🧠:<think>Okay, so I have this problem about a coffee shop owner tracking customers with a sinusoidal function. There are two parts: the first is to calculate the total number of customers over the first 28 days, and the second is to find the correlation coefficient between the coffee shop and a nearby restaurant's customer numbers. Hmm, let me tackle them one by one.Starting with part 1: The function given is ( C(t) = 50 + 30 sinleft(frac{pi t}{14}right) ). I need to find the total number of customers over the first 28 days. So, essentially, I need to sum up ( C(t) ) from ( t = 0 ) to ( t = 27 ) because day 28 would be the 28th day, but since it's the first 28 days, maybe it's up to ( t = 27 ). Wait, actually, in some contexts, the first day is ( t = 1 ), but here ( t ) is the number of days since the beginning, so day 1 is ( t = 1 ), day 2 is ( t = 2 ), etc., up to day 28, which is ( t = 28 ). Hmm, so I think the total number of customers is the sum from ( t = 1 ) to ( t = 28 ) of ( C(t) ). But actually, the function is defined for ( t ) as days since the beginning, so ( t = 0 ) would be day 0, which is before the year starts, so maybe we start at ( t = 1 ). Wait, the problem says "over the first 28 days of the year," so that would be ( t = 1 ) to ( t = 28 ). Hmm, but the function is defined for ( t ) as days since the beginning, so ( t = 0 ) is day 0, which is not part of the year. So, yeah, probably ( t = 1 ) to ( t = 28 ).But wait, the function ( C(t) ) is given for any ( t ), so maybe it's continuous? Or is it discrete? The problem says "the number of customers who visit his cafe each day," so it's discrete, right? So ( t ) is an integer from 1 to 28. So, to find the total number of customers, I need to compute the sum ( sum_{t=1}^{28} C(t) ).But ( C(t) = 50 + 30 sinleft(frac{pi t}{14}right) ). So, the sum becomes ( sum_{t=1}^{28} left(50 + 30 sinleft(frac{pi t}{14}right)right) ). I can split this into two sums: ( sum_{t=1}^{28} 50 + sum_{t=1}^{28} 30 sinleft(frac{pi t}{14}right) ).The first sum is straightforward: 50 added 28 times, so that's ( 50 times 28 ). Let me compute that: 50 times 28 is 1400.The second sum is 30 times the sum of ( sinleft(frac{pi t}{14}right) ) from ( t = 1 ) to ( t = 28 ). Hmm, so I need to compute ( sum_{t=1}^{28} sinleft(frac{pi t}{14}right) ).Wait, is there a formula for the sum of sine functions with equally spaced arguments? I think there is. The general formula for the sum of ( sin(a + (k-1)d) ) from ( k = 1 ) to ( n ) is ( frac{sinleft(frac{n d}{2}right) cdot sinleft(a + frac{(n - 1)d}{2}right)}{sinleft(frac{d}{2}right)} ). Let me verify that.Yes, the formula is:( sum_{k=0}^{n-1} sin(a + kd) = frac{sinleft(frac{n d}{2}right) cdot sinleft(a + frac{(n - 1)d}{2}right)}{sinleft(frac{d}{2}right)} )In our case, the sum is from ( t = 1 ) to ( t = 28 ), so ( n = 28 ). The argument of the sine is ( frac{pi t}{14} ), so ( a = frac{pi}{14} ) when ( t = 1 ), and the common difference ( d = frac{pi}{14} ) as each term increases by ( frac{pi}{14} ).So, applying the formula:( sum_{t=1}^{28} sinleft(frac{pi t}{14}right) = sum_{k=1}^{28} sinleft(frac{pi k}{14}right) )Let me adjust the formula to fit:Here, ( a = frac{pi}{14} ), ( d = frac{pi}{14} ), and ( n = 28 ).So, plugging into the formula:( frac{sinleft(frac{28 cdot frac{pi}{14}}{2}right) cdot sinleft(frac{pi}{14} + frac{(28 - 1) cdot frac{pi}{14}}{2}right)}{sinleft(frac{frac{pi}{14}}{2}right)} )Simplify step by step.First, compute ( frac{n d}{2} = frac{28 cdot frac{pi}{14}}{2} = frac{2 pi}{2} = pi ).Next, compute ( a + frac{(n - 1)d}{2} = frac{pi}{14} + frac{27 cdot frac{pi}{14}}{2} = frac{pi}{14} + frac{27 pi}{28} = frac{2 pi}{28} + frac{27 pi}{28} = frac{29 pi}{28} ).So, the numerator becomes ( sin(pi) cdot sinleft(frac{29 pi}{28}right) ).But ( sin(pi) = 0 ), so the entire numerator is 0. Therefore, the sum is 0.Wait, that's interesting. So, the sum of ( sinleft(frac{pi t}{14}right) ) from ( t = 1 ) to ( t = 28 ) is 0.Therefore, the second sum is 30 times 0, which is 0.So, the total number of customers is 1400 + 0 = 1400.But wait, is that correct? Let me think. The sine function over a full period sums to zero. Since the period of ( sinleft(frac{pi t}{14}right) ) is ( frac{2pi}{pi/14} = 28 ). So, over 28 days, which is exactly one full period, the sum of the sine terms will indeed cancel out to zero. So, that makes sense.Therefore, the total number of customers is 1400.Moving on to part 2: The owner also observes that the number of customers at a nearby restaurant ( R(t) ) is modeled by ( R(t) = 40 + 25 cosleft(frac{pi t}{14}right) ). We need to determine the correlation coefficient between ( C(t) ) and ( R(t) ) over the same 28-day period.Correlation coefficient, okay. So, correlation coefficient between two time series is given by the Pearson correlation coefficient, which is:( r = frac{sum (C_i - bar{C})(R_i - bar{R})}{sqrt{sum (C_i - bar{C})^2} sqrt{sum (R_i - bar{R})^2}} )Where ( bar{C} ) and ( bar{R} ) are the means of ( C(t) ) and ( R(t) ) respectively.But since both ( C(t) ) and ( R(t) ) are sinusoidal functions with the same frequency, their correlation can be found using properties of sinusoids.Alternatively, since we have expressions for both ( C(t) ) and ( R(t) ), we can compute the necessary sums.But let's see. First, let's recall that ( C(t) = 50 + 30 sinleft(frac{pi t}{14}right) ) and ( R(t) = 40 + 25 cosleft(frac{pi t}{14}right) ).Note that ( sin theta ) and ( cos theta ) are orthogonal functions over a full period, meaning their inner product (sum over a period) is zero. So, the cross-correlation between the sinusoidal parts will be zero. However, the constant terms will affect the means.So, let's compute the means first.For ( C(t) ), the mean ( bar{C} ) is the average over 28 days. Since the sine term averages out to zero over a full period, ( bar{C} = 50 ).Similarly, for ( R(t) ), the mean ( bar{R} ) is 40, since the cosine term also averages out to zero over a full period.Now, the numerator of the correlation coefficient is the covariance between ( C(t) ) and ( R(t) ):( sum_{t=1}^{28} (C(t) - bar{C})(R(t) - bar{R}) )Substituting the expressions:( sum_{t=1}^{28} left(30 sinleft(frac{pi t}{14}right)right) left(25 cosleft(frac{pi t}{14}right)right) )Which simplifies to:( 30 times 25 times sum_{t=1}^{28} sinleft(frac{pi t}{14}right) cosleft(frac{pi t}{14}right) )That's 750 times the sum of ( sintheta costheta ) over ( t = 1 ) to 28, where ( theta = frac{pi t}{14} ).But ( sintheta costheta = frac{1}{2} sin(2theta) ). So, the sum becomes:( 750 times frac{1}{2} sum_{t=1}^{28} sinleft(frac{2pi t}{14}right) = 375 sum_{t=1}^{28} sinleft(frac{pi t}{7}right) )Now, let's compute ( sum_{t=1}^{28} sinleft(frac{pi t}{7}right) ).Again, using the formula for the sum of sines with equally spaced arguments.Here, ( a = frac{pi}{7} ), ( d = frac{pi}{7} ), ( n = 28 ).So, applying the formula:( sum_{k=1}^{28} sinleft(frac{pi k}{7}right) = frac{sinleft(frac{28 cdot frac{pi}{7}}{2}right) cdot sinleft(frac{pi}{7} + frac{(28 - 1) cdot frac{pi}{7}}{2}right)}{sinleft(frac{frac{pi}{7}}{2}right)} )Simplify:( frac{sinleft(frac{28 cdot pi}{14}right) cdot sinleft(frac{pi}{7} + frac{27 pi}{14}right)}{sinleft(frac{pi}{14}right)} )Simplify each part:First, ( frac{28 cdot pi}{14} = 2pi ), so ( sin(2pi) = 0 ).Therefore, the entire numerator is 0, so the sum is 0.Thus, the covariance is 375 times 0, which is 0.Wait, so the covariance is zero? That would mean the correlation coefficient is zero. But that seems counterintuitive because both functions are sinusoidal with the same frequency, just shifted by 90 degrees (since one is sine and the other is cosine). But over a full period, their covariance is zero because they are orthogonal.But wait, let me think again. The functions ( sin(theta) ) and ( cos(theta) ) are orthogonal over a full period, meaning their inner product is zero. So, yes, their covariance is zero, hence the correlation coefficient is zero.But let me verify this by computing the denominator as well, just to be thorough.The denominator is the product of the standard deviations of ( C(t) ) and ( R(t) ).First, compute the variance of ( C(t) ):( text{Var}(C) = frac{1}{28} sum_{t=1}^{28} (C(t) - bar{C})^2 = frac{1}{28} sum_{t=1}^{28} left(30 sinleft(frac{pi t}{14}right)right)^2 )Which is:( frac{1}{28} times 900 sum_{t=1}^{28} sin^2left(frac{pi t}{14}right) )Similarly, the variance of ( R(t) ):( text{Var}(R) = frac{1}{28} sum_{t=1}^{28} (R(t) - bar{R})^2 = frac{1}{28} sum_{t=1}^{28} left(25 cosleft(frac{pi t}{14}right)right)^2 )Which is:( frac{1}{28} times 625 sum_{t=1}^{28} cos^2left(frac{pi t}{14}right) )Now, we know that ( sin^2theta + cos^2theta = 1 ), so ( sin^2theta = 1 - cos^2theta ). But over a full period, the average of ( sin^2theta ) and ( cos^2theta ) is the same, each being 0.5.Therefore, ( sum_{t=1}^{28} sin^2left(frac{pi t}{14}right) = 28 times 0.5 = 14 ).Similarly, ( sum_{t=1}^{28} cos^2left(frac{pi t}{14}right) = 14 ).So, variance of ( C(t) ):( frac{1}{28} times 900 times 14 = frac{900}{28} times 14 = frac{900}{2} = 450 )Similarly, variance of ( R(t) ):( frac{1}{28} times 625 times 14 = frac{625}{2} = 312.5 )Therefore, the standard deviations are:( sigma_C = sqrt{450} approx 21.2132 )( sigma_R = sqrt{312.5} approx 17.6777 )But since the covariance is zero, the correlation coefficient ( r = 0 / (sigma_C sigma_R) = 0 ).So, the correlation coefficient is zero.But wait, intuitively, since one is sine and the other is cosine, which are 90 degrees out of phase, their correlation should be zero, which matches our result.Therefore, the correlation coefficient is zero.So, summarizing:1. The total number of customers over the first 28 days is 1400.2. The correlation coefficient between the coffee shop and restaurant customers is 0.Final Answer1. The total number of customers is boxed{1400}.2. The correlation coefficient is boxed{0}.</think>

🧠:<think>Okay, so I have this problem about a digital artist archiving their artworks. The artist is using a lossless compression algorithm, which means the quality doesn't degrade, but the file sizes get reduced. There are two parts to the problem. Let me try to tackle them one by one.Starting with the first question: The original file size of each artwork follows a normal distribution with a mean of 500 MB and a standard deviation of 50 MB. The compression algorithm reduces the file size by a factor that follows an exponential decay model, f(t) = e^(-λt), where λ is the compression rate and t is time in hours. The compression rate λ is given as 0.1 per hour, and we need to find the expected file size after 24 hours.Hmm, okay. So, the original file size is normally distributed, N(500, 50^2). The compression factor is e^(-0.1*24). Let me compute that first.Calculating the exponent: 0.1 * 24 = 2.4. So, e^(-2.4). Let me compute e^(-2.4). I know that e^(-2) is approximately 0.1353, and e^(-0.4) is approximately 0.6703. So, e^(-2.4) would be e^(-2) * e^(-0.4) ≈ 0.1353 * 0.6703 ≈ 0.0907. So, about 0.0907.Therefore, the compression factor is approximately 0.0907. So, the expected file size after compression would be the original mean multiplied by this factor.Original mean is 500 MB. So, 500 * 0.0907 ≈ 45.35 MB. So, the expected file size after 24 hours is approximately 45.35 MB.Wait, but hold on. Is the compression factor applied multiplicatively? The problem says the compression reduces the file size by a factor that follows an exponential decay model. So, I think that means the file size is multiplied by e^(-λt). So, yes, 500 * e^(-0.1*24) ≈ 500 * 0.0907 ≈ 45.35 MB.So, that should be the expected file size. I think that's straightforward.Moving on to the second question: The artist plans to archive 1,000 digital artworks with a storage limit of 300 GB. We need to find the probability that the total storage required after 24 hours of compression will exceed 300 GB.Alright, so first, let's note that 300 GB is 300,000 MB. Each artwork is compressed to approximately 45.35 MB on average, but of course, each artwork's file size is a random variable. So, the total storage required is the sum of 1,000 such random variables.Given that each original file size is normally distributed, and then multiplied by a constant factor (the compression factor), the resulting distribution should still be normal, right? Because multiplying a normal variable by a constant scales the mean and variance accordingly.So, let's model this. Let X be the original file size of an artwork, which is N(500, 50^2). After compression, the file size Y = X * e^(-λt). So, Y is a scaled version of X. Therefore, Y is also normally distributed with mean μ_Y = μ_X * e^(-λt) and variance σ_Y^2 = (σ_X)^2 * (e^(-λt))^2.So, plugging in the numbers, μ_Y = 500 * e^(-2.4) ≈ 500 * 0.0907 ≈ 45.35 MB, as before. The standard deviation σ_Y = 50 * e^(-2.4) ≈ 50 * 0.0907 ≈ 4.535 MB.Therefore, each compressed file size Y_i is N(45.35, 4.535^2). Now, the total storage required is the sum of 1,000 such Y_i's. Let's denote the total storage as S = Y_1 + Y_2 + ... + Y_1000.Since each Y_i is normal, the sum S will also be normal, with mean μ_S = 1000 * μ_Y ≈ 1000 * 45.35 = 45,350 MB. The variance σ_S^2 = 1000 * σ_Y^2 ≈ 1000 * (4.535)^2 ≈ 1000 * 20.57 ≈ 20,570. Therefore, the standard deviation σ_S ≈ sqrt(20,570) ≈ 143.42 MB.Wait, but 45,350 MB is 45.35 GB, and the storage limit is 300 GB. Wait, hold on. 45,350 MB is 45.35 GB, which is way below 300 GB. That can't be right. Wait, perhaps I made a mistake in units.Wait, the original mean is 500 MB, so 1,000 artworks would be 500,000 MB, which is 500 GB. After compression, each is 45.35 MB, so 1,000 * 45.35 MB = 45,350 MB, which is 45.35 GB. So, the expected total storage is 45.35 GB, which is way below 300 GB. So, the probability that the total storage exceeds 300 GB is practically zero? That seems odd.Wait, maybe I misread the problem. Let me check again.The artist has a storage limit of 300 GB. So, 300 GB is 300,000 MB. The expected total storage is 45,350 MB, which is 45.35 GB. So, 45.35 GB is much less than 300 GB. Therefore, the probability that the total storage exceeds 300 GB is almost zero. But maybe I made a mistake in interpreting the compression factor.Wait, let me double-check the compression factor. The compression factor is e^(-λt). So, with λ = 0.1 per hour and t = 24 hours, it's e^(-2.4). As I calculated earlier, that's approximately 0.0907. So, the file size is reduced to about 9.07% of the original.So, 500 MB becomes 45.35 MB. So, 1,000 files would be 45,350 MB or 45.35 GB. So, 45.35 GB is way below 300 GB. So, the probability that the total storage exceeds 300 GB is practically zero. But maybe the question is about the total storage after compression, but perhaps the compression is not applied to each file individually, but rather the entire collection is compressed? Wait, the problem says "each digital artwork is stored as a high-resolution image file," and the compression is applied to each artwork. So, each artwork is compressed individually.Alternatively, perhaps the compression factor is additive rather than multiplicative? But the problem says it's a factor, so it should be multiplicative.Wait, maybe I misapplied the exponent. Let me recalculate e^(-2.4). Let me use a calculator for more precision.e^(-2.4) ≈ e^(-2) * e^(-0.4). e^(-2) ≈ 0.135335, e^(-0.4) ≈ 0.67032. Multiplying them: 0.135335 * 0.67032 ≈ 0.090718. So, approximately 0.0907, which is about 9.07%. So, 500 MB * 0.0907 ≈ 45.35 MB. So, that's correct.So, the expected total storage is 45.35 GB, which is way below 300 GB. Therefore, the probability that the total storage exceeds 300 GB is practically zero. But maybe the question is about the total storage required, considering the compression is applied over time, and perhaps the compression isn't completed yet? Wait, no, the problem says after 24 hours of compression.Wait, perhaps I misread the compression model. It says the compression reduces the file size by a factor that follows an exponential decay model f(t) = e^(-λt). So, is the compression factor f(t) = e^(-λt), meaning that the file size is multiplied by f(t)? Or is it that the reduction is f(t), so the remaining size is 1 - f(t)? Wait, the problem says "reduces the file size by a factor." So, I think it's multiplicative. So, the file size becomes original size multiplied by f(t). So, yes, 500 * e^(-0.1*24) ≈ 45.35 MB.Alternatively, maybe the compression factor is additive? Like, the file size is reduced by e^(-λt) each hour? But that would be a different model. The problem says "reduces the file size by a factor that follows an exponential decay model." So, factor implies multiplicative. So, I think my initial approach is correct.Therefore, the expected total storage is 45.35 GB, which is much less than 300 GB. So, the probability that the total storage exceeds 300 GB is effectively zero. But maybe I need to consider the distribution of the total storage and calculate the probability that S > 300,000 MB.Wait, 300 GB is 300,000 MB. The expected total storage is 45,350 MB, with a standard deviation of approximately 143.42 MB. So, the z-score for 300,000 MB would be (300,000 - 45,350) / 143.42 ≈ (254,650) / 143.42 ≈ 1,775. So, that's a z-score of about 1,775, which is way beyond any standard normal table. The probability of S > 300,000 MB is practically zero.But wait, that seems too straightforward. Maybe I made a mistake in calculating the variance. Let me double-check.Each Y_i is N(45.35, 4.535^2). So, the variance of each Y_i is (4.535)^2 ≈ 20.57. Then, the variance of the sum S is 1000 * 20.57 ≈ 20,570. So, the standard deviation is sqrt(20,570) ≈ 143.42 MB. That seems correct.But 1,000 artworks, each compressed to ~45 MB, so total is ~45,350 MB. 300 GB is 300,000 MB, which is 6.6 times larger than the expected total storage. So, the probability is practically zero. But maybe the question is about the total storage before compression? No, the question says after compression.Alternatively, perhaps the compression factor is applied per hour, and after 24 hours, the total compression is the sum of the factors? No, the problem says the compression factor is f(t) = e^(-λt), so it's a continuous decay over time. So, after 24 hours, it's e^(-2.4).Wait, maybe the compression is applied to the entire collection, not individually? But the problem says each artwork is stored as a high-resolution image file, and the compression is applied to each artwork. So, each is compressed individually.Alternatively, perhaps the compression factor is applied to the total storage? But that would be unusual, as compression is typically applied per file.Wait, let me read the problem again: "the compression algorithm reduces the file size of each artwork in a manner that follows a predictable pattern." So, each artwork's file size is reduced by the factor f(t). So, yes, each is compressed individually.Therefore, the total storage is the sum of 1,000 independent normal variables, each with mean 45.35 MB and variance 20.57 MB². So, the total storage S is N(45,350, 20,570). So, the probability that S > 300,000 MB is P(S > 300,000). Since 300,000 is way beyond the mean, the probability is effectively zero.But maybe I need to express it in terms of z-scores and use the standard normal distribution. Let's compute the z-score:z = (300,000 - 45,350) / 143.42 ≈ (254,650) / 143.42 ≈ 1,775.Looking up z = 1,775 in the standard normal table, but z-scores that high are essentially 1 in terms of probability. Wait, no, actually, for z-scores beyond about 3 or 4, the probability is already negligible. For z = 1,775, the probability that Z > 1,775 is effectively zero.Therefore, the probability that the total storage exceeds 300 GB is practically zero.But wait, maybe I made a mistake in interpreting the compression factor. Let me think again. If the compression factor is f(t) = e^(-λt), then the file size after compression is original size * f(t). So, yes, 500 * e^(-2.4) ≈ 45.35 MB.Alternatively, maybe the compression factor is 1 - e^(-λt), meaning that the file size is reduced by that factor. So, the remaining size is original * (1 - e^(-λt)). Wait, but the problem says "reduces the file size by a factor," which implies that the factor is the multiplier. So, if it's reduced by a factor of f(t), then the new size is original * f(t). So, I think my initial approach is correct.Alternatively, maybe the compression factor is additive, like the file size is reduced by f(t) each hour. But that would be a different model. The problem says "reduces the file size by a factor that follows an exponential decay model." So, factor implies multiplicative.Therefore, I think my calculations are correct. The expected total storage is 45.35 GB, and the probability of exceeding 300 GB is effectively zero.But just to be thorough, let me consider if the compression factor is applied per hour, and after 24 hours, the total compression is the product of hourly factors. But that would be the same as f(24) = e^(-0.1*24) = e^(-2.4), so it's the same result.Alternatively, if the compression factor is applied cumulatively, like each hour the file size is multiplied by e^(-0.1), then after 24 hours, it's (e^(-0.1))^24 = e^(-2.4), which is the same as before. So, same result.Therefore, I think my conclusion is correct. The probability is practically zero.But wait, maybe the problem expects a different approach. Let me think again. Maybe the compression factor is applied to the total storage, not per file. So, the total storage is 1,000 * 500 MB = 500,000 MB. Then, the compression factor is e^(-2.4), so the total storage becomes 500,000 * e^(-2.4) ≈ 500,000 * 0.0907 ≈ 45,350 MB, which is 45.35 GB. So, same result. So, the total storage is 45.35 GB, which is less than 300 GB. So, again, the probability is zero.But the problem says "the compression algorithm reduces the file size of each artwork," so it's per artwork, not the total. So, the total storage is the sum of compressed files, each compressed individually.Therefore, the total storage is 1,000 * 45.35 MB = 45,350 MB, which is 45.35 GB. So, the probability that 45.35 GB exceeds 300 GB is zero.Wait, but maybe the problem is considering that the compression is not perfect, and there's variability in the compression factor. So, each artwork's compressed size is a random variable, and the total storage is the sum of these variables. So, even though the expected total is 45.35 GB, there's a distribution around that mean, and we need to find the probability that the total exceeds 300 GB.But as I calculated earlier, the standard deviation of the total storage is about 143.42 MB, which is 0.14342 GB. So, the total storage is N(45.35, 0.14342^2) GB. So, the z-score for 300 GB is (300 - 45.35) / 0.14342 ≈ 254.65 / 0.14342 ≈ 1,775, which is way beyond any practical z-score. So, the probability is effectively zero.Therefore, the answer to the second question is that the probability is approximately zero.But maybe I need to express it more formally. Let me write down the steps:1. For each artwork, the compressed size Y_i = X_i * e^(-0.1*24), where X_i ~ N(500, 50^2).2. Therefore, Y_i ~ N(500 * e^(-2.4), (50 * e^(-2.4))^2) ≈ N(45.35, 4.535^2).3. The total storage S = sum_{i=1}^{1000} Y_i ~ N(1000*45.35, 1000*4.535^2) ≈ N(45,350, 20,570).4. Convert 300 GB to MB: 300,000 MB.5. Compute z-score: z = (300,000 - 45,350) / sqrt(20,570) ≈ (254,650) / 143.42 ≈ 1,775.6. The probability P(S > 300,000) is P(Z > 1,775), which is effectively zero.Therefore, the probability is approximately zero.So, summarizing:1. Expected file size after 24 hours: ~45.35 MB.2. Probability that total storage exceeds 300 GB: ~0.But let me check if I made any miscalculations. Let me recalculate e^(-2.4):Using a calculator: e^(-2.4) ≈ 0.090717953. So, 500 * 0.090717953 ≈ 45.3589765 MB. So, ~45.36 MB.Total storage: 1,000 * 45.36 ≈ 45,360 MB ≈ 45.36 GB.Standard deviation per file: 50 * 0.090717953 ≈ 4.5359 MB.Variance per file: (4.5359)^2 ≈ 20.57.Total variance: 1,000 * 20.57 ≈ 20,570.Total standard deviation: sqrt(20,570) ≈ 143.42 MB.So, z-score: (300,000 - 45,360) / 143.42 ≈ 254,640 / 143.42 ≈ 1,775.Yes, that's correct.Therefore, the answers are:1. The expected file size is approximately 45.36 MB.2. The probability is effectively zero.But maybe the problem expects a more precise answer, like using more decimal places or expressing it in terms of e^(-2.4). Let me see.Alternatively, maybe I should express the expected file size as 500 * e^(-2.4) MB, which is exact, rather than approximating it to 45.35 MB.Similarly, for the probability, perhaps it's better to express it in terms of the z-score and recognize that it's beyond the typical tables, so the probability is negligible.But in any case, the probability is so low that it's practically zero.So, to sum up:1. Expected file size after 24 hours: 500 * e^(-2.4) ≈ 45.36 MB.2. Probability of total storage exceeding 300 GB: Approximately 0.I think that's the solution.</think>

🧠:<think>Okay, so I have this problem about Dr. Smith, a retired surgeon, and his career modeled by the function ( S(t) = a cdot e^{bt} ). I need to find the constants ( a ) and ( b ) first, and then calculate the total number of surgeries he performed over 30 years using an integral. Let me try to break this down step by step.Starting with part 1: determining ( a ) and ( b ). The problem gives me two specific data points. In his first year, which would be ( t = 0 ), he performed 50 surgeries. Then, in his 10th year, ( t = 10 ), he performed 200 surgeries. So, I can set up two equations using these points.First, when ( t = 0 ):( S(0) = a cdot e^{b cdot 0} = a cdot e^0 = a cdot 1 = a ).And we know ( S(0) = 50 ), so that means ( a = 50 ). That was straightforward.Now, moving on to the second point when ( t = 10 ):( S(10) = 50 cdot e^{b cdot 10} = 200 ).So, plugging in the values, we have:( 50 cdot e^{10b} = 200 ).I need to solve for ( b ). Let me write that equation again:( 50e^{10b} = 200 ).First, divide both sides by 50 to simplify:( e^{10b} = 4 ).To solve for ( b ), take the natural logarithm of both sides:( ln(e^{10b}) = ln(4) ).Simplify the left side:( 10b = ln(4) ).Therefore, ( b = frac{ln(4)}{10} ).Let me compute ( ln(4) ) to see what value we get. I know that ( ln(4) ) is approximately 1.386294. So, dividing that by 10 gives ( b approx 0.1386294 ).So, summarizing part 1, ( a = 50 ) and ( b approx 0.1386 ). I can write ( b ) as ( frac{ln(4)}{10} ) exactly, which is better for calculations.Moving on to part 2: calculating the total number of surgeries over 30 years using the integral ( int_{0}^{30} S(t) , dt ).Given that ( S(t) = 50e^{bt} ) and ( b = frac{ln(4)}{10} ), let me substitute ( b ) into the function:( S(t) = 50e^{left(frac{ln(4)}{10}right)t} ).Simplify the exponent:( left(frac{ln(4)}{10}right)t = frac{t}{10} ln(4) = ln(4^{t/10}) ).Therefore, ( S(t) = 50 cdot e^{ln(4^{t/10})} = 50 cdot 4^{t/10} ).So, ( S(t) = 50 cdot 4^{t/10} ). Alternatively, since ( 4^{t/10} = (e^{ln(4)})^{t/10} = e^{(ln(4)/10)t} ), which is consistent with the original function.Now, I need to compute the integral ( int_{0}^{30} 50 cdot 4^{t/10} , dt ).Let me recall that the integral of ( a cdot b^{kt} ) with respect to ( t ) is ( frac{a}{k ln(b)} cdot b^{kt} + C ). So, applying that formula here.First, rewrite the integral:( int_{0}^{30} 50 cdot 4^{t/10} , dt ).Let me set ( k = frac{1}{10} ) and ( b = 4 ). So, the integral becomes:( 50 cdot int_{0}^{30} 4^{t/10} , dt = 50 cdot left[ frac{4^{t/10}}{(frac{1}{10}) ln(4)} right]_0^{30} ).Simplify the denominator:( frac{1}{frac{1}{10} ln(4)} = frac{10}{ln(4)} ).So, the integral becomes:( 50 cdot frac{10}{ln(4)} cdot left[4^{t/10}right]_0^{30} ).Compute the limits:At ( t = 30 ): ( 4^{30/10} = 4^3 = 64 ).At ( t = 0 ): ( 4^{0/10} = 4^0 = 1 ).Subtracting the lower limit from the upper limit:( 64 - 1 = 63 ).Putting it all together:Total surgeries = ( 50 cdot frac{10}{ln(4)} cdot 63 ).Compute this value step by step.First, compute ( frac{10}{ln(4)} ). Since ( ln(4) approx 1.386294 ), so ( frac{10}{1.386294} approx 7.213475 ).Then, multiply by 63:( 7.213475 times 63 approx ) Let me compute that.7.213475 * 60 = 432.80857.213475 * 3 = 21.640425Adding together: 432.8085 + 21.640425 ≈ 454.448925Now, multiply by 50:50 * 454.448925 ≈ 22,722.44625So, approximately 22,722.45 surgeries over 30 years.Wait, let me double-check my calculations because that seems a bit high, but considering exponential growth, it might make sense.Alternatively, maybe I can compute it more accurately.First, let's compute ( frac{10}{ln(4)} times 63 times 50 ).Compute ( frac{10}{ln(4)} times 63 times 50 ).Alternatively, compute ( 50 times frac{10}{ln(4)} times 63 ).Let me compute ( 50 times 10 = 500 ).Then, ( 500 times 63 = 31,500 ).Then, ( frac{31,500}{ln(4)} approx frac{31,500}{1.386294} approx ).Compute 31,500 divided by 1.386294.Let me compute 31,500 / 1.386294:First, approximate 1.386294 * 22,700 ≈ 31,500.Wait, 1.386294 * 22,700:Compute 1.386294 * 20,000 = 27,725.881.386294 * 2,700 = approx 1.386294 * 2,000 = 2,772.588; 1.386294 * 700 ≈ 970.4058. So total ≈ 2,772.588 + 970.4058 ≈ 3,742.9938.So, total 27,725.88 + 3,742.9938 ≈ 31,468.87, which is close to 31,500.So, 1.386294 * 22,700 ≈ 31,468.87, which is about 31,500. So, 31,500 / 1.386294 ≈ 22,700.Therefore, the total number of surgeries is approximately 22,700.Wait, but earlier I got 22,722.45, which is close to 22,700. So, that seems consistent.But let me check my earlier step.Wait, when I computed ( 50 times frac{10}{ln(4)} times 63 ), I think I might have miscalculated.Wait, actually, the integral is:Total = ( 50 times frac{10}{ln(4)} times (4^{3} - 1) ).Which is ( 50 times frac{10}{ln(4)} times 63 ).So, that's 50 * 10 * 63 / ln(4) = 31,500 / ln(4).Since ln(4) ≈ 1.386294, so 31,500 / 1.386294 ≈ 22,700.So, approximately 22,700 surgeries.But let me compute it more accurately.Compute 31,500 divided by 1.386294.Let me do this division step by step.1.386294 * 22,700 = ?Compute 1.386294 * 20,000 = 27,725.881.386294 * 2,700 = ?Compute 1.386294 * 2,000 = 2,772.5881.386294 * 700 = 970.4058So, 2,772.588 + 970.4058 = 3,742.9938So, total 27,725.88 + 3,742.9938 = 31,468.8738Which is approximately 31,468.87, which is 31,500 - 31,468.87 = 31.13 less.So, to get 31,500, we need to add a little more.Compute how much more:31,500 - 31,468.87 = 31.13So, 31.13 / 1.386294 ≈ 22.48So, total multiplier is 22,700 + 22.48 ≈ 22,722.48So, 31,500 / 1.386294 ≈ 22,722.48Therefore, the total number of surgeries is approximately 22,722.48.So, rounding to a reasonable number, maybe 22,722 or 22,723.But since we're dealing with whole surgeries, it should be an integer. So, approximately 22,722 surgeries.Wait, but let me check if I did the integral correctly.The integral of ( 50 cdot 4^{t/10} ) from 0 to 30 is:( 50 cdot left[ frac{4^{t/10}}{(ln(4)/10)} right]_0^{30} )Which is ( 50 cdot frac{10}{ln(4)} cdot (4^{3} - 1) )Yes, that's correct.So, ( 4^{3} = 64 ), so 64 - 1 = 63.Thus, ( 50 times frac{10}{ln(4)} times 63 ).Which is ( 50 times 10 times 63 / ln(4) = 31,500 / ln(4) approx 22,722.48 ).So, approximately 22,722 surgeries.But let me make sure I didn't make a mistake in the integral setup.The integral of ( e^{kt} ) is ( frac{e^{kt}}{k} ). So, in this case, ( S(t) = 50e^{bt} ), so the integral is ( frac{50}{b} e^{bt} ) evaluated from 0 to 30.But wait, in my earlier steps, I rewrote ( S(t) = 50 cdot 4^{t/10} ), which is equivalent to ( 50e^{bt} ) with ( b = ln(4)/10 ).So, integrating ( 50e^{bt} ) from 0 to 30 gives ( frac{50}{b}(e^{30b} - 1) ).Since ( b = ln(4)/10 ), then ( e^{30b} = e^{30*(ln4)/10} = e^{3 ln4} = (e^{ln4})^3 = 4^3 = 64 ).So, the integral becomes ( frac{50}{(ln4)/10} (64 - 1) = 50 * (10 / ln4) * 63 = 50 * 10 * 63 / ln4 = 31,500 / ln4 ≈ 22,722.48 ).Yes, that's correct.Alternatively, I can compute it using the exact value of ( ln(4) ).But for the purposes of this problem, I think 22,722 is a reasonable approximation.Wait, but let me check if I can express the answer in terms of exact exponentials.Alternatively, since ( ln(4) = 2ln(2) ), so ( frac{1}{ln(4)} = frac{1}{2ln(2)} ).So, 31,500 / (2 ln2) ≈ 31,500 / (2 * 0.693147) ≈ 31,500 / 1.386294 ≈ 22,722.48, which is the same as before.So, I think that's correct.Therefore, the total number of surgeries Dr. Smith performed over his 30-year career is approximately 22,722.But let me just think again: starting at 50 surgeries, growing exponentially with a growth rate such that in 10 years it becomes 200. So, each year, the number of surgeries is multiplied by ( 4^{1/10} ), which is approximately 1.1487, so about a 14.87% increase each year. Over 30 years, that's a significant growth, so the total being around 22,722 seems plausible.Alternatively, if I compute the sum of the surgeries each year, it's a geometric series with first term 50, ratio ( r = 4^{1/10} approx 1.1487 ), and 30 terms. The sum of a geometric series is ( S = a frac{r^n - 1}{r - 1} ).So, let me compute that as a check.Compute ( S = 50 times frac{(4^{3} - 1)}{(4^{1/10} - 1)} ).Wait, because ( r = 4^{1/10} ), so ( r^{30} = (4^{1/10})^{30} = 4^{3} = 64 ).So, the sum is ( 50 times frac{64 - 1}{4^{1/10} - 1} = 50 times frac{63}{4^{1/10} - 1} ).Compute ( 4^{1/10} ). Since ( 4^{1/10} = e^{(ln4)/10} ≈ e^{0.1386294} ≈ 1.14869835 ).So, ( 4^{1/10} - 1 ≈ 0.14869835 ).Therefore, ( frac{63}{0.14869835} ≈ 423.3108 ).Multiply by 50: 50 * 423.3108 ≈ 21,165.54.Wait, that's different from the integral result of approximately 22,722.Hmm, that's a discrepancy. So, why is there a difference?Because the integral assumes continuous growth, whereas the sum of the geometric series is discrete, assuming that each year's surgeries are a fixed multiple of the previous year. So, the integral is a continuous approximation, while the sum is the exact discrete sum.But in the problem statement, it says "Assume continuous surgery performance over the years for the purposes of integration." So, we are supposed to use the integral, not the sum.Therefore, the integral result of approximately 22,722 is the correct answer for the total number of surgeries.But just to understand why the sum is different: the integral is integrating the continuous function, which is always growing, whereas the discrete sum would be the sum of the function evaluated at integer points. Since the function is increasing, the integral will be greater than the sum of the left endpoints and less than the sum of the right endpoints.But since we are told to use the integral, we should go with that.So, to recap:1. Found ( a = 50 ) and ( b = ln(4)/10 approx 0.1386 ).2. Computed the integral ( int_{0}^{30} 50e^{bt} dt approx 22,722 ).Therefore, the total number of surgeries is approximately 22,722.But to express it more precisely, since we have exact expressions, let me write the exact value before approximating.The exact value is ( frac{31,500}{ln(4)} ).Since ( ln(4) = 2ln(2) ), we can write it as ( frac{31,500}{2ln(2)} = frac{15,750}{ln(2)} ).But unless the problem asks for an exact form, which it doesn't, we can leave it as approximately 22,722.Alternatively, using more decimal places for ln(4):ln(4) ≈ 1.3862943611.So, 31,500 / 1.3862943611 ≈ 22,722.446.So, approximately 22,722.45, which we can round to 22,722 or 22,723.But since the number of surgeries must be an integer, 22,722 or 22,723. Depending on rounding, it's 22,722.45, so 22,722 when rounded down, or 22,722 when rounded to the nearest whole number.But in any case, the answer is approximately 22,722.So, to summarize:1. ( a = 50 ), ( b = frac{ln(4)}{10} ).2. Total surgeries ≈ 22,722.I think that's it.Final Answer1. The values of ( a ) and ( b ) are ( boxed{50} ) and ( boxed{frac{ln(4)}{10}} ) respectively.2. The total number of surgeries Dr. Smith performed over his 30-year career is approximately ( boxed{22722} ).</think>

🧠:<think>Okay, so I have this problem about a recommender system's performance, and I need to calculate Precision, Recall, and then the F1 Score. Let me try to break this down step by step.First, the problem is divided into two parts. The first part is about a specific user, and the second part is about the overall system performance based on 100 users. I'll tackle them one by one.Starting with part 1: For a particular user, the system recommended 10 items. The user's relevant items are {A, B, C, D, E, F, G}, which is 7 items. The system recommended {A, C, E, G, H, I, J, K, L, M}. So, I need to find out how many of these recommended items are actually relevant to the user.Let me list the recommended items: A, C, E, G, H, I, J, K, L, M. Comparing this with the relevant items {A, B, C, D, E, F, G}, the overlapping items are A, C, E, G. So, that's 4 relevant items out of 10 recommended.Now, using the definitions given:Precision is the number of relevant items recommended divided by the total number of items recommended. So, that's 4 relevant items out of 10 recommended. So, Precision = 4/10 = 0.4.Recall is the number of relevant items recommended divided by the total number of relevant items. The user has 7 relevant items, and 4 were recommended. So, Recall = 4/7 ≈ 0.5714.Wait, let me double-check that. The relevant items are 7, and 4 were correctly recommended. So, yes, 4/7 is approximately 0.5714.So, for part 1, Precision is 0.4 and Recall is approximately 0.5714.Moving on to part 2: The representative has collected data from 100 users, and the average Precision is 0.75, and the average Recall is 0.6. They want to calculate the F1 Score, which is the harmonic mean of Precision and Recall.The formula given is F1 = 2 * (Precision * Recall) / (Precision + Recall). So, plugging in the average values:F1 = 2 * (0.75 * 0.6) / (0.75 + 0.6).First, calculate the numerator: 0.75 * 0.6 = 0.45.Then, the denominator: 0.75 + 0.6 = 1.35.So, F1 = 2 * (0.45 / 1.35).Calculating 0.45 / 1.35: That's the same as 45/135, which simplifies to 1/3 ≈ 0.3333.Then, multiplying by 2: 2 * 0.3333 ≈ 0.6666.So, the F1 Score is approximately 0.6667, which is 2/3.Wait, let me verify that again. 0.75 times 0.6 is 0.45. 0.75 plus 0.6 is 1.35. 0.45 divided by 1.35 is indeed 1/3. Multiply by 2 gives 2/3, which is approximately 0.6667.So, summarizing:1. For the specific user, Precision is 0.4 and Recall is approximately 0.5714.2. For the overall system, the F1 Score is approximately 0.6667.I think that's all. Let me just make sure I didn't make any calculation errors. For part 1, 4 relevant out of 10 recommended is 0.4 precision. 4 out of 7 relevant is about 0.5714 recall. That seems right.For part 2, using the harmonic mean formula, plugging in 0.75 and 0.6, the calculations check out. So, yes, F1 is 2/3 or approximately 0.6667.Final Answer1. Precision is boxed{0.4} and Recall is boxed{dfrac{4}{7}}.2. The F1 Score is boxed{dfrac{2}{3}}.</think>