🧠:<think>Alright, so I have two problems to solve here. Let me tackle them one by one.Starting with the first problem: the photographer is arranging 6 models in a hexagonal pattern. Each model is at a vertex of the hexagon, and the distance between any two adjacent models is exactly 5 meters. I need to calculate the total area enclosed by this hexagon.Hmm, okay. I remember that a regular hexagon can be divided into six equilateral triangles. Each side of these triangles is equal to the distance between the models, which is 5 meters. So, if I can find the area of one of these equilateral triangles and then multiply it by six, that should give me the total area of the hexagon.First, let me recall the formula for the area of an equilateral triangle. The area (A) of an equilateral triangle with side length (a) is given by:A = (√3 / 4) * a²So, plugging in the value of a, which is 5 meters:A = (√3 / 4) * (5)²A = (√3 / 4) * 25A = (25√3) / 4Okay, so that's the area of one equilateral triangle. Since the hexagon is made up of six such triangles, the total area (A_total) would be:A_total = 6 * (25√3 / 4)A_total = (150√3) / 4A_total = (75√3) / 2Wait, let me double-check that. 6 multiplied by 25 is 150, and 150 divided by 4 is 37.5, but since it's multiplied by √3, it's 37.5√3. Alternatively, 75√3 divided by 2 is the same as 37.5√3. Either way, that's the total area.But just to make sure I didn't make a mistake in the formula. Yes, the area of a regular hexagon can also be calculated using the formula:A = (3√3 / 2) * a²Let me try that formula with a = 5 meters:A = (3√3 / 2) * (5)²A = (3√3 / 2) * 25A = (75√3) / 2Yes, that's the same result as before. So, that confirms it. The total area enclosed by the hexagon is (75√3)/2 square meters.Moving on to the second problem: the photographer is using a high-resolution camera with 50 megapixels. Each image is stored as a raw file, and each megapixel requires 3 bytes of storage. I need to find the total storage required for 150 such images.Alright, let's break this down. First, each image is 50 megapixels. Each megapixel is 3 bytes. So, the storage per image would be:Storage per image = 50 megapixels * 3 bytes/megapixelStorage per image = 150 megabytesWait, hold on. Is that correct? Wait, 50 megapixels times 3 bytes per megapixel. But actually, 1 megapixel is 1 million pixels. So, 50 megapixels is 50 million pixels. Each pixel is 3 bytes, so the total storage per image is:Storage per image = 50,000,000 pixels * 3 bytes/pixelStorage per image = 150,000,000 bytesNow, converting bytes to megabytes. Since 1 megabyte is 1,000,000 bytes, so:Storage per image = 150,000,000 bytes / 1,000,000 bytes/megabyteStorage per image = 150 megabytesOkay, so each image is 150 megabytes. Now, for 150 images, the total storage required would be:Total storage = 150 images * 150 megabytes/imageTotal storage = 22,500 megabytesHmm, 22,500 megabytes. But sometimes storage is measured in gigabytes. Let me convert that to gigabytes for better understanding.Since 1 gigabyte is 1,000 megabytes, so:Total storage = 22,500 / 1,000Total storage = 22.5 gigabytesBut the question doesn't specify the unit, just asks for total storage. So, 22,500 megabytes is the answer, but it's also good to note that it's 22.5 gigabytes.Wait, let me verify the calculations again. 50 megapixels per image, each megapixel is 3 bytes. So, 50 * 3 = 150 megabytes per image. 150 images would be 150 * 150 = 22,500 megabytes. Yes, that seems correct.Alternatively, if I think in terms of bytes:Each image: 50,000,000 pixels * 3 bytes = 150,000,000 bytes150 images: 150 * 150,000,000 = 22,500,000,000 bytesConvert to gigabytes: 22,500,000,000 / 1,000,000,000 = 22.5 gigabytesYes, that's consistent. So, either 22,500 megabytes or 22.5 gigabytes. Since the question didn't specify, but in storage, gigabytes are more commonly used for such quantities, so 22.5 GB is a more understandable measure.But to stick strictly to the question, it just asks for total storage, so both are correct, but maybe the question expects the answer in megabytes. Let me check the problem statement again.It says, "what is the total storage required for 150 such images?" It doesn't specify the unit, but since each megapixel is 3 bytes, and the initial calculation is in megapixels, maybe the answer is expected in megabytes. So, 22,500 megabytes.But sometimes, in computing, 1 megabyte is 1,048,576 bytes, but in storage, it's often approximated as 1,000,000 bytes. The problem doesn't specify, but since it's a photography context, they might be using the decimal definition where 1 megabyte is 1 million bytes. So, 22,500 megabytes is appropriate.Alternatively, if we use the binary definition, 1 megabyte is 1,048,576 bytes, so let's see:Total bytes: 22,500,000,000 bytesConvert to binary gigabytes: 22,500,000,000 / 1,073,741,824 ≈ 20.97 GBBut since the problem didn't specify, I think it's safer to go with the decimal definition, so 22.5 GB or 22,500 MB.But the problem says "each megapixel requires 3 bytes of storage." So, 50 megapixels * 3 bytes = 150 MB per image. So, 150 images * 150 MB = 22,500 MB. So, 22,500 MB is the answer.I think that's the way to go.So, summarizing:1. The area of the regular hexagon is (75√3)/2 square meters.2. The total storage required is 22,500 megabytes.But let me write them in the required format.For the first problem, the area is (75√3)/2 m², which is approximately 64.95 m², but since the question didn't specify rounding, I should present the exact value.For the second problem, 22,500 MB.Wait, but in the first problem, the area is (75√3)/2, which is equal to 37.5√3. Maybe I should write it as 75√3/2 or 37.5√3? Both are correct, but perhaps 75√3/2 is more standard.Alternatively, maybe factor it differently, but I think either is fine.So, final answers:1. The area is (75√3)/2 square meters.2. The total storage is 22,500 megabytes.Final Answer1. The total area enclosed by the hexagon is boxed{dfrac{75sqrt{3}}{2}} square meters.2. The total storage required is boxed{22500} megabytes.</think>

🧠:<think>Okay, so I have this problem about a government spokesperson and a journalist who frequently clash during press conferences. The problem has two parts, and I need to figure out both. Let me take them one at a time.Starting with part 1: It says that the probability the spokesperson makes a controversial statement in a given press conference follows a Poisson distribution with an average rate of λ statements per conference. Then, the probability that a journalist will confront the spokesperson on any given controversial statement is p. I need to derive the probability that in a particular press conference, the spokesperson makes exactly k controversial statements and is confronted by the journalist on exactly m of those statements.Hmm, okay. So, first, the number of controversial statements follows a Poisson distribution. The Poisson probability mass function is given by P(K = k) = (λ^k e^{-λ}) / k! where K is the number of events (controversial statements here). So, that's the first part.Now, given that there are k controversial statements, each of these statements can be confronted by the journalist with probability p. So, for each statement, it's a Bernoulli trial with success probability p (success meaning the journalist confronts). Therefore, the number of confrontations given k statements should follow a binomial distribution. The binomial probability mass function is P(M = m | K = k) = C(k, m) p^m (1 - p)^{k - m}, where C(k, m) is the combination of k things taken m at a time.So, to find the joint probability that there are exactly k statements and exactly m confrontations, I think I need to multiply these two probabilities together because the number of statements and the number of confrontations given the statements are independent. So, the joint probability P(K = k, M = m) should be P(K = k) * P(M = m | K = k).Putting it together, that would be:P(K = k, M = m) = [ (λ^k e^{-λ}) / k! ] * [ C(k, m) p^m (1 - p)^{k - m} ]Simplify that expression. Let me write it out:= (λ^k e^{-λ} / k!) * (k! / (m! (k - m)!)) p^m (1 - p)^{k - m}Wait, the k! in the denominator cancels with the k! in the numerator from the combination. So, we get:= (λ^k e^{-λ}) * (1 / (m! (k - m)!)) p^m (1 - p)^{k - m}Hmm, that seems a bit complicated. Is there a better way to write this? Maybe factor out terms.Let me see: (λ^k) * (p^m) * (1 - p)^{k - m} can be written as (λ p)^m * (λ (1 - p))^{k - m} * (1 / (m! (k - m)!)) ) * e^{-λ}Wait, that looks like the product of two Poisson terms? Or maybe a bivariate Poisson distribution?But actually, since the number of statements is Poisson and the number of confrontations given that is binomial, the joint distribution is a Poisson-Binomial distribution. But perhaps we can express it in terms of factorial moments or something else.Alternatively, maybe we can write it as:= e^{-λ} * (λ p)^m / m! * (λ (1 - p))^{k - m} / (k - m)! * [1 / (k - m)! ] Wait, no, that doesn't seem right.Wait, let's think differently. The joint distribution is the product of Poisson for K and binomial for M given K. So, perhaps it's a Poisson-Binomial distribution, but I don't recall the exact form.Alternatively, maybe we can express it as a product of two terms:= e^{-λ} * (λ p)^m / m! * e^{-λ (1 - p)} * (λ (1 - p))^{k - m} / (k - m)! ) * something?Wait, that might not be correct because the original Poisson is for K, and then the binomial is conditional on K.Alternatively, perhaps the joint distribution can be expressed as a product of independent Poisson variables, but I don't think so because M is dependent on K.Wait, maybe it's better to leave it as the product of the Poisson and binomial probabilities. So, the final expression is:P(K = k, M = m) = (λ^k e^{-λ} / k!) * (C(k, m) p^m (1 - p)^{k - m})Which simplifies to:= (λ^k e^{-λ} / k!) * (k! / (m! (k - m)!)) p^m (1 - p)^{k - m}Which further simplifies to:= (λ^k e^{-λ} p^m (1 - p)^{k - m}) / (m! (k - m)!))So, that's the expression. Alternatively, we can write it as:= e^{-λ} (λ p)^m (λ (1 - p))^{k - m} / (m! (k - m)!))But I don't think that's a standard distribution, so perhaps that's as simplified as it gets.Wait, but actually, if we think of M as the number of confrontations, which is the sum of independent Bernoulli trials with probability p, given K, which is Poisson, then M follows a Poisson-Binomial distribution. But in this case, since each trial is identical, it's actually a Poisson distribution thinned by p. So, the number of confrontations M would have a Poisson distribution with parameter λ p. But wait, that's the marginal distribution, not the joint.So, the joint distribution is as above, but the marginal distribution of M is Poisson(λ p). But the question is about the joint probability, so I think the expression I have is correct.So, to recap, the joint probability is the product of the Poisson probability for K = k and the binomial probability for M = m given K = k. So, that gives us:P(K = k, M = m) = (λ^k e^{-λ} / k!) * (k! / (m! (k - m)!)) p^m (1 - p)^{k - m}Simplifying, the k! cancels, so:= (λ^k e^{-λ} p^m (1 - p)^{k - m}) / (m! (k - m)!))Alternatively, we can factor out the terms:= e^{-λ} (λ p)^m (λ (1 - p))^{k - m} / (m! (k - m)!))But I don't think that's a standard form, so maybe that's the final expression.Okay, so that's part 1.Moving on to part 2: Given that the total number of press conferences attended by both the spokesperson and the journalist over a year is N, and the average number of controversial statements made per conference remains λ, calculate the expected number of confrontations between the spokesperson and the journalist over the year. Assume the probability p remains constant for each confrontation.So, we need to find the expected number of confrontations over N conferences.First, in each conference, the number of confrontations is a random variable, say M_i for the i-th conference. Then, the total number of confrontations over N conferences is M = M_1 + M_2 + ... + M_N.Therefore, the expected number of confrontations is E[M] = E[M_1 + M_2 + ... + M_N] = E[M_1] + E[M_2] + ... + E[M_N] because expectation is linear.Since each conference is independent and identically distributed, E[M_i] is the same for all i. So, E[M] = N * E[M_i].So, we just need to find E[M_i], the expected number of confrontations in a single conference, and then multiply by N.From part 1, we know that in a single conference, the number of confrontations M is a random variable. What's E[M]?Well, in each conference, the number of controversial statements K is Poisson(λ), and given K, the number of confrontations M is Binomial(K, p). Therefore, the expected number of confrontations is E[M] = E[ E[M | K] ].Since given K, M is Binomial(K, p), E[M | K] = K p. Therefore, E[M] = E[K p] = p E[K].But K is Poisson(λ), so E[K] = λ. Therefore, E[M] = p λ.Therefore, the expected number of confrontations in one conference is p λ. So, over N conferences, it's N p λ.So, the expected number of confrontations over the year is N λ p.Wait, that seems straightforward. Let me double-check.Alternatively, think of each statement as an independent event with probability p of confrontation. So, the total number of statements over N conferences is Poisson(N λ), since each conference has Poisson(λ) statements, and they're independent. Then, the number of confrontations is a thinning of this Poisson process with probability p, so the total confrontations would be Poisson(N λ p). Therefore, the expected number is N λ p.Yes, that makes sense.So, putting it all together, the expected number is N λ p.Okay, so that's part 2.So, summarizing:1. The joint probability is P(K = k, M = m) = (λ^k e^{-λ} / k!) * (C(k, m) p^m (1 - p)^{k - m}) which simplifies to (λ^k e^{-λ} p^m (1 - p)^{k - m}) / (m! (k - m)!)).2. The expected number of confrontations over N conferences is N λ p.I think that's it. Let me just make sure I didn't make any mistakes in the algebra.In part 1, starting from Poisson and binomial, multiplying them gives the joint distribution. The simplification cancels the k! in the denominator with the k! in the combination, leaving us with the expression divided by m! (k - m)!.Yes, that seems correct.And in part 2, using linearity of expectation, we find that the expected number of confrontations per conference is λ p, so over N conferences, it's N λ p. That also makes sense.So, I think I'm confident with these answers.Final Answer1. The probability is boxed{dfrac{e^{-lambda} (lambda p)^m (lambda (1 - p))^{k - m}}{m! (k - m)!}}.2. The expected number of confrontations is boxed{N lambda p}.</think>

🧠:<think>Okay, so I have these two math problems to solve, both related to mail sorting optimization. Let me take them one at a time.Starting with the first problem: City A's mail sorter has a function ( f(x) = frac{x^3 - 3x + 2}{x^2 - 1} ). City B wants to use a similar model but needs to make sure the function is well-defined for all postal code indices from 1 to 10. I need to find the set of postal code indices where the function is undefined and explain why.Alright, so a function is undefined where the denominator is zero because division by zero is undefined. So, I need to find the values of ( x ) in the range 1 to 10 where ( x^2 - 1 = 0 ).Let me solve ( x^2 - 1 = 0 ). That factors into ( (x - 1)(x + 1) = 0 ), so the solutions are ( x = 1 ) and ( x = -1 ). But since the postal code indices range from 1 to 10, ( x = -1 ) isn't in that range. So, the only value in 1 to 10 where the function is undefined is ( x = 1 ).Wait, but maybe I should check if the numerator also becomes zero at ( x = 1 ). If both numerator and denominator are zero, it might be a removable discontinuity instead of a vertical asymptote. Let me plug ( x = 1 ) into the numerator: ( 1^3 - 3(1) + 2 = 1 - 3 + 2 = 0 ). So, yes, both numerator and denominator are zero at ( x = 1 ). That means there's a hole at ( x = 1 ), but the function is still undefined there because you can't divide by zero. So, ( x = 1 ) is excluded from the domain.Therefore, the set of postal code indices where the function is undefined is just {1}.Moving on to the second problem: City B models the volume of mails with a probability density function ( g(x) = kx(10 - x) ) for ( 0 leq x leq 10 ). The expected volume is 25 units, and I need to find the constant ( k ).Hmm, okay. So, for a probability density function, the total area under the curve from 0 to 10 should be 1. But here, they mention the expected volume is 25 units. Wait, is this a probability density function or a volume density function? Because if it's a probability density function, the integral over the interval should be 1, but if it's a volume density function, the integral would give the total volume, which is 25 units.Let me read the problem again: "model the volume of mails for each zone by a probability density function ( g(x) = kx(10 - x) )". Hmm, so it's a probability density function, but the expected volume is 25 units. That seems a bit confusing because usually, the expected value is calculated from the probability density function, not the other way around.Wait, maybe the expected value here is 25 units, meaning that the mean of the distribution is 25. But hold on, the function is defined from 0 to 10, so x ranges from 0 to 10. If the expected value is 25, that doesn't make sense because x can't be more than 10. Maybe I'm misunderstanding.Alternatively, perhaps the expected value is 25 units, meaning that the average volume is 25. So, the expected value E[x] is 25? But x is only up to 10, so that can't be. Alternatively, maybe the expected volume is 25, meaning that the integral of x times g(x) from 0 to 10 is 25.Wait, let's think step by step.First, if ( g(x) ) is a probability density function, then the total area under the curve from 0 to 10 should be 1. So, we can find k by integrating ( g(x) ) from 0 to 10 and setting it equal to 1.But the problem says the expected volume is 25 units. So, maybe the expected value E[x] is 25? But x is only defined up to 10, so that's impossible. Alternatively, maybe the expected volume is 25, which would mean that the integral of g(x) times x from 0 to 10 is 25. So, E[x] = 25.Wait, but if the function is a probability density function, then E[x] is calculated as the integral of x*g(x) dx from 0 to 10. So, if E[x] = 25, then:( int_{0}^{10} x cdot g(x) dx = 25 )But also, since it's a PDF, we have:( int_{0}^{10} g(x) dx = 1 )So, we have two equations:1. ( int_{0}^{10} kx(10 - x) dx = 1 )2. ( int_{0}^{10} kx^2(10 - x) dx = 25 )So, we can solve for k using the first equation, and then verify if the second equation holds or if we need to adjust.Wait, but if the expected value is 25, which is outside the range of x (which is 0 to 10), that seems impossible. So, maybe I'm misinterpreting the problem.Alternatively, perhaps the expected volume is 25 units, meaning that the average value of the function g(x) over the interval is 25. But that also doesn't make much sense because the average value would be the integral of g(x) over 0 to 10 divided by 10, which would be 1/10 if it's a PDF.Wait, this is confusing. Let me read the problem again:"model the volume of mails for each zone by a probability density function ( g(x) = kx(10 - x) ) for ( 0 leq x leq 10 ), where ( k ) is a constant. Suppose the expected volume of mails for a zone is 25 units. Calculate the value of ( k )."Hmm, so maybe "expected volume" refers to the expected value of the volume, which would be the integral of x * g(x) dx from 0 to 10, which is 25.But as I thought earlier, if x is only from 0 to 10, the expected value can't be 25. So, perhaps the problem is misstated, or maybe I'm misunderstanding.Alternatively, maybe the volume is modeled by ( g(x) ), and the expected volume is 25, meaning that the integral of g(x) from 0 to 10 is 25. So, total volume is 25, not probability.Wait, that might make sense. If it's a volume density function, then the integral over the interval would give the total volume. So, in that case, we don't need to set the integral to 1, but instead to 25.So, let's try that approach.So, if ( g(x) ) is a volume density function, then:( int_{0}^{10} g(x) dx = 25 )So, ( int_{0}^{10} kx(10 - x) dx = 25 )So, let's compute that integral.First, expand the integrand:( kx(10 - x) = k(10x - x^2) )So, the integral becomes:( k int_{0}^{10} (10x - x^2) dx )Compute the integral:Integral of 10x is 5x^2, integral of x^2 is (1/3)x^3.So, evaluating from 0 to 10:( k [5(10)^2 - (1/3)(10)^3 - (5(0)^2 - (1/3)(0)^3)] )Simplify:( k [5*100 - (1/3)*1000 - 0] = k [500 - 1000/3] )Compute 500 - 1000/3:500 is 1500/3, so 1500/3 - 1000/3 = 500/3.So, the integral is ( k*(500/3) = 25 ).Therefore, ( k = 25 * (3/500) = (75)/500 = 3/20 = 0.15 ).So, k is 3/20.Wait, but let me check if that makes sense. If the integral of g(x) is 25, then k is 3/20. Let me verify the calculations.Compute ( int_{0}^{10} kx(10 - x) dx ):= k * ∫0^10 (10x - x²) dx= k * [5x² - (1/3)x³] from 0 to 10At x=10: 5*(100) - (1/3)*1000 = 500 - 1000/3 = (1500 - 1000)/3 = 500/3At x=0: 0 - 0 = 0So, integral is k*(500/3) = 25Thus, k = 25 * 3 / 500 = 75/500 = 3/20.Yes, that seems correct.So, k is 3/20.But wait, the problem says it's a probability density function. If it's a PDF, the integral should be 1, but here the integral is 25. So, that contradicts. Therefore, perhaps the problem is misworded, and it's not a PDF but a volume density function.Alternatively, maybe the expected value is 25, which would require integrating x*g(x) dx = 25, but as I thought earlier, that would be impossible because x is only up to 10.Wait, let me try that approach just in case.If E[x] = 25, then:( int_{0}^{10} x * g(x) dx = 25 )But since g(x) is a PDF, we also have:( int_{0}^{10} g(x) dx = 1 )So, let's compute both integrals.First, find k such that ( int_{0}^{10} kx(10 - x) dx = 1 )We already did that earlier, which gave k = 3/500.Wait, no, earlier when we set the integral equal to 25, we found k = 3/20. But if it's a PDF, the integral should be 1, so k = 3/500.Wait, let me recast.If g(x) is a PDF, then:( int_{0}^{10} g(x) dx = 1 )So, ( k int_{0}^{10} x(10 - x) dx = 1 )We computed that integral as 500/3, so:k*(500/3) = 1 => k = 3/500.Then, the expected value E[x] would be:( int_{0}^{10} x * g(x) dx = int_{0}^{10} x * (3/500)x(10 - x) dx = (3/500) int_{0}^{10} x^2(10 - x) dx )Compute that integral:Expand x^2(10 - x) = 10x^2 - x^3Integrate term by term:Integral of 10x^2 is (10/3)x^3Integral of x^3 is (1/4)x^4So, evaluating from 0 to 10:(10/3)(1000) - (1/4)(10000) = (10000/3) - (10000/4)Compute 10000/3 ≈ 3333.333 and 10000/4 = 2500So, 3333.333 - 2500 ≈ 833.333Thus, the integral is approximately 833.333So, E[x] = (3/500)*833.333 ≈ (3/500)*833.333 ≈ (2500/500) = 5Wait, 833.333 is 2500/3, so:E[x] = (3/500)*(2500/3) = 2500/500 = 5.So, the expected value is 5, not 25.But the problem says the expected volume is 25 units. So, this suggests that either the problem is misworded, or I'm misunderstanding.Alternatively, perhaps the expected volume is 25, meaning that the integral of g(x) over 0 to 10 is 25, which would make it a volume density function, not a probability density function.In that case, we don't set the integral to 1, but to 25, so k = 3/20 as I found earlier.But the problem explicitly says it's a probability density function. So, that's conflicting.Wait, maybe the expected volume is 25, meaning that the average volume per unit x is 25. But that would be the integral of g(x) over 0 to 10 divided by 10, which would be 25. So, total volume is 250.Wait, that might be another interpretation.If the average volume is 25, then:( frac{1}{10} int_{0}^{10} g(x) dx = 25 )Thus, ( int_{0}^{10} g(x) dx = 250 )So, then:( k int_{0}^{10} x(10 - x) dx = 250 )We know the integral is 500/3, so:k*(500/3) = 250 => k = 250 * 3 / 500 = 750 / 500 = 1.5So, k = 3/2.But then, if k = 3/2, then the function g(x) = (3/2)x(10 - x) is a volume density function with total volume 250, and average volume 25.But the problem says it's a probability density function, which is confusing because then the total should be 1.Alternatively, maybe the expected value of the volume is 25, meaning E[volume] = 25, but volume is a separate variable.Wait, perhaps I'm overcomplicating. Let me go back.The problem says: "model the volume of mails for each zone by a probability density function ( g(x) = kx(10 - x) ) for ( 0 leq x leq 10 ), where ( k ) is a constant. Suppose the expected volume of mails for a zone is 25 units. Calculate the value of ( k )."So, perhaps "expected volume" is referring to the expected value of x, which is 25. But x is only up to 10, so that can't be.Alternatively, maybe the volume is a separate variable, say V, and V is modeled by g(x), but that seems unclear.Wait, maybe the volume is proportional to g(x), and the expected volume is 25. So, if g(x) is a probability density function, then the expected value of x is 5, as we calculated earlier. But the expected volume is 25, which is 5 times 5. Hmm, not sure.Alternatively, perhaps the volume is given by another function, but the problem says it's modeled by g(x). So, maybe the total volume is 25, meaning the integral of g(x) is 25, so k = 3/20.But since the problem says it's a probability density function, that would require the integral to be 1, which conflicts with the expected volume being 25.I think the problem might have a typo or misstatement. But given the information, if we assume that the expected volume is 25, meaning the integral of g(x) is 25, then k = 3/20.Alternatively, if it's a PDF and the expected value is 25, which is impossible because x only goes up to 10, then perhaps the problem is wrong.But given the problem as stated, I think the intended interpretation is that the total volume is 25, so we set the integral of g(x) from 0 to 10 equal to 25, leading to k = 3/20.So, I'll go with that.Final Answer1. The function is undefined at postal code index boxed{1}.2. The value of ( k ) is boxed{dfrac{3}{20}}.</think>

🧠:<think>Okay, so I have two probability problems to solve here. Let me take them one by one.Problem 1: Poisson DistributionThe first problem is about a snack shop where the daily customer arrival rate follows a Poisson distribution with a mean of 120 customers per day. During lunch hours, from 11:30 AM to 1:30 PM, the arrival rate doubles. I need to find the probability that the shop will receive at least 50 customers during the lunch hours on any given day.Alright, so first, let's understand the Poisson distribution. It's used to model the number of events happening in a fixed interval of time or space, given the average rate of occurrence. The formula for the Poisson probability mass function is:[ P(k) = frac{lambda^k e^{-lambda}}{k!} ]where ( lambda ) is the average rate (mean), ( k ) is the number of occurrences, and ( e ) is the base of the natural logarithm.In this case, the mean arrival rate is 120 customers per day. But during lunch hours, the rate doubles. So, I need to figure out what the rate is during lunch.First, let's figure out how much time is considered lunch hours. It's from 11:30 AM to 1:30 PM, which is 2 hours. So, the entire day is 24 hours, but lunch is 2 hours. Therefore, the proportion of the day that is lunch time is ( frac{2}{24} = frac{1}{12} ).But wait, the arrival rate during lunch hours doubles. So, the mean arrival rate during lunch hours is double the usual rate for that time period.Wait, hold on. Let me think again.The overall mean is 120 customers per day. So, the rate per hour is ( frac{120}{24} = 5 ) customers per hour.During lunch hours, the arrival rate doubles, so it becomes ( 5 times 2 = 10 ) customers per hour.Since lunch is 2 hours, the mean number of customers during lunch hours is ( 10 times 2 = 20 ) customers.So, the Poisson parameter ( lambda ) during lunch hours is 20.We need the probability that the shop will receive at least 50 customers during lunch. So, ( P(X geq 50) ).But wait, 50 customers in 2 hours when the mean is 20? That seems quite high. The Poisson distribution is skewed, but 50 is way above the mean. So, the probability of getting at least 50 customers would be very low.But let me confirm if I calculated the rate correctly.Wait, the mean arrival rate is 120 per day. So, per hour, it's 5. During lunch, it's double, so 10 per hour. Over 2 hours, that's 20. So, yes, ( lambda = 20 ).But 50 customers in 2 hours when the mean is 20 is 2.5 times the mean. That's quite high. So, the probability should be very small.But let me think about whether the arrival rate is 120 per day, so during lunch, it's double that rate for the 2-hour period. So, is it 240 per day? Wait, no, because it's only double during lunch.Wait, perhaps I misinterpreted the problem. Let me read it again."The daily customer arrival rate at the shop follows a Poisson distribution with a mean of 120 customers per day. Due to the high demand during lunch hours (11:30 AM to 1:30 PM), the arrival rate doubles during this period."So, the daily rate is 120, but during lunch, the arrival rate is double. So, perhaps the arrival rate during lunch is double the usual rate for that time period.So, the usual rate for the entire day is 120. So, the usual rate per hour is 5. During lunch, it's double, so 10 per hour. So, over 2 hours, it's 20. So, ( lambda = 20 ).Therefore, the number of customers during lunch is Poisson distributed with ( lambda = 20 ).We need ( P(X geq 50) ). Since Poisson can be approximated by a normal distribution when ( lambda ) is large, maybe we can use that.But 20 is not that large, but maybe it's manageable.Alternatively, we can use the Poisson formula, but calculating ( P(X geq 50) ) directly would require summing from 50 to infinity, which is impractical.Alternatively, we can use the normal approximation.So, for Poisson distribution, the mean ( mu = lambda = 20 ), and the variance ( sigma^2 = lambda = 20 ), so ( sigma = sqrt{20} approx 4.4721 ).Using the normal approximation, we can calculate ( P(X geq 50) ).But since we're approximating a discrete distribution with a continuous one, we should apply continuity correction. So, ( P(X geq 50) ) is approximated by ( P(X geq 49.5) ).So, we can calculate the z-score:[ z = frac{49.5 - mu}{sigma} = frac{49.5 - 20}{4.4721} approx frac{29.5}{4.4721} approx 6.596 ]Looking up this z-score in the standard normal distribution table, a z-score of 6.596 is way beyond the typical tables. The probability of z > 6.596 is practically zero.So, the probability of having at least 50 customers during lunch is approximately zero.But wait, is this correct? Because 50 is way beyond the mean of 20. So, yes, it's extremely unlikely.Alternatively, maybe I made a mistake in calculating the rate.Wait, another thought: the arrival rate is 120 per day. During lunch, the arrival rate doubles. So, is the rate during lunch 240 per day? No, because it's only for 2 hours.Wait, perhaps the rate is 120 per day, so per hour, it's 5. During lunch, it's double, so 10 per hour. So, over 2 hours, it's 20. So, yes, ( lambda = 20 ).Alternatively, maybe the arrival rate during lunch is double the overall rate. So, if the overall rate is 120 per day, during lunch, it's 240 per day? But that doesn't make sense because lunch is only 2 hours.Wait, perhaps the arrival rate during lunch is double the rate per hour. So, if normally it's 5 per hour, during lunch, it's 10 per hour. So, over 2 hours, 20.Yes, that seems correct.So, I think my initial calculation is correct. Therefore, the probability is practically zero.But let me check with another approach. Maybe using the Poisson formula.The Poisson probability of at least 50 customers is 1 minus the probability of less than 50.But calculating ( P(X geq 50) = 1 - P(X leq 49) ).But calculating ( P(X leq 49) ) for ( lambda = 20 ) is still a huge computation. Maybe using software or tables, but since I don't have that, I can rely on the normal approximation.Given that the z-score is about 6.6, which is way beyond the typical range, the probability is effectively zero.So, I think the answer is approximately zero.Problem 2: Normal Distribution and Total ProfitThe second problem is about the snack shop selling an average of 3 different types of snacks per customer. Each type of snack has a profit margin following a normal distribution with mean μ = 5 yuan and standard deviation σ = 2 yuan. If the shop serves 200 customers in a day, what is the probability that the total profit for the day will exceed 3000 yuan?Alright, let's break this down.Each customer buys an average of 3 snacks. Each snack's profit is normally distributed with μ = 5 and σ = 2.So, for each customer, the total profit is the sum of 3 independent normal variables.The sum of independent normal variables is also normal, with mean equal to the sum of the means and variance equal to the sum of the variances.So, for one customer:- Mean profit: 3 * 5 = 15 yuan- Variance: 3 * (2)^2 = 12- Standard deviation: sqrt(12) ≈ 3.4641 yuanNow, for 200 customers, the total profit is the sum of 200 independent variables, each with mean 15 and variance 12.So, the total profit:- Mean: 200 * 15 = 3000 yuan- Variance: 200 * 12 = 2400- Standard deviation: sqrt(2400) ≈ 48.9898 yuanWe need the probability that the total profit exceeds 3000 yuan, i.e., ( P(X > 3000) ).But the mean is 3000, so we're looking for the probability that a normal variable exceeds its mean. For a normal distribution, this is 0.5, or 50%.Wait, that seems too straightforward. Let me double-check.Each customer contributes a profit that is normal with mean 15. So, the total profit is normal with mean 3000 and standard deviation ~48.99.So, ( P(X > 3000) = 0.5 ).But wait, is that correct? Because the total profit is exactly the mean. So, the probability of exceeding the mean is 0.5.But the problem says "exceed 3000 yuan." So, yes, it's 0.5.But let me think again. Is there any catch here?Wait, each customer buys an average of 3 snacks. So, the number of snacks per customer is fixed at 3? Or is it a random variable?Wait, the problem says "sells an average of 3 different types of snacks per customer." So, it's an average, implying that it's a random variable. So, perhaps the number of snacks per customer is variable, but the average is 3.Wait, but the problem says "each type of snack has a profit margin following a normal distribution." So, perhaps each customer buys exactly 3 snacks, each with profit margin ~N(5, 2^2). So, the total profit per customer is sum of 3 normals, which is N(15, 12).Alternatively, if the number of snacks per customer is variable, say, a random variable, then the total profit would be a bit more complicated. But the problem says "sells an average of 3 different types of snacks per customer." So, maybe it's fixed at 3.Given that, then each customer contributes exactly 3 snacks, each with profit ~N(5,4). So, total per customer is N(15, 12). Then, for 200 customers, total profit is N(3000, 2400). So, standard deviation ~48.99.Therefore, ( P(X > 3000) = 0.5 ).But wait, the problem says "the probability that the total profit for the day will exceed 3000 yuan." So, it's exactly the mean, so 50%.But maybe I'm missing something. Let me think again.Wait, the problem says "the snack shop sells an average of 3 different types of snacks per customer." So, maybe the number of snacks per customer is a random variable with mean 3, but the profit per snack is normal.So, if the number of snacks per customer is variable, say, N ~ Poisson or something else, then the total profit would be a compound distribution.But the problem doesn't specify the distribution of the number of snacks per customer, only that it's an average of 3. So, perhaps we can assume that each customer buys exactly 3 snacks, making the total profit per customer fixed as 3 * profit per snack.But wait, each snack's profit is normal, so the total per customer is sum of 3 normals, which is normal.Alternatively, if the number of snacks is variable, but the average is 3, then the total profit would be a bit more involved.But since the problem doesn't specify, I think it's safe to assume that each customer buys exactly 3 snacks, each with profit ~N(5,4). Therefore, total profit per customer is N(15, 12), and total for 200 customers is N(3000, 2400).Therefore, the probability of exceeding 3000 is 0.5.But wait, let me think again. If the total profit is exactly the mean, then the probability of exceeding is 0.5.But maybe the problem expects a different approach. Let me see.Alternatively, maybe the total profit is calculated as 200 customers * 3 snacks each * profit per snack.But the profit per snack is normal, so the total profit is 200*3=600 snacks, each with profit ~N(5,4). So, total profit is sum of 600 normals, which is N(600*5, 600*4) = N(3000, 2400). So, same result.Therefore, ( P(X > 3000) = 0.5 ).But wait, maybe the problem is considering that each customer buys 3 snacks, but the number of snacks is fixed, so the total profit is sum of 3*200=600 normals, which is N(3000, 2400). So, same as before.Therefore, the probability is 0.5.But let me think again. If the total profit is exactly the mean, then the probability of exceeding is 0.5.But maybe the problem is considering that the number of snacks per customer is variable, but the average is 3. So, the total number of snacks is 200*3=600 on average, but variable.But without knowing the distribution of the number of snacks per customer, we can't model the total profit accurately.But the problem says "sells an average of 3 different types of snacks per customer." So, perhaps it's fixed at 3 per customer, making the total number of snacks 600, each with profit ~N(5,4). So, total profit is N(3000, 2400), and the probability of exceeding 3000 is 0.5.Alternatively, if the number of snacks per customer is variable, say, a Poisson distribution with mean 3, then the total number of snacks would be Poisson(600), but the profit per snack is normal. So, the total profit would be a compound distribution, but that's more complex.But since the problem doesn't specify the distribution of the number of snacks per customer, I think it's safe to assume that each customer buys exactly 3 snacks, making the total number of snacks 600, each with profit ~N(5,4). Therefore, total profit is N(3000, 2400), and the probability of exceeding 3000 is 0.5.But wait, let me think again. If the number of snacks per customer is variable, but the average is 3, then the total number of snacks is 200*3=600 on average, but with some variance. However, since the profit per snack is normal, the total profit would still be approximately normal due to the Central Limit Theorem, with mean 3000 and variance 600*(2^2)=2400, same as before. So, regardless of whether the number of snacks per customer is fixed or variable, as long as the average is 3, the total profit's mean and variance remain the same.Therefore, the probability is 0.5.But wait, the problem says "the probability that the total profit for the day will exceed 3000 yuan." So, exactly the mean, so 0.5.But maybe I'm missing something. Let me think about the units.Each snack has a profit margin of 5 yuan on average. So, 3 snacks per customer would be 15 yuan per customer on average. 200 customers would be 3000 yuan. So, the expected total profit is 3000 yuan. Therefore, the probability of exceeding 3000 is 0.5.Yes, that makes sense.So, the answer is 0.5, or 50%.But let me think again. Is there a possibility that the total profit is not exactly the mean? Wait, no, because we're calculating the probability of exceeding the mean, which is 0.5 for a symmetric distribution like normal.Therefore, I think the answer is 0.5.But wait, let me think about the variance again. If each snack's profit is N(5,4), then the total profit is sum of 600 such variables, so variance is 600*4=2400, standard deviation ~48.99.So, the distribution is N(3000, 2400). So, the probability of being above 3000 is 0.5.Yes, that's correct.So, summarizing:1. The probability of at least 50 customers during lunch is approximately 0.2. The probability that total profit exceeds 3000 yuan is 0.5.But wait, for the first problem, I think it's better to express it as 0, but in reality, it's a very small number, but for the sake of the problem, maybe it's acceptable to say 0.Alternatively, if I use the Poisson formula, the exact probability can be calculated, but it's computationally intensive.But since the z-score is about 6.6, which is beyond the typical tables, the probability is effectively 0.So, I think that's the answer.Final Answer1. The probability is boxed{0}.2. The probability is boxed{0.5}.</think>