second street. So, the joint probability is e^(-5) * e^(-5) = e^(-10), which is the same as before.So, that's consistent. So, the probability is e^(-10) ≈ 0.0000454, which is approximately 0.0000 when rounded to four decimal places.But wait, 0.0000454 is 0.0000 when rounded to four decimal places because the fifth decimal is 4, which is less than 5, so we don't round up. So, 0.0000.But that seems extremely low. Is that correct? Let me think about it.If the average number of cars in 30 seconds is 10, the chance of zero cars is indeed very low. Because if you expect 10 cars, the chance of none is minuscule. So, 0.000045 is correct.But just to make sure, let me compute e^(-10) more accurately.Using a calculator, e^(-10) is approximately 0.00004539993.So, 0.00004539993, which is approximately 0.0000454. So, 0.0000 when rounded to four decimal places.Alternatively, maybe the question is asking for the probability that no cars arrive from either street in a 30-second interval, but perhaps it's considering each street separately? Wait, no, the parent and child are crossing, so they have to consider both streets. So, the combined rate is 10 cars per 30 seconds.Alternatively, maybe I'm supposed to calculate the probability that no cars arrive from each street in 30 seconds, which would be the product of the two probabilities.Wait, but that's what I did. Because for each street, the probability of no cars is e^(-5), and since the streets are independent, the joint probability is e^(-5) * e^(-5) = e^(-10). So, that's correct.Therefore, the answer is approximately 0.0000.But wait, 0.0000454 is 0.0000 when rounded to four decimal places, but sometimes, in some contexts, people might write it as 0.0000 or 0.0000. Alternatively, maybe we need to write it as 0.0000, but it's actually 0.0000.But let me check if the question says "no cars arrive from either street". So, it's the probability that neither street has any cars. So, yes, that's correct.So, moving on to the second question: After reading the book, the parent devises a safety plan where they will only cross when they can see a gap of at least 15 seconds with no cars arriving from either street. Given the Poisson distribution of car arrivals, calculate the expected number of such 15-second gaps in a 10-minute period.Hmm, okay. So, they want the expected number of 15-second intervals where no cars arrive from either street in a 10-minute period.First, let's figure out the rate for a 15-second interval.Again, each street has a rate of 10 cars per minute, so in 15 seconds, which is 15/60 = 0.25 minutes, the rate is 10 * 0.25 = 2.5 cars per 15 seconds per street.Since there are two streets, the combined rate is 2.5 + 2.5 = 5 cars per 15 seconds.So, the probability that no cars arrive from either street in a 15-second interval is e^(-5), similar to the first part but with λ = 5.Wait, no. Wait, actually, for each street, the rate is 2.5, so the probability of no cars from one street is e^(-2.5). Since the two streets are independent, the joint probability is e^(-2.5) * e^(-2.5) = e^(-5). So, yes, that's correct.So, the probability of a 15-second gap with no cars is e^(-5). Let me compute that.e^(-5) is approximately 0.006737947.So, approximately 0.006737947.Now, the question is about the expected number of such 15-second gaps in a 10-minute period.First, let's figure out how many 15-second intervals are in 10 minutes.10 minutes is 600 seconds. Each interval is 15 seconds, so 600 / 15 = 40 intervals.So, there are 40 possible 15-second intervals in 10 minutes.But wait, actually, in a 10-minute period, how many non-overlapping 15-second intervals are there? It's 40, as above.But wait, actually, in a Poisson process, the number of events in non-overlapping intervals are independent, so the expected number of gaps is the number of intervals multiplied by the probability of a gap in each interval.But wait, actually, the expected number of gaps is the sum over all intervals of the probability that each interval is a gap.So, if we have 40 intervals, each with probability p = e^(-5) of being a gap, then the expected number of gaps is 40 * p.So, 40 * e^(-5) ≈ 40 * 0.006737947 ≈ 0.26951788.So, approximately 0.2695, which is about 0.27 when rounded to two decimal places, but the question doesn't specify rounding, just to calculate the expected number.But wait, let me think again. Is that correct?Wait, actually, in a Poisson process, the number of events in non-overlapping intervals are independent, so the expected number of gaps is indeed the number of intervals multiplied by the probability of a gap in each interval.But wait, another way to think about it is that the expected number of gaps is equal to the total time divided by the expected time between gaps. But I think the first approach is correct.Alternatively, perhaps the question is considering overlapping intervals? Because in reality, you can have overlapping 15-second windows. For example, starting at second 0, 15 seconds, 30 seconds, etc., but also starting at second 1, 16 seconds, etc. But in that case, the number of intervals would be much larger, but the events are not independent.But the question says "a gap of at least 15 seconds with no cars arriving from either street". So, it's about the occurrence of such gaps, regardless of where they are in the 10-minute period.Wait, actually, in a Poisson process, the distribution of the time between events is exponential. So, the probability that there is a gap of at least 15 seconds is equal to the probability that no cars arrive in that 15-second interval, which is e^(-λt), where λ is the rate.But in this case, the combined rate is 5 cars per 15 seconds, so λ = 5. So, the probability is e^(-5).But the expected number of such gaps in 10 minutes is a bit tricky. Because the occurrence of gaps can overlap, and they are not independent events.Wait, perhaps it's better to model this as a renewal process, where we're looking for the expected number of renewals (gaps) in a given time.But I might be overcomplicating it.Alternatively, perhaps the question is simpler. It just wants the expected number of 15-second intervals with no cars, considering the entire 10-minute period as 40 non-overlapping 15-second intervals.So, in that case, the expected number is 40 * e^(-5) ≈ 0.2695.But let me check if that's the correct interpretation.The question says: "calculate the expected number of such 15-second gaps in a 10-minute period."So, if we consider the 10-minute period as 40 non-overlapping 15-second intervals, then the expected number is 40 * e^(-5).But another interpretation is that we're looking for the expected number of times a 15-second gap occurs, regardless of where it is in the 10 minutes. In that case, the number of gaps is not just 40, because a gap can start at any second, not just every 15 seconds.But in that case, the problem becomes more complex because the events are overlapping and not independent.Wait, but in a Poisson process, the number of gaps of at least a certain length can be calculated using the formula for the expected number of such gaps.I recall that in a Poisson process with rate λ, the expected number of gaps of length at least t in a time interval T is (T - t) * e^(-λ t).But I'm not entirely sure. Let me think.Alternatively, the expected number of times a gap of at least t occurs is equal to the integral over the entire period of the probability that a gap starts at time s and lasts for t.But that might be more complicated.Wait, perhaps it's better to model the process as a Poisson process with rate λ = 10 cars per minute from each street, so combined rate is 20 cars per minute.Wait, no, actually, the combined rate is 10 + 10 = 20 cars per minute, which is 20 cars per minute.Wait, but earlier, for a 15-second interval, we had a combined rate of 5 cars per 15 seconds, which is 20 cars per minute.Wait, hold on, 15 seconds is 0.25 minutes, so 10 cars per minute * 0.25 minutes = 2.5 cars per 15 seconds per street, so combined 5 cars per 15 seconds.So, the rate is 5 cars per 15 seconds.So, the probability that a 15-second interval has no cars is e^(-5).Now, to find the expected number of such intervals in a 10-minute period.But if we consider non-overlapping intervals, it's 40 intervals, each with probability e^(-5), so expected number is 40 * e^(-5).But if we consider overlapping intervals, the expected number is different.Wait, in the case of overlapping intervals, the expected number of gaps can be calculated as (T - t + 1) * e^(-λ t), but I'm not sure.Wait, actually, in a Poisson process, the expected number of gaps of at least t in a time period T is (T - t) * e^(-λ t).But I need to verify this.Alternatively, I can think of it as a sliding window. For each possible starting time s in [0, T - t], the probability that the interval [s, s + t] has no cars is e^(-λ t). Since these intervals are overlapping, the events are not independent, but the expectation is linear regardless of independence.So, the expected number of such intervals is the integral from s = 0 to s = T - t of e^(-λ t) ds.Which is (T - t) * e^(-λ t).So, in this case, T is 10 minutes, which is 600 seconds, t is 15 seconds.So, T - t = 600 - 15 = 585 seconds.But wait, actually, in terms of units, we need to make sure that λ is in the same units as t.Earlier, we had λ = 5 cars per 15 seconds. So, λ t would be 5 * 15 seconds? Wait, no, λ is already per 15 seconds.Wait, perhaps it's better to express everything in seconds.The combined rate is 20 cars per minute, which is 20 / 60 = 1/3 cars per second.So, λ = 1/3 cars per second.Then, for a 15-second interval, the expected number of cars is λ * t = (1/3) * 15 = 5 cars.So, the probability of no cars in 15 seconds is e^(-5).Now, the expected number of 15-second gaps in a 10-minute period.If we consider the entire 10-minute period as 600 seconds, the number of possible starting points for a 15-second gap is 600 - 15 + 1 = 586.But since we're dealing with continuous time, it's actually an integral from 0 to 600 - 15 = 585 seconds.But in expectation, the number of gaps is the integral over s from 0 to 585 of the probability that the interval [s, s + 15] has no cars.Since the process is stationary, this probability is the same for all s, which is e^(-5).Therefore, the expected number is (585) * e^(-5).But wait, 585 seconds is 9.75 minutes. Wait, but 600 - 15 = 585 seconds.But 585 * e^(-5) ≈ 585 * 0.006737947 ≈ 585 * 0.006737947.Calculating that:585 * 0.006737947 ≈ Let's compute 500 * 0.006737947 = 3.368973585 * 0.006737947 ≈ 0.572725995Adding together: 3.3689735 + 0.572725995 ≈ 3.9417So, approximately 3.9417.But wait, that can't be right because earlier, with non-overlapping intervals, it was about 0.2695, but with overlapping, it's 3.9417.But that seems contradictory.Wait, actually, in the non-overlapping case, we had 40 intervals, each contributing an expectation of e^(-5), so total expectation 40 * e^(-5) ≈ 0.2695.But in the overlapping case, considering all possible starting points, the expectation is (600 - 15) * e^(-5) ≈ 585 * e^(-5) ≈ 3.9417.But which one is correct?Wait, the question says: "calculate the expected number of such 15-second gaps in a 10-minute period."It doesn't specify whether the gaps are non-overlapping or not. So, in reality, the number of gaps can overlap, so the expected number is higher.But wait, in the Poisson process, the expected number of gaps of at least t in a time period T is (T - t) * e^(-λ t).But let me confirm this formula.Yes, in a Poisson process, the expected number of intervals of length t with no events in a time period T is (T - t + 1) * e^(-λ t). But actually, in continuous time, it's (T - t) * e^(-λ t).Wait, let me think about it.The expected number of such intervals is the integral from 0 to T - t of the probability that the interval [s, s + t] has no events, which is e^(-λ t) for each s.Since the process is stationary, the probability is constant, so the integral is (T - t) * e^(-λ t).Therefore, in this case, T = 600 seconds, t = 15 seconds, λ = 1/3 cars per second.So, the expected number is (600 - 15) * e^(- (1/3)*15 ) = 585 * e^(-5) ≈ 585 * 0.006737947 ≈ 3.9417.So, approximately 3.9417.But let me check the units again.Wait, λ is 1/3 cars per second, so λ * t = (1/3) * 15 = 5, which is correct.Therefore, the expected number is 585 * e^(-5) ≈ 3.9417.So, approximately 3.9417.But wait, that seems high because in 10 minutes, expecting almost 4 gaps of 15 seconds each with no cars. But given that the average number of cars per 15 seconds is 5, the probability of a gap is low, but since we're considering all possible starting points, the expectation is higher.Alternatively, let me think about it differently. The expected number of gaps is the integral over all possible starting times of the probability that a gap starts at that time.Since each gap is 15 seconds, the number of possible starting times is T - t, which is 585 seconds.Each starting time has a probability e^(-5) of being a gap.Therefore, the expected number is 585 * e^(-5) ≈ 3.9417.So, that seems correct.Therefore, the expected number is approximately 3.9417, which is about 3.9417.But let me compute it more accurately.First, e^(-5) ≈ 0.006737947.585 * 0.006737947.Let me compute 500 * 0.006737947 = 3.368973585 * 0.006737947 ≈ 0.572725995Adding together: 3.3689735 + 0.572725995 ≈ 3.9416995So, approximately 3.9417.Therefore, the expected number is approximately 3.9417.But the question says "calculate the expected number of such 15-second gaps in a 10-minute period."So, that's approximately 3.9417.But let me think again if this is correct.Alternatively, perhaps the question is considering the number of non-overlapping gaps. But the question doesn't specify, so I think the correct approach is to consider all possible starting points, which gives us the higher expectation.Therefore, the expected number is approximately 3.9417.But let me check if there's another way to calculate it.Another approach is to model the process as a Poisson process and use the formula for the expected number of runs or gaps.In a Poisson process, the number of gaps of length at least t is a Poisson random variable with parameter (T - t) * e^(-λ t).But wait, no, that's not quite right. The number of gaps is not Poisson distributed, but the expectation is (T - t) * e^(-λ t).So, yes, that's consistent with what I calculated earlier.Therefore, the expected number is approximately 3.9417.So, rounding to four decimal places, it's 3.9417.But the question doesn't specify rounding, just to calculate the expected number.Alternatively, maybe I should express it as a fraction or something, but 3.9417 is already precise enough.Wait, but let me check the exact value.585 * e^(-5) = 585 * 0.006737947 ≈ 3.9417.Yes, that's correct.Therefore, the expected number is approximately 3.9417.But let me think again about the interpretation.If we consider the 10-minute period as 600 seconds, and we're looking for the expected number of 15-second intervals with no cars, regardless of where they are, then yes, it's 585 * e^(-5).But another way to think about it is that the expected number of such gaps is equal to the expected number of times the process has a run of at least 15 seconds without any cars.In a Poisson process, the distribution of the lengths of intervals between events is exponential. So, the probability that the interval between two consecutive cars is at least 15 seconds is e^(-λ t), where λ is the rate.But in this case, we're looking for gaps of at least 15 seconds, so the expected number of such gaps can be calculated as the expected number of inter-arrival intervals that are at least 15 seconds.But that's a different approach.Wait, in a Poisson process with rate λ, the expected number of events in time T is λ T.The number of inter-arrival intervals in T is equal to the number of events, which is Poisson distributed with parameter λ T.But the lengths of these intervals are exponential with parameter λ.So, the expected number of intervals longer than t is equal to the expected number of events, which is λ T, multiplied by the probability that an interval is longer than t, which is e^(-λ t).Wait, no, that's not correct.Wait, actually, the expected number of intervals longer than t is equal to the expected number of events, which is λ T, multiplied by the probability that a given interval is longer than t, which is e^(-λ t).But wait, that would be λ T * e^(-λ t).But in our case, λ is 20 cars per minute, which is 1/3 cars per second.So, λ = 1/3 per second.T = 600 seconds.t = 15 seconds.So, the expected number of intervals longer than 15 seconds is λ T * e^(-λ t) = (1/3) * 600 * e^(- (1/3)*15 ) = 200 * e^(-5) ≈ 200 * 0.006737947 ≈ 1.3475894.So, approximately 1.3476.Wait, that's different from the previous result.Hmm, so which one is correct?I think the confusion arises from whether we're counting the number of gaps between events or the number of intervals in the entire period.In the first approach, we considered all possible starting points, leading to (T - t) * e^(-λ t) ≈ 3.9417.In the second approach, considering the expected number of inter-arrival intervals longer than t, we get λ T * e^(-λ t) ≈ 1.3476.So, which one is correct?I think the second approach is correct when considering the number of gaps between events, but the first approach counts all possible intervals, including those that don't necessarily correspond to actual gaps between events.Wait, for example, if two cars arrive back-to-back with less than 15 seconds between them, the interval between them is less than 15 seconds, but there might be other intervals elsewhere that are longer than 15 seconds, even if they don't correspond to gaps between events.Wait, no, actually, in a Poisson process, the entire time is divided into intervals by the arrival times. So, the number of gaps longer than t is equal to the number of inter-arrival intervals longer than t.But in that case, the expected number is λ T * e^(-λ t).But in our case, the process is two independent Poisson processes (from each street), so the combined process is also Poisson with rate λ = 20 cars per minute.Therefore, the expected number of inter-arrival intervals longer than 15 seconds is λ T * e^(-λ t) = (20 cars per minute) * (10 minutes) * e^(-20 * (15/60)).Wait, hold on, let me clarify.Wait, λ is 20 cars per minute, so in 10 minutes, the expected number of cars is 200.The expected number of inter-arrival intervals is equal to the expected number of cars, which is 200.Each interval has an exponential distribution with rate λ = 20 cars per minute, so the probability that an interval is longer than 15 seconds is e^(-λ t), where t is in minutes.15 seconds is 0.25 minutes.So, e^(-20 * 0.25) = e^(-5).Therefore, the expected number of intervals longer than 15 seconds is 200 * e^(-5) ≈ 200 * 0.006737947 ≈ 1.3475894.So, approximately 1.3476.But this contradicts the first approach.Wait, so which one is correct?I think the confusion is about what exactly constitutes a "gap". If a gap is defined as a period between two consecutive cars where no cars arrive for at least 15 seconds, then the expected number is 1.3476.But if a gap is defined as any 15-second interval within the 10-minute period where no cars arrive, regardless of whether it's between two cars or not, then the expected number is 3.9417.So, the question says: "they will only cross when they can see a gap of at least 15 seconds with no cars arriving from either street."So, the parent is looking for any 15-second interval where no cars arrive, regardless of where it is. So, it's not necessarily between two cars, but any 15-second window.Therefore, the correct approach is the first one, where we consider all possible 15-second intervals in the 10-minute period, leading to an expected number of approximately 3.9417.But let me think again.In the Poisson process, the number of events in any interval of length t is Poisson distributed with parameter λ t.The expected number of such intervals with no events is (T - t + 1) * e^(-λ t) in discrete time, but in continuous time, it's (T - t) * e^(-λ t).Therefore, in this case, T = 600 seconds, t = 15 seconds, λ = 1/3 per second.So, (600 - 15) * e^(- (1/3)*15 ) = 585 * e^(-5) ≈ 3.9417.Therefore, the expected number is approximately 3.9417.So, that's the answer.But to make sure, let me think about the units again.λ is 1/3 cars per second.t is 15 seconds.So, λ t = 5, which is correct.Therefore, the expected number is 585 * e^(-5) ≈ 3.9417.So, the final answer is approximately 3.9417.But let me compute it more accurately.585 * e^(-5) = 585 * 0.006737947 ≈ 585 * 0.006737947.Calculating 500 * 0.006737947 = 3.368973585 * 0.006737947 ≈ 0.572725995Adding together: 3.3689735 + 0.572725995 ≈ 3.9416995So, approximately 3.9417.Therefore, the expected number is approximately 3.9417.So, rounding to four decimal places, it's 3.9417.But the question doesn't specify rounding, so maybe we can leave it as is.Alternatively, if we need to present it as a decimal, it's approximately 3.9417.So, summarizing:1. The probability that no cars arrive from either street in a 30-second interval is e^(-10) ≈ 0.0000454, which is 0.0000 when rounded to four decimal places.2. The expected number of 15-second gaps in a 10-minute period is approximately 3.9417.But wait, let me check the first answer again.The first answer was e^(-10) ≈ 0.0000454, which is 0.0000 when rounded to four decimal places.But 0.0000454 is 0.0000 when rounded to four decimal places because the fifth decimal is 4, which is less than 5, so we don't round up.But sometimes, people might write it as 0.0000, but it's actually 0.0000.Alternatively, if we consider significant figures, 0.0000454 is 4.54 x 10^(-5), which is approximately 0.0000 when rounded to four decimal places.Therefore, the answers are:1. 0.00002. Approximately 3.9417But let me check if the second answer is correct.Wait, another way to think about it is that the expected number of gaps is the integral over the entire period of the probability that a gap starts at each point.So, for each second s in [0, 600 - 15], the probability that the interval [s, s + 15] has no cars is e^(-5).Therefore, the expected number is (600 - 15) * e^(-5) = 585 * e^(-5) ≈ 3.9417.Yes, that seems correct.Therefore, the final answers are:1. 0.00002. Approximately 3.9417But let me write them as per the instructions.For the first question, the answer is 0.0000.For the second question, the answer is approximately 3.9417.But let me check if the second answer should be an integer or not.Wait, the expected number doesn't have to be an integer, so 3.9417 is fine.Alternatively, maybe we can write it as a fraction.But 3.9417 is approximately 3.9417.Alternatively, if we compute it more accurately, 585 * e^(-5) ≈ 585 * 0.006737947 ≈ 3.9417.So, that's correct.Therefore, the answers are:1. 0.00002. 3.9417But let me check if the second answer is correct in terms of units.Yes, because we're considering 10 minutes as 600 seconds, and t = 15 seconds, so the calculation is consistent.Therefore, I think these are the correct answers.

answer：First, I need to determine the total time spent on each component of a single event. The storytelling session takes 45 minutes, the digital art workshop takes 60 minutes, and the Q&A session takes 30 minutes. Adding these together gives a total of 135 minutes per event.Next, since there are 4 events planned each month, I multiply the time per event by the number of events: 135 minutes multiplied by 4 equals 540 minutes.Finally, to convert the total minutes into hours, I divide by 60. So, 540 minutes divided by 60 equals 9 hours.Therefore, the artist spends a total of 9 hours on these events in one month.

answer：First, I need to determine the total time Emily spent performing. She played 4 songs, each lasting 3 minutes. Next, I'll calculate the total break time. After each song, there was a 2-minute break, and since there were 4 songs, there were 3 breaks in between them.Finally, I'll add the total performance time and the total break time to find the overall time Emily spent on stage.

answer：First, I need to calculate the total amount of reparations across all three case studies. In the first case, 500,000 was distributed to 250 individuals. In the second case, 750,000 was awarded to 300 people. In the third case, 450,000 was given to 150 individuals.Adding these amounts together: 500,000 + 750,000 + 450,000 equals 1,700,000.Next, I need to find the total number of individuals who received reparations. Adding the number of individuals from each case: 250 + 300 + 150 equals 700.Finally, to find the average amount received per individual, I divide the total reparations by the total number of individuals: 1,700,000 divided by 700 equals 2,428.57.Therefore, the average amount received per individual across all three case studies is 2,428.57.

answer：First, I need to determine the number of throws each grandchild makes per basket. Each basket is designed to be completed in 3 throws on average, and each grandchild makes 2 extra throws per basket. So, the total throws per basket per grandchild would be 3 plus 2, which equals 5 throws.Next, there are 5 baskets in the course. Therefore, each grandchild will make 5 throws per basket multiplied by 5 baskets, resulting in 25 throws per grandchild.Since there are 3 grandchildren, the total number of throws for all of them combined would be 25 throws per grandchild multiplied by 3 grandchildren, totaling 75 throws.

answer：First, I need to determine the cost of each fabric by multiplying the number of yards Jamie buys by the cost per yard.For the plaid fabric, Jamie buys 2 yards at 5 per yard, which totals 10.For the floral fabric, Jamie buys 2 yards at 7 per yard, totaling 14.For the striped fabric, Jamie buys 2 yards at 4 per yard, which amounts to 8.Finally, I'll add up the costs of all three fabrics to find the total amount Jamie spends on fabric.