🧠:<think>First, I need to determine the actual original price of the laptop before any inflation. The store advertised the price as 2,000 after inflating it by 25%. To find the original price, I'll divide the advertised price by 1.25, which accounts for the 25% increase.Next, I'll calculate the final price at the other store that offers a 20% discount on the genuine original price. Using the actual original price found in the first step, I'll apply the 20% discount to determine the final amount the customer would pay.</think>

🧠:<think>Okay, so I have these two statistics problems to solve, both related to Ms. Rodriguez's study on teaching methods and bilingual students' math performance. Let me take them one at a time.Starting with the first problem: I need to find the probability that a randomly selected student from Group B scores higher than a randomly selected student from Group T. Both groups have normally distributed scores, with Group T having a mean of 75 and standard deviation of 10, and Group B having a mean of 80 and standard deviation of 8.Hmm, okay. So, I remember that when dealing with two independent normal distributions, the difference between two random variables from each distribution is also normally distributed. So, if I let X be a score from Group T and Y be a score from Group B, then Y - X should be normally distributed. The mean of Y - X would be μ_B - μ_T, which is 80 - 75 = 5. The variance of Y - X would be the sum of the variances of Y and X since they are independent. So, Var(Y - X) = σ_B² + σ_T² = 8² + 10² = 64 + 100 = 164. Therefore, the standard deviation is sqrt(164). Let me calculate that: sqrt(164) is approximately 12.806.So, Y - X ~ N(5, 12.806²). Now, I need the probability that Y > X, which is the same as the probability that Y - X > 0. So, I can standardize this to a Z-score. The Z-score is (0 - 5)/12.806 ≈ -0.3906. So, I need the probability that Z > -0.3906, which is the same as 1 - Φ(-0.3906), where Φ is the standard normal CDF.Looking up Φ(-0.39) in the Z-table, I find that Φ(-0.39) is approximately 0.3483. So, 1 - 0.3483 = 0.6517. Therefore, the probability that a randomly selected student from Group B scores higher than one from Group T is approximately 65.17%.Wait, let me double-check my calculations. The difference in means is 5, and the standard deviation of the difference is sqrt(64 + 100) = sqrt(164) ≈ 12.806. So, Z = (0 - 5)/12.806 ≈ -0.3906. Yes, that seems right. And the probability that Z is greater than -0.39 is indeed about 0.6517. So, I think that's correct.Moving on to the second problem: Ms. Rodriguez wants to perform a two-sample t-test to compare the means of the two groups. The sample sizes are n_T = 40 and n_B = 35. She wants to test whether the difference in means is statistically significant at the α = 0.05 level.First, I need to state the null and alternative hypotheses. The null hypothesis, H0, is that there is no difference in the means, so μ_B = μ_T. The alternative hypothesis, H1, is that there is a difference, so μ_B ≠ μ_T. Since she's interested in whether the bilingual support is more effective, maybe it's a one-tailed test? But the problem doesn't specify, so I think it's safer to assume a two-tailed test unless stated otherwise. Wait, actually, the problem says "whether the difference in means is statistically significant," which is typically two-tailed. So, H0: μ_B = μ_T vs. H1: μ_B ≠ μ_T.Next, I need to perform the two-sample t-test. Since the sample sizes are 40 and 35, which are both greater than 30, we can use the t-test assuming equal variances or unequal variances. But since the population variances are not given, we might have to use the Welch's t-test, which doesn't assume equal variances.Wait, actually, in the first part, we were given the population means and standard deviations, but in the second part, it's about performing a hypothesis test on the sample data. So, do we have the sample means and variances, or are we supposed to use the population parameters? Hmm, the problem says she collects data from two groups, so I think we can assume that the given μ and σ are the sample means and standard deviations. Wait, no, actually, the first part says the performance scores follow normal distributions with given μ and σ, so maybe those are the population parameters. But in the second part, she's performing a hypothesis test on the sample data, so perhaps we need to use the sample means and variances.Wait, the problem doesn't specify the sample means and variances, only the population parameters. Hmm, that's confusing. Let me re-read the problem.In the first part, it says: "Given the following data from her study: Group T: μ_T = 75, σ_T = 10; Group B: μ_B = 80, σ_B = 8." So, these are the parameters of the distributions. Then, in the second part, it says: "Assuming independent samples of sizes n_T = 40 and n_B = 35 from Group T and Group B respectively, perform a two-sample t-test..."Wait, so if the groups are normally distributed with the given μ and σ, and she takes samples of size 40 and 35, then the sample means would be normally distributed with means μ_T and μ_B and standard errors σ_T/sqrt(n_T) and σ_B/sqrt(n_B). So, perhaps the t-test is not necessary because we can use the Z-test since the population variances are known. But the problem says to perform a two-sample t-test, so maybe we have to proceed as if we don't know the population variances and estimate them from the sample.But wait, the problem doesn't give us the sample means and variances, only the population parameters. So, perhaps we can assume that the sample means are equal to the population means, and the sample variances are equal to the population variances. That might be a stretch, but maybe that's the case.Alternatively, perhaps the problem expects us to use the given μ and σ as the sample statistics. So, sample mean for T is 75, sample mean for B is 80, sample standard deviations are 10 and 8, with sample sizes 40 and 35. Then, we can perform the t-test.So, let's proceed with that assumption. So, we have:Group T: n_T = 40, mean_T = 75, std_T = 10Group B: n_B = 35, mean_B = 80, std_B = 8We need to perform a two-sample t-test. Since the population variances are unknown and we're using sample variances, we'll use the t-test. We can use Welch's t-test because the sample sizes are different and we don't know if the variances are equal.First, calculate the difference in sample means: mean_B - mean_T = 80 - 75 = 5.Next, calculate the standard error (SE) of the difference. The formula for SE when variances are unequal is sqrt[(s_T²/n_T) + (s_B²/n_B)]. Plugging in the numbers:s_T² = 10² = 100, s_B² = 8² = 64So, SE = sqrt[(100/40) + (64/35)] = sqrt[2.5 + 1.8286] = sqrt[4.3286] ≈ 2.0806Then, the t-statistic is (mean_B - mean_T)/SE = 5 / 2.0806 ≈ 2.403Now, we need to find the degrees of freedom (df) for Welch's t-test. The formula is:df = [(s_T²/n_T + s_B²/n_B)²] / [(s_T²/n_T)²/(n_T - 1) + (s_B²/n_B)²/(n_B - 1)]Plugging in the numbers:Numerator: (100/40 + 64/35)² = (2.5 + 1.8286)² ≈ (4.3286)² ≈ 18.73Denominator: (100/40)²/(40 - 1) + (64/35)²/(35 - 1) = (2.5)²/39 + (1.8286)²/34 ≈ (6.25)/39 + (3.344)/34 ≈ 0.1603 + 0.0983 ≈ 0.2586So, df ≈ 18.73 / 0.2586 ≈ 72.44We can round this down to 72 degrees of freedom.Now, with a t-statistic of approximately 2.403 and 72 degrees of freedom, we can find the p-value. Since it's a two-tailed test, the p-value is the probability that |t| > 2.403.Looking up t-tables or using a calculator, the critical t-value for α=0.05 and df=72 is approximately ±1.99. Since 2.403 > 1.99, the p-value is less than 0.05. To find the exact p-value, we can use a t-distribution calculator. The p-value for t=2.403 with df=72 is approximately 0.018 (two-tailed).So, the p-value is approximately 0.018, which is less than α=0.05. Therefore, we reject the null hypothesis and conclude that there is a statistically significant difference in the means of the two groups.Wait, let me double-check the calculations. The difference in means is 5, SE is sqrt(100/40 + 64/35) ≈ sqrt(2.5 + 1.8286) ≈ sqrt(4.3286) ≈ 2.0806. So, t ≈ 5 / 2.0806 ≈ 2.403. Degrees of freedom calculation: numerator is (100/40 + 64/35)^2 ≈ (4.3286)^2 ≈ 18.73. Denominator is (100/40)^2/(39) + (64/35)^2/(34) ≈ (6.25)/39 + (3.344)/34 ≈ 0.1603 + 0.0983 ≈ 0.2586. So, df ≈ 18.73 / 0.2586 ≈ 72.44, which we round to 72. Using a t-table or calculator, t=2.403 with df=72 gives a p-value around 0.018. Yes, that seems correct.So, to summarize:1. The probability that a randomly selected student from Group B scores higher than one from Group T is approximately 65.17%.2. The two-sample t-test yields a t-statistic of approximately 2.403, with a p-value of approximately 0.018, which is less than α=0.05. Therefore, we reject the null hypothesis and conclude that there is a statistically significant difference in the means.I think that's it. I should make sure I didn't mix up any formulas, especially for the t-test. Welch's t-test is appropriate here because the sample sizes and variances are different. The degrees of freedom calculation is a bit involved, but I think I did it correctly. The p-value calculation is approximate, but it's sufficient for the conclusion.</think>

🧠:<think>Alright, so I've got this problem about a team manager and their friend who's helping analyze the team's performance. It's split into two sub-problems. Let me tackle them one by one.Starting with Sub-problem 1. The performance is modeled by the function ( P(t) = A sin(Bt + C) + D ). I need to find the constants A, B, C, and D. The given information is that the performance peaks every 6 months, the maximum performance is 100 units, the minimum is 40 units, and at time t=0, the performance is 70 units.Okay, let's break this down. First, the general form of a sine function is ( A sin(Bt + C) + D ). Here, A is the amplitude, B affects the period, C is the phase shift, and D is the vertical shift.Given that the performance peaks every 6 months, that tells me about the period of the sine function. The period of a sine function is ( frac{2pi}{B} ). So if the period is 6 months, then:( frac{2pi}{B} = 6 )Solving for B:( B = frac{2pi}{6} = frac{pi}{3} )So B is ( frac{pi}{3} ).Next, the maximum and minimum performance values. The maximum performance is 100, and the minimum is 40. In a sine function, the maximum value is ( D + A ) and the minimum is ( D - A ). So:( D + A = 100 )( D - A = 40 )If I add these two equations together, I get:( 2D = 140 ) => ( D = 70 )Then, substituting back into one of the equations:( 70 + A = 100 ) => ( A = 30 )So A is 30 and D is 70.Now, we need to find C. We know that at t=0, the performance P(0) is 70 units. Let's plug t=0 into the function:( P(0) = A sin(B*0 + C) + D = 70 )We already know A=30, D=70, so:( 30 sin(C) + 70 = 70 )Subtract 70 from both sides:( 30 sin(C) = 0 )Divide both sides by 30:( sin(C) = 0 )So, C is an angle where sine is zero. The solutions are C = 0, π, 2π, etc. But since sine is periodic, we can choose the principal value, which is 0. So C=0.Wait, but let me think. If C=0, then the sine function starts at 0 when t=0. But the performance at t=0 is 70, which is equal to D. So that makes sense because the sine term is zero, so P(0)=D=70. So yes, that works.So, summarizing:A = 30B = π/3C = 0D = 70So that's Sub-problem 1 done.Moving on to Sub-problem 2. The friend wants to find the best time to implement a new strategy when the rate of change of performance is at its maximum. So we need to find the time t when the derivative of P(t) is at its maximum.First, let's find the derivative of P(t). Since P(t) = 30 sin( (π/3)t + 0 ) + 70, the derivative is:( frac{dP}{dt} = 30 * frac{pi}{3} cos( frac{pi}{3} t ) )Simplify:( frac{dP}{dt} = 10pi cos( frac{pi}{3} t ) )So the rate of change is ( 10pi cos( frac{pi}{3} t ) ). We need to find the time t when this rate of change is at its maximum.The maximum of the cosine function is 1, so the maximum rate of change is ( 10pi * 1 = 10pi ). But we need to find the time t when this occurs.So, set the derivative equal to its maximum value:( 10pi cos( frac{pi}{3} t ) = 10pi )Divide both sides by 10π:( cos( frac{pi}{3} t ) = 1 )When is cosine equal to 1? At multiples of 2π:( frac{pi}{3} t = 2pi n ), where n is an integer.Solving for t:( t = 2pi n / ( pi / 3 ) = 6n )So t = 6n, where n is an integer.But since we're talking about time in months, and the first peak is at t=0, the next maximum rate of change occurs at t=6, 12, 18, etc. months.But wait, let me think again. The derivative ( dP/dt ) is 10π cos( (π/3)t ). The maximum rate of change is 10π, which occurs when cos( (π/3)t ) = 1, which is at t=0, 6, 12, etc.But the problem says "the best time to implement a new strategy. The strategy will be most effective when the rate of change of the team's performance is at its maximum."So, the maximum rate of change occurs at t=0, 6, 12, etc. But t=0 is the starting point, so the next time would be t=6 months.But wait, let me double-check. The function P(t) is a sine function with period 6 months. So the derivative, which is a cosine function, will have the same period, 6 months. The maximum rate of change occurs at the points where the cosine is 1, which are at t=0, 6, 12, etc.But in the context of implementing a strategy, t=0 is the starting point, so the next best time would be at t=6 months. However, let me think about the behavior of the function.At t=0, the performance is 70, which is the average. The performance then increases to a peak at t=3 months (since the period is 6 months, the peak is at half the period). Wait, no, the period is 6 months, so the sine function completes a full cycle every 6 months. So the maximum occurs at t=3 months, because sin( (π/3)*3 ) = sin(π) = 0? Wait, no, wait.Wait, the function is P(t) = 30 sin( (π/3)t ) + 70. So the maximum occurs when sin( (π/3)t ) = 1, which is when (π/3)t = π/2 + 2π n.So solving for t:(π/3)t = π/2 + 2π nt = (π/2 + 2π n) * 3/π = (3/π)(π/2 + 2π n) = 3/2 + 6nSo the maximum performance occurs at t=3/2, 15/2, 27/2, etc. months, which is 1.5, 7.5, 13.5 months, etc.Similarly, the minimum occurs when sin( (π/3)t ) = -1, which is at (π/3)t = 3π/2 + 2π n, so t= (3π/2 + 2π n)*3/π = 9/2 + 6n, which is 4.5, 10.5, etc.So the performance peaks at t=1.5, 7.5, 13.5, etc., and troughs at t=4.5, 10.5, etc.But the derivative, which is the rate of change, is 10π cos( (π/3)t ). The maximum rate of change is 10π, which occurs when cos( (π/3)t )=1, i.e., at t=0, 6, 12, etc.So at t=0, the rate of change is maximum, but the performance is at 70, which is the average. Then, the performance starts increasing, reaching a peak at t=1.5 months, then decreasing to a trough at t=4.5 months, then increasing again to a peak at t=7.5 months, and so on.But the maximum rate of change occurs at t=0, 6, 12, etc. So the times when the rate of change is maximum are at the start of each period, i.e., t=0, 6, 12, etc.But in the context of implementing a strategy, t=0 is the starting point, so the next time would be at t=6 months. However, let me think about whether the maximum rate of change occurs at the start of the period or somewhere else.Wait, the derivative is 10π cos( (π/3)t ). The maximum of the cosine function is 1, which occurs at t=0, 6, 12, etc. So yes, those are the points where the rate of change is maximum.But wait, the rate of change is maximum at t=0, which is the starting point. So if the manager wants to implement a strategy when the rate of change is maximum, it would be at t=0, but that's the initial time. So the next time would be at t=6 months.But let me think about the behavior. At t=0, the performance is 70, and it's about to increase. The rate of change is maximum positive at t=0, meaning the performance is increasing the fastest at that point. So implementing a strategy at t=0 would be when the team is about to start improving the most.But perhaps the friend is looking for the next time after t=0 when the rate of change is maximum. So t=6 months.Alternatively, maybe the maximum rate of change in terms of the magnitude, but since the derivative is 10π cos( (π/3)t ), the maximum positive rate is 10π, and the maximum negative rate is -10π. But the problem says "the rate of change is at its maximum", which I think refers to the maximum positive rate, i.e., when the function is increasing the fastest.So, the times when the rate of change is maximum (positive) are at t=0, 6, 12, etc.But since t=0 is the starting point, the next time would be at t=6 months.Alternatively, maybe the friend is considering the maximum absolute rate of change, but the problem specifies "the rate of change is at its maximum", which is likely the maximum positive rate.So, the answer is t=6n months, where n is an integer. But since we're looking for the best time, it's likely the first time after t=0, which is t=6 months.Wait, but let me think again. The function P(t) is a sine wave with period 6 months. The derivative is a cosine wave with the same period. The maximum rate of change occurs at the points where the cosine is 1, which are at t=0, 6, 12, etc. So yes, t=6 months is the next time after t=0 when the rate of change is maximum.So, the time t when the rate of change is at its maximum is t=6 months.But wait, let me double-check the derivative. The derivative is 10π cos( (π/3)t ). The maximum value is 10π, which occurs when cos( (π/3)t )=1. So solving for t:(π/3)t = 2π n => t=6n, where n is integer.So yes, t=0, 6, 12, etc.Therefore, the best time to implement the strategy is at t=6 months, 12 months, etc. But since the problem doesn't specify a range, the general answer is t=6n months, but if we're to give a specific time, it's t=6 months.Wait, but let me think about the context. The manager is looking to implement a strategy when the rate of change is maximum. So, if the strategy is implemented at t=6 months, that's when the team's performance is about to start increasing again after a trough. Wait, no, let me see.Wait, the performance peaks at t=1.5 months, then decreases to a trough at t=4.5 months, then increases again to a peak at t=7.5 months, and so on. So at t=6 months, the performance is on its way up from the trough at t=4.5 months. So the rate of change is maximum positive at t=6 months, meaning the performance is increasing the fastest at that point.But wait, let me plot the function mentally. At t=0, P=70, derivative=10π, so increasing. At t=1.5, P=100, derivative=0. At t=3, P=70, derivative=-10π. At t=4.5, P=40, derivative=0. At t=6, P=70, derivative=10π.So yes, at t=6 months, the performance is back to 70, and the rate of change is maximum positive again. So that's when the strategy should be implemented.Therefore, the answer is t=6 months.</think>

🧠:<think>Okay, so I need to solve these two sub-problems about the demographic changes in the African American population in a region from 1800 to 1860. Let me take it step by step.First Sub-problem: Determine the annual growth rate r from 1800 to 1820.Alright, the population in 1800 was 20,000, and it increased to 30,000 by 1820. The time period here is 20 years. The growth is exponential, so I can use the formula for exponential growth:[ P(t) = P_0 times (1 + r)^t ]Where:- ( P(t) ) is the population after t years,- ( P_0 ) is the initial population,- ( r ) is the annual growth rate (in decimal),- ( t ) is the time in years.Given:- ( P_0 = 20,000 ),- ( P(20) = 30,000 ),- ( t = 20 ).Plugging in the values:[ 30,000 = 20,000 times (1 + r)^{20} ]Let me simplify this equation. First, divide both sides by 20,000:[ frac{30,000}{20,000} = (1 + r)^{20} ][ 1.5 = (1 + r)^{20} ]Now, I need to solve for ( r ). To do this, I can take the natural logarithm of both sides:[ ln(1.5) = lnleft( (1 + r)^{20} right) ][ ln(1.5) = 20 times ln(1 + r) ]Now, divide both sides by 20:[ frac{ln(1.5)}{20} = ln(1 + r) ]Calculate the left side:First, compute ( ln(1.5) ). I know that ( ln(1.5) ) is approximately 0.4055.So,[ frac{0.4055}{20} approx 0.020275 ]Therefore,[ ln(1 + r) approx 0.020275 ]Now, exponentiate both sides to solve for ( 1 + r ):[ e^{0.020275} approx 1 + r ]Compute ( e^{0.020275} ). I remember that ( e^{0.02} ) is approximately 1.0202, so 0.020275 is slightly more. Let me compute it more accurately.Using the Taylor series expansion for ( e^x ) around 0:[ e^x approx 1 + x + frac{x^2}{2} + frac{x^3}{6} ]Let me plug in ( x = 0.020275 ):First term: 1Second term: 0.020275Third term: ( (0.020275)^2 / 2 approx 0.000411 )Fourth term: ( (0.020275)^3 / 6 approx 0.0000138 )Adding them up:1 + 0.020275 = 1.0202751.020275 + 0.000411 = 1.0206861.020686 + 0.0000138 ≈ 1.0207So, ( e^{0.020275} approx 1.0207 )Therefore,[ 1 + r approx 1.0207 ][ r approx 0.0207 ]Convert this to a percentage:[ r approx 2.07% ]So, the annual growth rate is approximately 2.07%.Wait, let me check if this makes sense. If we have a 2.07% growth rate over 20 years, starting from 20,000, does it reach 30,000?Let me compute ( 20,000 times (1.0207)^{20} ).First, compute ( (1.0207)^{20} ).I know that ( (1.02)^{20} ) is approximately 1.4859, and 1.0207 is slightly higher, so the result should be a bit higher than 1.4859.Alternatively, using logarithms:Compute ( ln(1.0207) approx 0.0204 )Multiply by 20: 0.0204 * 20 = 0.408Exponentiate: ( e^{0.408} approx 1.503 )So, ( 20,000 * 1.503 approx 30,060 ), which is close to 30,000. So, the growth rate is approximately 2.07%.Alternatively, maybe I can use a calculator for more precision, but since I don't have one, 2.07% seems reasonable.So, I think the annual growth rate r is approximately 2.07%.Second Sub-problem: Find constants A and B in the logistic growth function, then estimate the population in 1860 with a 2% annual growth rate.Given:- The logistic growth function is ( P(t) = frac{K}{1 + A e^{-Bt}} )- Carrying capacity ( K = 150,000 )- Population in 1850 was 100,000First, we need to find A and B.But wait, the logistic growth function is given from 1820 to 1850. So, t is measured in years since 1820.So, in 1850, t = 30 years.Given that in 1850, the population was 100,000.So, plug into the logistic function:[ 100,000 = frac{150,000}{1 + A e^{-B times 30}} ]Simplify:Multiply both sides by denominator:[ 100,000 (1 + A e^{-30B}) = 150,000 ]Divide both sides by 100,000:[ 1 + A e^{-30B} = 1.5 ]Subtract 1:[ A e^{-30B} = 0.5 ]So, equation (1): ( A e^{-30B} = 0.5 )But we need another equation to solve for A and B. Since the logistic function is used from 1820 to 1850, we know the population in 1820 as well.Wait, in the first sub-problem, the population in 1820 was 30,000.So, in 1820, t = 0, so:[ P(0) = frac{150,000}{1 + A e^{0}} = frac{150,000}{1 + A} = 30,000 ]So,[ frac{150,000}{1 + A} = 30,000 ]Multiply both sides by (1 + A):[ 150,000 = 30,000 (1 + A) ]Divide both sides by 30,000:[ 5 = 1 + A ][ A = 4 ]So, A is 4.Now, plug A = 4 into equation (1):[ 4 e^{-30B} = 0.5 ]Divide both sides by 4:[ e^{-30B} = 0.125 ]Take natural logarithm:[ -30B = ln(0.125) ]Compute ( ln(0.125) ). Since 0.125 is 1/8, and ( ln(1/8) = -ln(8) approx -2.079 )So,[ -30B = -2.079 ][ 30B = 2.079 ][ B = frac{2.079}{30} approx 0.0693 ]So, B is approximately 0.0693 per year.Therefore, the logistic growth function is:[ P(t) = frac{150,000}{1 + 4 e^{-0.0693 t}} ]Now, we need to estimate the population in 1860, which is 10 years after 1850. So, from 1850 to 1860, the population is growing at an annual rate of 2%.Wait, but the logistic function is only up to 1850. So, from 1850 to 1860, we have a different growth model, which is exponential with a 2% annual growth rate.So, first, let me confirm: in 1850, the population is 100,000. From 1850 to 1860, it grows at 2% per year. So, we can model this as:[ P(1860) = 100,000 times (1 + 0.02)^{10} ]Compute this.First, compute ( (1.02)^{10} ). I remember that ( (1.02)^{10} ) is approximately 1.21899.So,[ P(1860) approx 100,000 times 1.21899 = 121,899 ]So, approximately 121,900.But wait, let me verify if the logistic function at t=30 (1850) gives 100,000, which we already used, and then from 1850 onwards, it's a different model.Alternatively, is there a chance that the logistic function is extended beyond 1850? The problem says that from 1820 to 1850, it's logistic, and from 1850 to 1860, it's exponential with rate g=2%.So, yes, we just need to compute the exponential growth from 1850 to 1860.So, 100,000 growing at 2% annually for 10 years is approximately 121,900.But let me compute it more accurately.Compute ( (1.02)^{10} ):We can use the formula:[ (1 + r)^n = e^{n ln(1 + r)} ]So,[ ln(1.02) approx 0.0198026 ][ 10 times 0.0198026 approx 0.198026 ][ e^{0.198026} approx 1.21899 ]So, yes, 1.21899, so 100,000 * 1.21899 ≈ 121,899.Rounding to the nearest whole number, 121,899, which is approximately 121,900.Alternatively, if we want to be precise, it's 121,899, but we can write it as 121,900.So, the estimated population in 1860 is approximately 121,900.But wait, let me think again. The logistic function was used up to 1850, so in 1850, the population is 100,000. Then, from 1850 to 1860, it's growing at 2% per year. So, yes, that's correct.Alternatively, is there a chance that the logistic function is used beyond 1850? The problem says from 1820 to 1850, so I think not. So, we just model the 1850-1860 period as exponential growth.Therefore, the population in 1860 is approximately 121,900.Wait, but let me check if the logistic function at t=40 (which would be 1860) gives a different result. Maybe the problem expects us to use the logistic function beyond 1850? Let me check the problem statement again."From 1820 to 1850, the population increase can be modeled by the logistic growth function... From 1850 to 1860, the population growth rate changes due to new laws and social conditions, and the population increases by an annual rate of g%."So, it's two separate models: logistic from 1820 to 1850, and exponential from 1850 to 1860. So, we don't need to use the logistic function beyond 1850. So, my initial approach is correct.Therefore, the population in 1860 is approximately 121,900.But let me also compute it step by step for each year to ensure accuracy, although it's time-consuming.Alternatively, use the formula:[ P = P_0 times (1 + r)^t ][ P = 100,000 times (1.02)^{10} ]We can compute this as:Year 1850: 100,0001851: 100,000 * 1.02 = 102,0001852: 102,000 * 1.02 = 104,0401853: 104,040 * 1.02 = 106,120.81854: 106,120.8 * 1.02 ≈ 108,243.221855: 108,243.22 * 1.02 ≈ 110,408.081856: 110,408.08 * 1.02 ≈ 112,616.241857: 112,616.24 * 1.02 ≈ 114,868.561858: 114,868.56 * 1.02 ≈ 117,165.931859: 117,165.93 * 1.02 ≈ 119,409.251860: 119,409.25 * 1.02 ≈ 121,797.43So, approximately 121,797, which is about 121,800. Earlier, using the exponentiation, I got 121,899. The slight difference is due to rounding errors in each step. So, both methods give approximately 121,800 to 121,900.Therefore, the estimated population in 1860 is approximately 121,800.But since the question says to estimate, and the exact value is around 121,899, we can write it as approximately 121,900.Alternatively, if we use more precise calculations, maybe 121,899, but for the purposes of this problem, 121,900 is acceptable.So, summarizing:1. The annual growth rate r from 1800 to 1820 is approximately 2.07%.2. The constants A and B in the logistic function are 4 and approximately 0.0693, respectively. The estimated population in 1860 is approximately 121,900.Final Answer1. The annual growth rate ( r ) is boxed{2.07%}.2. The estimated population in 1860 is boxed{121900}.</think>