🧠:<think>Okay, so João is this college student entrepreneur who's developing an app to connect other students with business people for mentorship opportunities. He needs to understand how the app is being used over time and make strategic decisions based on data. There are two main problems here that I need to solve.First, João has a function that models the number of new users signing up each month, which is N(t) = 100e^{0.05t}, where t is the time in months since the launch. He wants to find the total number of users who signed up during the first year. Hmm, so that would be from t=0 to t=12 months, right? I remember that when you have a rate function, like the number of new users per month, to find the total number over a period, you need to integrate that function over that time interval. So, the total number of users, let's call it U(t), should be the integral of N(t) from 0 to 12. Let me write that down: U = ∫₀¹² 100e^{0.05t} dt. To solve this integral, I can factor out the 100, so it becomes 100 times the integral of e^{0.05t} dt. The integral of e^{kt} dt is (1/k)e^{kt} + C, so applying that here, it would be 100 * (1/0.05)e^{0.05t} evaluated from 0 to 12. Calculating 1/0.05, that's 20. So, U = 100 * 20 [e^{0.05*12} - e^{0}]. Simplify that: 100*20 is 2000. Then, e^{0.6} minus e^{0}. I know e^0 is 1, and e^{0.6} is approximately... let me recall, e^0.5 is about 1.6487, so e^0.6 should be a bit more. Maybe around 1.8221? Let me double-check that. Alternatively, I can calculate it more precisely using a calculator, but since I don't have one, I'll go with 1.8221.So, 2000*(1.8221 - 1) = 2000*(0.8221) = 1644.2. So, approximately 1644 users in the first year. But wait, let me make sure I didn't make a mistake in the exponent. 0.05*12 is 0.6, correct. So, that part is right. And the integral setup is correct because we're summing up all the new users each month over the year.Moving on to the second problem. João wants to optimize the response time of mentors to student requests. He modeled the time of response, T, as an exponential random variable with a mean of 3 hours. He wants at least 80% of the responses to be given in less than 5 hours. So, I need to find the cumulative distribution function (CDF) F(t) and check if F(5) is at least 0.8.First, recalling that the exponential distribution has the CDF F(t) = 1 - e^{-λt}, where λ is the rate parameter, which is 1 over the mean. Since the mean is 3 hours, λ = 1/3 ≈ 0.3333.So, F(t) = 1 - e^{-(1/3)t}. Therefore, F(5) = 1 - e^{-5/3}. Let me compute that. 5/3 is approximately 1.6667. So, e^{-1.6667} is about... e^{-1} is 0.3679, e^{-1.6} is around 0.2019, and e^{-1.6667} is a bit less than that. Maybe approximately 0.1889?So, F(5) ≈ 1 - 0.1889 = 0.8111. That's about 81.11%, which is more than 80%. So, the condition is satisfied. Wait, let me verify the calculation for e^{-5/3}. 5/3 is 1.6667. Let me compute e^{-1.6667} more accurately. I know that ln(2) is about 0.6931, so e^{-1.6667} is equal to 1 / e^{1.6667}. Let me compute e^{1.6667}.We know that e^1 = 2.71828, e^1.6 is approximately 4.953, e^1.6667 is a bit more. Let me compute it step by step.We can write 1.6667 as 1 + 0.6667. So, e^{1.6667} = e^1 * e^{0.6667}. e^{0.6667} is approximately e^{2/3}. I remember that e^{0.6} is about 1.8221, e^{0.7} is about 2.0138. So, 2/3 is approximately 0.6667, which is between 0.6 and 0.7. Let's interpolate.The difference between e^{0.6} and e^{0.7} is about 2.0138 - 1.8221 = 0.1917 over an interval of 0.1. So, per 0.01 increase, it's about 0.1917/0.1 = 1.917 per 0.01. So, from 0.6 to 0.6667 is 0.0667, so 0.0667 * 1.917 ≈ 0.1278. So, e^{0.6667} ≈ 1.8221 + 0.1278 ≈ 1.9499.Therefore, e^{1.6667} ≈ e^1 * e^{0.6667} ≈ 2.71828 * 1.9499 ≈ Let's compute that.2.71828 * 1.9499:First, 2 * 1.9499 = 3.89980.7 * 1.9499 ≈ 1.36490.01828 * 1.9499 ≈ approximately 0.0357Adding them up: 3.8998 + 1.3649 = 5.2647 + 0.0357 ≈ 5.3004.So, e^{1.6667} ≈ 5.3004, so e^{-1.6667} ≈ 1 / 5.3004 ≈ 0.1887.Therefore, F(5) = 1 - 0.1887 ≈ 0.8113, which is approximately 81.13%. So, that's more than 80%, so the condition is satisfied.Wait, but let me think again. The exponential distribution is memoryless, so the CDF is indeed F(t) = 1 - e^{-λt}, correct. And since the mean is 3, λ is 1/3, so that's correct.So, putting it all together, the first part is integrating the function N(t) over 12 months, which gives approximately 1644 users. The second part is calculating the CDF at 5 hours, which is about 81.13%, which is above 80%, so the condition is met.I think that's it. I don't see any mistakes in my reasoning, but let me just recap to make sure.For the first problem, integrating the exponential function correctly gives the total number of users. The integral of e^{kt} is (1/k)e^{kt}, so that's correct. Plugging in the limits, 0 to 12, gives the total.For the second problem, understanding that the exponential distribution's CDF is 1 - e^{-λt}, calculating λ as 1/3, then evaluating at t=5 gives the probability that T is less than 5, which is about 81%, which is above 80%. So, João's condition is satisfied.Yeah, I think that's solid.Final Answer1. The total number of users in the first year is boxed{1644}.2. The function of the cumulative distribution is ( F(t) = 1 - e^{-frac{t}{3}} ) and the condition is satisfied since ( F(5) approx 81.13% geq 80% ), so the final answer is boxed{F(t) = 1 - e^{-frac{t}{3}}} and the condition is met.</think>

🧠:<think>Alright, so I need to estimate the parameter λ for the Poisson distribution that models the number of runs scored by the Sacramento State Hornets in each game. They've given me the distribution of runs over 100 games, so I can use that data to find the MLE estimate of λ.First, I remember that the Maximum Likelihood Estimation for the Poisson distribution is simply the sample mean. That makes sense because the Poisson distribution is characterized by its mean, and the MLE estimator is the average of the observed data. So, my task is to calculate the average number of runs per game based on the given data.Looking at the data:- 0 runs: 5 games- 1 run: 12 games- 2 runs: 20 games- 3 runs: 25 games- 4 runs: 18 games- 5 runs: 12 games- 6 runs: 5 games- 7 or more runs: 3 gamesI need to compute the total number of runs scored over these 100 games and then divide by 100 to get the average.Let me calculate the total runs:For 0 runs: 0 * 5 = 0 runsFor 1 run: 1 * 12 = 12 runsFor 2 runs: 2 * 20 = 40 runsFor 3 runs: 3 * 25 = 75 runsFor 4 runs: 4 * 18 = 72 runsFor 5 runs: 5 * 12 = 60 runsFor 6 runs: 6 * 5 = 30 runsFor 7 or more runs: The problem says 7 or more, but it doesn't specify the exact number. Hmm, this might be a bit tricky. Since we don't have the exact number of runs for these 3 games, we might have to make an assumption. One common approach is to assume that these games scored exactly 7 runs each, but that might not be accurate. Alternatively, we could treat them as 7 or more, but without more data, it's hard to get an exact total. However, since the number of games here is small (only 3 games), the impact on the overall average might be minimal. Maybe I can proceed by assuming each of these games scored exactly 7 runs, so 7 * 3 = 21 runs.Alternatively, if I don't make that assumption, I might have to leave it as an unknown, but since the problem is asking for an estimate, I think it's acceptable to make this assumption for the sake of calculation.So, adding up all these runs:0 + 12 + 40 + 75 + 72 + 60 + 30 + 21 = Let's compute step by step:Start with 0.Add 12: total is 12.Add 40: total is 52.Add 75: total is 127.Add 72: total is 199.Add 60: total is 259.Add 30: total is 289.Add 21: total is 310.So, the total number of runs is 310 over 100 games.Therefore, the sample mean λ is 310 / 100 = 3.1.Wait, but hold on, I assumed that the 7 or more runs were exactly 7. What if they were higher? For example, if some games had 8, 9, etc., the total runs would be higher, which would make λ larger. But since we don't have exact numbers, 7 is a reasonable assumption. Alternatively, if we consider that 7 or more could be 7, 8, 9, etc., but without knowing the exact distribution, it's hard to say. However, given that it's only 3 games, even if they scored, say, 10 runs each, that would add 30 runs, making the total 310 + 30 = 340, which would make λ = 3.4. But that's a big jump. Alternatively, if they scored 7, 8, 9, that would be 24 runs, which is similar to 21. So, maybe 3.1 is a bit low, but without exact data, I think 3.1 is the best estimate.Alternatively, another approach is to compute the expected value using the given frequencies. Since the data is grouped, we can treat the 7 or more as a category. However, without knowing the exact distribution within that category, it's difficult. But maybe we can use the midpoint or something? Wait, in Poisson, the mean is just the average, so perhaps we can compute it as:Total runs = sum over x * frequency(x)But for x >=7, we don't know x. So, perhaps we can approximate it by considering that the average for x >=7 is 7, 8, 9, etc., but without knowing, it's hard. Alternatively, maybe we can ignore the 7 or more and just compute the average for the rest, but that would undercount.Wait, actually, in the data, the last category is 7 or more, which is 3 games. So, perhaps we can consider that the average for that category is 7, as a lower bound, or maybe higher. But since we don't have exact numbers, perhaps the best we can do is to compute the total runs as 310, as I did before, assuming each of those 3 games had 7 runs. So, λ = 3.1.Alternatively, if we don't make that assumption, we can compute the total runs as:Sum from x=0 to x=6 of x * frequency(x) + sum from x=7 to infinity of x * frequency(x). But since we don't know the exact x for the last category, we can't compute that sum. So, perhaps the best approach is to use the data we have and compute the average up to x=6, and then for the remaining 3 games, we can assume that they contribute an average of 7 runs each, which is a reasonable assumption.Therefore, I think λ = 3.1 is a reasonable estimate.Now, moving on to part 2. Using this estimated λ = 3.1, I need to calculate the probability that the Hornets score at least 5 runs in a given game. That is, P(X >=5).As the note suggests, this can be calculated as 1 - P(X <5). So, I need to compute P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4), and then subtract that sum from 1.The Poisson probability mass function is given by:P(X = k) = (e^{-λ} * λ^k) / k!So, I need to compute this for k=0,1,2,3,4 and sum them up, then subtract from 1.Let me compute each term step by step.First, compute e^{-λ} where λ=3.1.e^{-3.1} ≈ e^{-3} * e^{-0.1} ≈ 0.0498 * 0.9048 ≈ 0.04517.Alternatively, I can use a calculator for e^{-3.1} ≈ 0.04517.Now, compute each term:For k=0:P(X=0) = e^{-3.1} * (3.1)^0 / 0! = 0.04517 * 1 / 1 = 0.04517For k=1:P(X=1) = e^{-3.1} * (3.1)^1 / 1! = 0.04517 * 3.1 / 1 ≈ 0.04517 * 3.1 ≈ 0.1399For k=2:P(X=2) = e^{-3.1} * (3.1)^2 / 2! = 0.04517 * 9.61 / 2 ≈ 0.04517 * 4.805 ≈ 0.2171For k=3:P(X=3) = e^{-3.1} * (3.1)^3 / 3! = 0.04517 * 29.791 / 6 ≈ 0.04517 * 4.965 ≈ 0.2243For k=4:P(X=4) = e^{-3.1} * (3.1)^4 / 4! = 0.04517 * 92.3521 / 24 ≈ 0.04517 * 3.848 ≈ 0.1737Now, let's sum these up:P(X=0) ≈ 0.04517P(X=1) ≈ 0.1399P(X=2) ≈ 0.2171P(X=3) ≈ 0.2243P(X=4) ≈ 0.1737Adding them together:0.04517 + 0.1399 = 0.185070.18507 + 0.2171 = 0.402170.40217 + 0.2243 = 0.626470.62647 + 0.1737 = 0.80017So, P(X <5) ≈ 0.80017Therefore, P(X >=5) = 1 - 0.80017 ≈ 0.19983, or approximately 0.20.Wait, let me double-check my calculations because 0.19983 seems a bit low. Let me recalculate each term more accurately.First, e^{-3.1} ≈ 0.04517For k=0: 0.04517For k=1: 0.04517 * 3.1 = 0.1399 (correct)For k=2: (3.1)^2 = 9.61, divided by 2! = 2, so 9.61/2 = 4.805. Then, 0.04517 * 4.805 ≈ 0.04517 * 4.8 = 0.2168, plus 0.04517 * 0.005 ≈ 0.00022585, so total ≈ 0.2170.For k=3: (3.1)^3 = 29.791, divided by 3! = 6, so 29.791/6 ≈ 4.965166667. Then, 0.04517 * 4.965166667 ≈ Let's compute 0.04517 * 4 = 0.18068, 0.04517 * 0.965166667 ≈ 0.0436. So total ≈ 0.18068 + 0.0436 ≈ 0.22428.For k=4: (3.1)^4 = 3.1 * 29.791 ≈ 92.3521. Divided by 4! = 24, so 92.3521 /24 ≈ 3.848. Then, 0.04517 * 3.848 ≈ Let's compute 0.04517 * 3 = 0.13551, 0.04517 * 0.848 ≈ 0.0383. So total ≈ 0.13551 + 0.0383 ≈ 0.17381.Now, summing up:0.04517 (k=0)+0.1399 (k=1) = 0.18507+0.2170 (k=2) = 0.40207+0.22428 (k=3) = 0.62635+0.17381 (k=4) = 0.80016So, P(X <5) ≈ 0.80016, so P(X >=5) ≈ 1 - 0.80016 ≈ 0.19984, which is approximately 0.20.But wait, let me check if I can compute this more accurately using a calculator or more precise exponentials.Alternatively, perhaps I can use the Poisson cumulative distribution function (CDF) for λ=3.1 and k=4, and then subtract from 1.Using a calculator or software, the CDF for Poisson(3.1) at k=4 is approximately 0.8001, so 1 - 0.8001 ≈ 0.1999, which is about 0.20.Therefore, the probability that the Hornets score at least 5 runs in a given game is approximately 0.20 or 20%.But let me verify this with more precise calculations.Alternatively, I can use the formula for each term with more decimal places.Compute e^{-3.1}:e^{-3} = 0.049787068e^{-0.1} ≈ 0.904837418So, e^{-3.1} = e^{-3} * e^{-0.1} ≈ 0.049787068 * 0.904837418 ≈ Let's compute:0.049787068 * 0.9 = 0.04480836120.049787068 * 0.004837418 ≈ Approximately 0.049787068 * 0.0048 ≈ 0.000238978So total ≈ 0.0448083612 + 0.000238978 ≈ 0.045047339So, e^{-3.1} ≈ 0.045047339Now, compute each term:k=0: 0.045047339k=1: 0.045047339 * 3.1 ≈ 0.13964675k=2: (3.1)^2 = 9.61, divided by 2! = 2, so 9.61/2 = 4.805. Then, 0.045047339 * 4.805 ≈ Let's compute:0.045047339 * 4 = 0.1801893560.045047339 * 0.805 ≈ 0.045047339 * 0.8 = 0.036037871, plus 0.045047339 * 0.005 ≈ 0.000225237, total ≈ 0.036037871 + 0.000225237 ≈ 0.036263108So total for k=2 ≈ 0.180189356 + 0.036263108 ≈ 0.216452464k=3: (3.1)^3 = 29.791, divided by 3! = 6, so 29.791/6 ≈ 4.965166667. Then, 0.045047339 * 4.965166667 ≈ Let's compute:0.045047339 * 4 = 0.1801893560.045047339 * 0.965166667 ≈ 0.045047339 * 0.9 = 0.040542605, plus 0.045047339 * 0.065166667 ≈ 0.002934873So total ≈ 0.040542605 + 0.002934873 ≈ 0.043477478Thus, total for k=3 ≈ 0.180189356 + 0.043477478 ≈ 0.223666834k=4: (3.1)^4 = 3.1 * 29.791 ≈ 92.3521, divided by 4! = 24, so 92.3521 /24 ≈ 3.848. Then, 0.045047339 * 3.848 ≈ Let's compute:0.045047339 * 3 = 0.1351420170.045047339 * 0.848 ≈ 0.045047339 * 0.8 = 0.036037871, plus 0.045047339 * 0.048 ≈ 0.002162272So total ≈ 0.036037871 + 0.002162272 ≈ 0.038200143Thus, total for k=4 ≈ 0.135142017 + 0.038200143 ≈ 0.17334216Now, summing up all terms:k=0: 0.045047339k=1: 0.13964675k=2: 0.216452464k=3: 0.223666834k=4: 0.17334216Adding them:Start with 0.045047339 + 0.13964675 = 0.184694089+0.216452464 = 0.401146553+0.223666834 = 0.624813387+0.17334216 = 0.798155547So, P(X <5) ≈ 0.798155547Therefore, P(X >=5) = 1 - 0.798155547 ≈ 0.201844453, which is approximately 0.2018 or 20.18%.So, rounding to four decimal places, it's approximately 0.2018.Therefore, the probability is approximately 20.18%.But let me check if I can find a more precise value using a calculator or Poisson CDF table.Alternatively, using the Poisson CDF formula:P(X <= k) = e^{-λ} * Σ_{i=0}^k (λ^i / i!)So, for λ=3.1 and k=4, we have:P(X <=4) = e^{-3.1} * [1 + 3.1 + (3.1)^2/2 + (3.1)^3/6 + (3.1)^4/24]We already computed e^{-3.1} ≈ 0.045047339Now, compute the sum inside:1 + 3.1 + (9.61)/2 + (29.791)/6 + (92.3521)/24Compute each term:1 = 13.1 = 3.19.61 / 2 = 4.80529.791 / 6 ≈ 4.96516666792.3521 /24 ≈ 3.848Now, sum these:1 + 3.1 = 4.14.1 + 4.805 = 8.9058.905 + 4.965166667 ≈ 13.87016666713.870166667 + 3.848 ≈ 17.718166667So, the sum inside is approximately 17.718166667Therefore, P(X <=4) ≈ 0.045047339 * 17.718166667 ≈ Let's compute:0.045047339 * 17 = 0.7658047630.045047339 * 0.718166667 ≈ 0.045047339 * 0.7 = 0.031533137, plus 0.045047339 * 0.018166667 ≈ 0.000818So total ≈ 0.031533137 + 0.000818 ≈ 0.032351137Thus, total P(X <=4) ≈ 0.765804763 + 0.032351137 ≈ 0.7981559Which matches our earlier calculation of approximately 0.798155547.Therefore, P(X >=5) ≈ 1 - 0.7981559 ≈ 0.2018441, which is approximately 0.2018 or 20.18%.So, rounding to four decimal places, it's 0.2018.Alternatively, if we want to express it as a percentage, it's approximately 20.18%.But since the question doesn't specify the format, I think 0.2018 is sufficient.Alternatively, if we use more precise calculations, perhaps using a calculator or software, the exact value might be slightly different, but for our purposes, 0.2018 is a good estimate.Therefore, the estimated λ is 3.1, and the probability of scoring at least 5 runs is approximately 0.2018.</think>

🧠:<think>Alright, so I'm trying to tackle these two linear algebra problems related to a spoken word poet's process of transforming guest stories into performances. Let me go through each problem step by step.Starting with problem 1: The poet has a matrix ( A ) which is symmetric, constructed from guest stories. Each story is a vector ( mathbf{x}_i ) in ( mathbb{R}^m ), and ( A ) is an ( n times m ) matrix. Wait, hold on—if each row is a story vector, then ( A ) should be ( n times m ), but it's given as symmetric. Hmm, symmetric matrices are square, so that means ( n = m ). So actually, ( A ) is an ( n times n ) symmetric matrix. That makes sense because for a matrix to be symmetric, it must be square.Now, the question is about the conditions under which ( A ) is positive semi-definite. I remember that for symmetric matrices, positive semi-definiteness can be characterized in a few ways. One of the main criteria is that all the eigenvalues of ( A ) are non-negative. Another way is that for any non-zero vector ( mathbf{x} ), the quadratic form ( mathbf{x}^T A mathbf{x} geq 0 ).But wait, since ( A ) is constructed from the stories, which are vectors in ( mathbb{R}^n ), is there another way to think about this? Maybe in terms of the matrix being a Gram matrix? Because if ( A ) is a Gram matrix, meaning it's of the form ( A = X^T X ) where ( X ) is some matrix, then ( A ) is automatically positive semi-definite. But in this case, ( A ) is constructed by having each row as a story vector. So if each row is a vector in ( mathbb{R}^n ), then ( A ) is an ( n times n ) matrix. If we think of ( A ) as ( X^T X ), where ( X ) is an ( n times n ) matrix, then ( A ) would be positive semi-definite. However, if ( A ) is just any symmetric matrix, it's positive semi-definite if all its eigenvalues are non-negative.So, putting it together, the condition for ( A ) being positive semi-definite is that all its eigenvalues are greater than or equal to zero. Alternatively, if ( A ) can be expressed as ( X^T X ) for some matrix ( X ), then it's positive semi-definite.Moving on to problem 2: The poet transforms matrix ( A ) into ( B = P^{-1} A P ), where ( P ) is an invertible matrix representing a stylistic transformation. We need to find the eigenvalues of ( B ) given that ( A ) has eigenvalues ( lambda_1, lambda_2, ldots, lambda_n ).I recall that similar matrices have the same eigenvalues. Since ( B ) is similar to ( A ) (because ( B = P^{-1} A P )), they must share the same set of eigenvalues. So, regardless of the invertible matrix ( P ), the eigenvalues of ( B ) are the same as those of ( A ), which are ( lambda_1, lambda_2, ldots, lambda_n ).But let me verify this. If ( A ) has eigenvalues ( lambda_i ), then there exists a non-zero vector ( mathbf{v}_i ) such that ( A mathbf{v}_i = lambda_i mathbf{v}_i ). Now, consider ( B = P^{-1} A P ). Let's see if ( lambda_i ) is also an eigenvalue of ( B ). Suppose ( mathbf{w}_i = P mathbf{v}_i ). Then,( B mathbf{w}_i = P^{-1} A P mathbf{w}_i = P^{-1} A (P mathbf{v}_i) = P^{-1} (A mathbf{v}_i) = P^{-1} (lambda_i mathbf{v}_i) = lambda_i P^{-1} mathbf{v}_i ).Wait, that doesn't directly give ( lambda_i mathbf{w}_i ). Hmm, maybe I made a mistake. Let me try again.Let ( mathbf{w}_i = P^{-1} mathbf{v}_i ). Then,( B mathbf{w}_i = P^{-1} A P mathbf{w}_i = P^{-1} A (P mathbf{w}_i) = P^{-1} A mathbf{v}_i = P^{-1} (lambda_i mathbf{v}_i) = lambda_i P^{-1} mathbf{v}_i = lambda_i mathbf{w}_i ).Ah, that works. So ( mathbf{w}_i ) is an eigenvector of ( B ) corresponding to eigenvalue ( lambda_i ). Therefore, ( B ) has the same eigenvalues as ( A ).So, summarizing, the eigenvalues of ( B ) are the same as those of ( A ), which are ( lambda_1, lambda_2, ldots, lambda_n ).Final Answer1. The matrix ( A ) is positive semi-definite if and only if all its eigenvalues are non-negative. Therefore, the condition is that (boxed{lambda_i geq 0}) for all (i).2. The eigenvalues of matrix ( B ) are the same as those of matrix ( A ). Thus, the eigenvalues of ( B ) are (boxed{lambda_1, lambda_2, ldots, lambda_n}).</think>

🧠:<think>Alright, so I have this problem about modeling the number of books written about the Vietnam War over different decades using the function ( B(t) = A cdot e^{kt} + C ). The constants ( A ), ( k ), and ( C ) are given, except for ( A ) and ( C ), which I need to find. The total number of books in the 1970s is 500, and in the 2000s, it's 1200. The value of ( k ) is 0.05. First, I need to understand what the function ( B(t) ) represents. It's an exponential function plus a constant. So, the number of books grows exponentially over time with a rate ( k ), plus a constant number of books ( C ) each year. The total number of books in a decade would be the integral of ( B(t) ) over that decade. Let me write down the given information:- Total books in the 1970s (1970-1979): 500- Total books in the 2000s (2000-2009): 1200- ( k = 0.05 )I need to find ( A ) and ( C ). To do this, I'll set up two equations based on the integrals of ( B(t) ) over the two decades. Let me denote the starting year of the 1970s as ( t_1 = 1970 ) and the starting year of the 2000s as ( t_2 = 2000 ). The integral of ( B(t) ) from ( t = a ) to ( t = b ) is:[int_{a}^{b} B(t) , dt = int_{a}^{b} (A e^{kt} + C) , dt]I can split this integral into two parts:[int_{a}^{b} A e^{kt} , dt + int_{a}^{b} C , dt]Calculating each integral separately:1. The integral of ( A e^{kt} ) with respect to ( t ) is ( frac{A}{k} e^{kt} ).2. The integral of ( C ) with respect to ( t ) is ( C t ).So, putting it all together, the integral from ( a ) to ( b ) is:[left[ frac{A}{k} e^{kt} + C t right]_{a}^{b} = left( frac{A}{k} e^{kb} + C b right) - left( frac{A}{k} e^{ka} + C a right)]Simplify this expression:[frac{A}{k} (e^{kb} - e^{ka}) + C (b - a)]Since each decade is 10 years, ( b - a = 10 ). Therefore, the integral simplifies to:[frac{A}{k} (e^{k cdot 10} - 1) + 10C]Now, let's apply this to both decades.For the 1970s (1970-1979):Let ( a = 1970 ), ( b = 1980 ). But since we're dealing with the same function, the starting point ( t ) is relative. Wait, actually, in the integral, the starting year doesn't matter because we're subtracting the same term. So, actually, the integral from 1970 to 1980 is:[frac{A}{k} (e^{k cdot 10} - 1) + 10C = 500]Similarly, for the 2000s (2000-2009):The integral from 2000 to 2010 is:[frac{A}{k} (e^{k cdot 10} - 1) + 10C = 1200]Wait, hold on. That can't be right because both integrals would be equal, but they are given as 500 and 1200. That suggests that my approach is missing something. Wait, no, actually, the function ( B(t) ) is defined for each year ( t ), so the integral over 1970-1979 is from ( t = 1970 ) to ( t = 1980 ), and the integral over 2000-2009 is from ( t = 2000 ) to ( t = 2010 ). But in the integral expression, the difference ( e^{kb} - e^{ka} ) depends on the specific years. So, for the 1970s, ( a = 1970 ), ( b = 1980 ), so ( e^{k cdot 1980} - e^{k cdot 1970} ). Similarly, for the 2000s, ( a = 2000 ), ( b = 2010 ), so ( e^{k cdot 2010} - e^{k cdot 2000} ).Therefore, the two integrals are:1. For 1970s:[frac{A}{k} (e^{k cdot 1980} - e^{k cdot 1970}) + 10C = 500]2. For 2000s:[frac{A}{k} (e^{k cdot 2010} - e^{k cdot 2000}) + 10C = 1200]Now, I can write these as two equations:Equation 1:[frac{A}{0.05} (e^{0.05 cdot 1980} - e^{0.05 cdot 1970}) + 10C = 500]Equation 2:[frac{A}{0.05} (e^{0.05 cdot 2010} - e^{0.05 cdot 2000}) + 10C = 1200]Let me compute the exponents first.Compute ( e^{0.05 cdot 1970} ):0.05 * 1970 = 98.5So, ( e^{98.5} ). That's a huge number. Similarly, ( e^{0.05 cdot 1980} = e^{99} ), which is even larger.Wait, this doesn't make sense because the number of books can't be that large. Maybe I'm misunderstanding the function. Perhaps ( t ) is not the actual year, but the time elapsed since a certain point, like since 1970?Wait, the problem says "the number of books written in year ( t )", so ( t ) is the year. So, 1970, 1971, etc. So, the function is defined for each year ( t ), and the integral over a decade is the sum of books over those 10 years.But integrating from 1970 to 1980 would involve ( e^{0.05 cdot t} ) where ( t ) is in the thousands, which would make ( e^{kt} ) extremely large. That can't be right because the total books are only 500 and 1200.Wait, perhaps ( t ) is not the actual year, but the number of years since a certain starting point. Maybe the problem is using ( t ) as the number of years since 1900 or something. But the problem doesn't specify. Hmm.Wait, let me read the problem again:"The number of books, ( B(t) ), written in year ( t ) can be modeled by a function of the form ( B(t) = A cdot e^{kt} + C )."So, ( t ) is the year. So, for example, ( t = 1970 ), ( t = 1971 ), etc. So, the function is defined for each year ( t ), and the integral over a decade is the sum of books over those 10 years.But integrating from 1970 to 1980 would involve ( e^{0.05 cdot 1970} ), which is a gigantic number. That can't be right because the total books are only 500 and 1200. So, perhaps the function is not intended to be used with ( t ) as the actual year, but rather as the number of years since a certain point, like since 1970.Alternatively, maybe ( t ) is in decades since 1970. Hmm, but the problem says "year ( t )", so I think ( t ) is the actual year.Wait, perhaps the model is only valid for a certain range of years, and ( t ) is relative. Maybe the function is defined such that ( t = 0 ) corresponds to 1970. Let me check the problem statement again.It says: "the number of books written in year ( t )". So, ( t ) is the year. So, 1970, 1971, etc. So, the function is defined for each year ( t ), and the integral over a decade is the sum of books over those 10 years.But integrating ( B(t) ) from 1970 to 1980 would be:[int_{1970}^{1980} (A e^{0.05 t} + C) dt]Which is:[left[ frac{A}{0.05} e^{0.05 t} + C t right]_{1970}^{1980}]Which is:[frac{A}{0.05} (e^{0.05 cdot 1980} - e^{0.05 cdot 1970}) + C (1980 - 1970)]Simplify:[frac{A}{0.05} (e^{99} - e^{98.5}) + 10C = 500]Similarly, for the 2000s:[frac{A}{0.05} (e^{0.05 cdot 2010} - e^{0.05 cdot 2000}) + 10C = 1200]But ( e^{99} ) and ( e^{98.5} ) are astronomically large numbers, which would make ( A ) have to be a negative number with a very small magnitude to counteract that, but then adding ( C ) would have to result in a total of 500 and 1200. This seems impossible because ( e^{99} ) is way too large.Wait, maybe I'm misinterpreting the function. Perhaps ( t ) is not the actual year, but the number of years since 1970. So, for the 1970s, ( t ) would range from 0 to 9, and for the 2000s, ( t ) would range from 30 to 39 (since 2000 - 1970 = 30). That would make more sense because then the exponents wouldn't be so large. Let me test this interpretation.If ( t ) is the number of years since 1970, then:- For the 1970s (1970-1979), ( t ) goes from 0 to 9.- For the 2000s (2000-2009), ( t ) goes from 30 to 39.So, the integral for the 1970s would be:[int_{0}^{9} (A e^{0.05 t} + C) dt = 500]And for the 2000s:[int_{30}^{39} (A e^{0.05 t} + C) dt = 1200]This makes much more sense because the exponents would be manageable. Let me proceed with this interpretation.So, let's redefine ( t ) as the number of years since 1970. Therefore, for the 1970s, ( t ) is from 0 to 9, and for the 2000s, ( t ) is from 30 to 39.Now, let's compute the integrals.First, the integral of ( B(t) ) from ( a ) to ( b ):[int_{a}^{b} (A e^{0.05 t} + C) dt = left[ frac{A}{0.05} e^{0.05 t} + C t right]_{a}^{b}]Which simplifies to:[frac{A}{0.05} (e^{0.05 b} - e^{0.05 a}) + C (b - a)]Since each decade is 10 years, ( b - a = 10 ). So, for both integrals, the second term is ( 10C ).Now, let's compute for the 1970s (a=0, b=9):Wait, no, if ( t ) is the number of years since 1970, then the 1970s would be ( t = 0 ) to ( t = 9 ), which is 10 years. Similarly, the 2000s would be ( t = 30 ) to ( t = 39 ).So, the integral for 1970s:[frac{A}{0.05} (e^{0.05 cdot 9} - e^{0.05 cdot 0}) + 10C = 500]Simplify:[frac{A}{0.05} (e^{0.45} - 1) + 10C = 500]Similarly, for the 2000s:[frac{A}{0.05} (e^{0.05 cdot 39} - e^{0.05 cdot 30}) + 10C = 1200]Simplify:[frac{A}{0.05} (e^{1.95} - e^{1.5}) + 10C = 1200]Now, let's compute the numerical values of the exponentials.First, compute ( e^{0.45} ):( e^{0.45} approx e^{0.4} cdot e^{0.05} approx 1.4918 cdot 1.0513 approx 1.5683 )But more accurately, using a calculator:( e^{0.45} approx 1.5683 )Similarly, ( e^{1.5} approx 4.4817 )( e^{1.95} approx e^{2} cdot e^{-0.05} approx 7.3891 cdot 0.9512 approx 7.0255 )Wait, let me check:Actually, ( e^{1.95} ) is approximately:1.95 is close to 2, so ( e^{1.95} approx e^{2 - 0.05} = e^2 cdot e^{-0.05} approx 7.3891 cdot 0.9512 approx 7.0255 )Similarly, ( e^{1.5} approx 4.4817 )So, let's plug these values back into the equations.Equation 1 (1970s):[frac{A}{0.05} (1.5683 - 1) + 10C = 500]Simplify:[frac{A}{0.05} (0.5683) + 10C = 500]Compute ( frac{A}{0.05} times 0.5683 ):( frac{A}{0.05} = 20A ), so:( 20A times 0.5683 = 11.366A )Thus, Equation 1 becomes:[11.366A + 10C = 500 quad (1)]Equation 2 (2000s):[frac{A}{0.05} (7.0255 - 4.4817) + 10C = 1200]Simplify:[frac{A}{0.05} (2.5438) + 10C = 1200]Again, ( frac{A}{0.05} = 20A ), so:( 20A times 2.5438 = 50.876A )Thus, Equation 2 becomes:[50.876A + 10C = 1200 quad (2)]Now, we have a system of two equations:1. ( 11.366A + 10C = 500 )2. ( 50.876A + 10C = 1200 )Let me subtract Equation 1 from Equation 2 to eliminate ( C ):( (50.876A + 10C) - (11.366A + 10C) = 1200 - 500 )Simplify:( 50.876A - 11.366A = 700 )( 39.51A = 700 )Solve for ( A ):( A = frac{700}{39.51} approx 17.72 )Now, plug ( A ) back into Equation 1 to find ( C ):( 11.366 times 17.72 + 10C = 500 )First, compute ( 11.366 times 17.72 ):11.366 * 17.72 ≈ Let's compute 11 * 17.72 = 194.92, and 0.366 * 17.72 ≈ 6.48So total ≈ 194.92 + 6.48 ≈ 201.4Thus:201.4 + 10C = 500Subtract 201.4:10C = 500 - 201.4 = 298.6So, ( C = 298.6 / 10 = 29.86 )Therefore, ( A approx 17.72 ) and ( C approx 29.86 )Let me check these values in Equation 2 to ensure consistency.Compute ( 50.876A + 10C ):50.876 * 17.72 ≈ Let's compute 50 * 17.72 = 886, and 0.876 * 17.72 ≈ 15.52Total ≈ 886 + 15.52 ≈ 901.52Then, 10C = 10 * 29.86 = 298.6So total ≈ 901.52 + 298.6 ≈ 1200.12, which is approximately 1200. So, it checks out.Therefore, the constants are approximately:( A approx 17.72 )( C approx 29.86 )But let me compute more accurately.First, compute ( A = 700 / 39.51 ):39.51 * 17 = 671.6739.51 * 17.7 = 39.51 * 17 + 39.51 * 0.7 = 671.67 + 27.657 = 699.32739.51 * 17.72 = 699.327 + 39.51 * 0.02 = 699.327 + 0.7902 ≈ 700.1172So, 39.51 * 17.72 ≈ 700.1172, which is very close to 700. So, ( A ≈ 17.72 )Similarly, compute ( 11.366 * 17.72 ):11.366 * 17 = 193.22211.366 * 0.72 ≈ 8.183Total ≈ 193.222 + 8.183 ≈ 201.405So, 201.405 + 10C = 500Thus, 10C = 500 - 201.405 = 298.595So, ( C = 29.8595 approx 29.86 )Therefore, the constants are:( A ≈ 17.72 )( C ≈ 29.86 )Now, moving on to part 2: Calculate the rate of change of ( B(t) ) at the beginning of the 1970s and the beginning of the 2000s. How do these rates compare?The rate of change of ( B(t) ) is the derivative ( B'(t) ).Given ( B(t) = A e^{kt} + C ), the derivative is:( B'(t) = A k e^{kt} )So, at the beginning of the 1970s, which is ( t = 0 ) (since we redefined ( t ) as years since 1970), the rate of change is:( B'(0) = A k e^{0} = A k )Similarly, at the beginning of the 2000s, which is ( t = 30 ), the rate of change is:( B'(30) = A k e^{k cdot 30} )Compute both:First, ( B'(0) = A k = 17.72 * 0.05 = 0.886 )Second, ( B'(30) = 17.72 * 0.05 * e^{0.05 * 30} )Compute ( e^{1.5} approx 4.4817 )So, ( B'(30) = 0.886 * 4.4817 ≈ 4.0 )Therefore, the rate of change at the beginning of the 1970s is approximately 0.886 books per year, and at the beginning of the 2000s, it's approximately 4.0 books per year.Comparing these, the rate of change increased significantly from the start of the 1970s to the start of the 2000s. Specifically, it increased by a factor of approximately 4.0 / 0.886 ≈ 4.516 times.So, the rate of change more than quadrupled over the decades.Final Answer1. The constants are ( A approx boxed{17.72} ) and ( C approx boxed{29.86} ).2. The rate of change at the beginning of the 1970s is approximately ( boxed{0.89} ) books per year, and at the beginning of the 2000s, it is approximately ( boxed{4.0} ) books per year. The rate of change increased significantly over the decades.</think>