answer：First, I need to determine the number of people who migrated each month.In the first month, 120 people migrated.In the second month, the number of migrants increased by 50% compared to the first month. To calculate this, I'll take 120 and add 50% of 120, which is 60. This gives a total of 180 people in the second month.In the third month, twice as many people migrated as in the second month. So, I'll multiply the second month's total of 180 by 2, resulting in 360 people.Finally, to find the total number of people who migrated over the three months, I'll add the numbers from each month: 120 + 180 + 360, which equals 660.

answer：First, I need to determine how many songs Jamie will add to her playlist from each album. She has four albums with the following number of songs: 14, 12, 16, and 10.Jamie decides to select 3 songs from each album. To find the total number of songs in the playlist, I will multiply the number of albums by the number of songs selected from each album.So, the calculation is 4 albums multiplied by 3 songs per album, which equals 12 songs in total.Therefore, Jamie's playlist will contain 12 songs.

answer：First, I need to determine the total investment the angel investor is making in both projects. The first project receives 12,000. The second project is 50% more than the first, so I calculate 50% of 12,000, which is 6,000, and add that to the original amount, resulting in 18,000 for the second project. Adding both investments together gives a total of 30,000.Next, I calculate the total carbon emissions reduction. Each project reduces carbon emissions by 150 tons annually. Since there are two projects, I multiply 150 tons by 2, resulting in a total reduction of 300 tons of carbon emissions per year.

answer：First, I need to identify the number of tickets Mr. Thompson needs to purchase for each category: adult, youth, and senior.Mr. Thompson and his wife are adults, so that's 2 adult tickets.Their two children, aged 8 and 12, qualify for youth tickets, making it 2 youth tickets.His grandparents, aged 70 and 82, are seniors, so that's 2 senior tickets.Next, I'll calculate the cost for each category by multiplying the number of tickets by their respective prices.For adult tickets: 2 tickets × 25 = 50.For youth tickets: 2 tickets × 18 = 36.For senior tickets: 2 tickets × 20 = 40.Finally, I'll add up the costs of all the tickets to find the total amount Mr. Thompson needs to pay.50 (adults) + 36 (youth) + 40 (seniors) = 126.

answer：First, I need to determine the initial number of chairs set up for the school assembly. There are 9 rows, and each row has 15 chairs. So, I'll multiply 9 by 15 to find the total number of chairs initially.Next, some students added 8 more chairs at the sides. I'll add these 8 chairs to the initial total to find the overall number of chairs available for the assembly.Finally, by performing these calculations, I'll arrive at the total number of chairs.

answer：Okay, so I have this problem about two villages, A and B, with different population growth models. Village A has a linear growth model, and Village B has a logistic growth model. The first part asks me to find the time ( t ) when both villages will have the same population. The second part is about how an increase in carrying capacity affects that time.Starting with the first part. The population of Village A is given by ( P_A(t) = 150t + 1000 ). That's straightforward, linear growth. So every year, the population increases by 150 people.Village B's population is modeled by a logistic function: ( P_B(t) = frac{5000}{1 + 9e^{-0.3t}} ). I remember that logistic growth models have an S-shape, starting with exponential growth and then leveling off as it approaches the carrying capacity. The carrying capacity here is 5000 because as ( t ) approaches infinity, the denominator approaches 1, so ( P_B(t) ) approaches 5000.So, I need to find ( t ) when ( P_A(t) = P_B(t) ). That means I need to solve the equation:( 150t + 1000 = frac{5000}{1 + 9e^{-0.3t}} )Hmm, this looks like a transcendental equation, which probably can't be solved algebraically. I might need to use numerical methods or graphing to find the solution. But let me see if I can manipulate it a bit first.Let me denote ( y = e^{-0.3t} ). Then, the equation becomes:( 150t + 1000 = frac{5000}{1 + 9y} )Let me solve for ( y ):Multiply both sides by ( 1 + 9y ):( (150t + 1000)(1 + 9y) = 5000 )Expand the left side:( 150t + 1000 + 1350ty + 9000y = 5000 )Bring all terms to one side:( 150t + 1000 + 1350ty + 9000y - 5000 = 0 )Simplify:( 150t - 4000 + 1350ty + 9000y = 0 )Factor terms with ( y ):( 150t - 4000 + y(1350t + 9000) = 0 )Now, solve for ( y ):( y(1350t + 9000) = -150t + 4000 )So,( y = frac{-150t + 4000}{1350t + 9000} )But remember that ( y = e^{-0.3t} ), which is always positive. Therefore, the right-hand side must also be positive. So,( frac{-150t + 4000}{1350t + 9000} > 0 )Which implies that both numerator and denominator have the same sign.Denominator: ( 1350t + 9000 ). Since ( t ) is time, it's non-negative, so denominator is always positive.Therefore, numerator must also be positive:( -150t + 4000 > 0 )( -150t > -4000 )Multiply both sides by -1 (inequality sign flips):( 150t < 4000 )( t < frac{4000}{150} )( t < frac{80}{3} approx 26.6667 ) years.So, the solution must be before approximately 26.67 years.Now, going back to ( y = e^{-0.3t} ):( e^{-0.3t} = frac{-150t + 4000}{1350t + 9000} )Take natural logarithm on both sides:( -0.3t = lnleft( frac{-150t + 4000}{1350t + 9000} right) )So,( t = -frac{1}{0.3} lnleft( frac{-150t + 4000}{1350t + 9000} right) )This still looks complicated. Maybe I can use numerical methods like Newton-Raphson to approximate the solution.Alternatively, let me plug in some values of ( t ) to see where ( P_A(t) ) and ( P_B(t) ) intersect.First, let's compute ( P_A(t) ) and ( P_B(t) ) at different ( t ):At ( t = 0 ):( P_A = 1000 )( P_B = 5000 / (1 + 9e^0) = 5000 / 10 = 500 )So, ( P_A > P_B )At ( t = 10 ):( P_A = 150*10 + 1000 = 2500 )( P_B = 5000 / (1 + 9e^{-3}) )Compute ( e^{-3} approx 0.0498 )So denominator: 1 + 9*0.0498 ≈ 1 + 0.448 ≈ 1.448Thus, ( P_B ≈ 5000 / 1.448 ≈ 3456 )So, ( P_A = 2500 < P_B ≈ 3456 )So between t=0 and t=10, ( P_A ) goes from 1000 to 2500, while ( P_B ) goes from 500 to ~3456. So, they cross somewhere between t=0 and t=10.Wait, but at t=0, ( P_A = 1000 ), ( P_B = 500 ). So, ( P_A > P_B ). At t=10, ( P_A = 2500 < P_B ≈ 3456 ). So, somewhere between t=0 and t=10, ( P_A ) overtakes ( P_B ). Wait, no, actually, ( P_A ) is increasing faster than ( P_B ) initially. Wait, but at t=0, ( P_A = 1000 ), ( P_B = 500 ). So, ( P_A ) is higher. At t=10, ( P_A ) is 2500, ( P_B ) is ~3456. So, ( P_B ) becomes larger than ( P_A ) at some point between t=0 and t=10.Wait, but the question is when they are equal. So, maybe they cross once between t=0 and t=10, and then since ( P_A ) is linear and ( P_B ) is logistic, which eventually plateaus, ( P_A ) will surpass ( P_B ) again? Wait, no, because ( P_A ) is linear, it will keep increasing, while ( P_B ) approaches 5000. So, after some point, ( P_A ) will surpass 5000, but ( P_B ) can't go beyond 5000. So, maybe they only cross once?Wait, but when t increases, ( P_A(t) = 150t + 1000 ). So, when does ( P_A(t) = 5000 )?Solve 150t + 1000 = 5000150t = 4000t = 4000 / 150 ≈ 26.6667 years.So, at t ≈26.6667, ( P_A(t) = 5000 ). But ( P_B(t) ) approaches 5000 as t approaches infinity. So, before t=26.6667, ( P_A(t) ) is less than 5000, and ( P_B(t) ) is increasing towards 5000. So, maybe they cross only once?Wait, at t=0: ( P_A = 1000 > 500 = P_B )At t=10: ( P_A = 2500 < 3456 ≈ P_B )So, they cross somewhere between t=0 and t=10.Then, as t increases beyond that crossing point, ( P_A(t) ) continues to increase linearly, while ( P_B(t) ) continues to increase but at a decreasing rate, approaching 5000.So, ( P_A(t) ) will eventually surpass ( P_B(t) ) again when ( P_A(t) ) reaches 5000 at t≈26.6667. But since ( P_B(t) ) is approaching 5000, maybe they meet again at t≈26.6667? Let me check.At t=26.6667, ( P_A(t) = 5000 ). What is ( P_B(t) )?Compute ( P_B(26.6667) = 5000 / (1 + 9e^{-0.3*26.6667}) )Compute exponent: 0.3*26.6667 ≈ 8So, e^{-8} ≈ 0.000335Thus, denominator ≈ 1 + 9*0.000335 ≈ 1 + 0.003015 ≈ 1.003015So, ( P_B ≈ 5000 / 1.003015 ≈ 4985.06 )So, at t≈26.6667, ( P_A = 5000 ), ( P_B ≈4985 ). So, ( P_A > P_B ). Therefore, after the crossing point between t=0 and t=10, ( P_A(t) ) remains below ( P_B(t) ) until t≈26.6667, where ( P_A(t) ) surpasses ( P_B(t) ). Wait, no, because ( P_A(t) ) is increasing linearly, while ( P_B(t) ) is increasing but approaching 5000. So, after the initial crossing, ( P_A(t) ) is less than ( P_B(t) ) until t≈26.6667, where ( P_A(t) ) becomes 5000, which is slightly higher than ( P_B(t) ) at that point.Wait, but does that mean they cross again? Or is it just that ( P_A(t) ) overtakes ( P_B(t) ) at t≈26.6667?Wait, let's compute ( P_B(t) ) at t=20:( P_B(20) = 5000 / (1 + 9e^{-6}) )e^{-6} ≈ 0.002479Denominator ≈1 + 9*0.002479 ≈1 + 0.0223 ≈1.0223So, ( P_B(20) ≈5000 /1.0223 ≈4887 )( P_A(20) =150*20 +1000=3000 +1000=4000 )So, ( P_A(20)=4000 < P_B(20)=4887 )At t=25:( P_A=150*25 +1000=3750 +1000=4750 )( P_B(25)=5000/(1 +9e^{-7.5}) )e^{-7.5}≈0.000553Denominator≈1 +9*0.000553≈1 +0.004977≈1.004977( P_B≈5000 /1.004977≈4975.12 )So, ( P_A=4750 < P_B≈4975 )At t=26.6667:( P_A=5000 )( P_B≈4985 )So, ( P_A=5000 > P_B≈4985 )So, between t=25 and t=26.6667, ( P_A(t) ) goes from 4750 to 5000, while ( P_B(t) ) goes from ~4975 to ~4985. So, ( P_A(t) ) crosses ( P_B(t) ) again somewhere between t=25 and t=26.6667.Wait, so does that mean there are two intersection points? One between t=0 and t=10, and another between t=25 and t=26.6667?But that contradicts my earlier thought that ( P_A(t) ) is linear and ( P_B(t) ) is logistic, so they might only cross once. Hmm.Wait, let me plot the functions mentally.At t=0: ( P_A=1000 ), ( P_B=500 ). So, ( P_A > P_B ).At t=10: ( P_A=2500 ), ( P_B≈3456 ). So, ( P_A < P_B ).So, they cross once between t=0 and t=10.Then, as t increases, ( P_A(t) ) continues to increase linearly, while ( P_B(t) ) approaches 5000. So, ( P_A(t) ) will eventually surpass 5000 at t≈26.6667, while ( P_B(t) ) is approaching 5000 from below.So, at t≈26.6667, ( P_A(t)=5000 ), ( P_B(t)≈4985 ). So, ( P_A(t) ) is higher.But between t=10 and t=26.6667, ( P_A(t) ) is increasing from 2500 to 5000, while ( P_B(t) ) is increasing from ~3456 to ~4985.So, is there another crossing point? Let's see.At t=10: ( P_A=2500 < P_B≈3456 )At t=20: ( P_A=4000 < P_B≈4887 )At t=25: ( P_A=4750 < P_B≈4975 )At t=26.6667: ( P_A=5000 > P_B≈4985 )So, between t=25 and t=26.6667, ( P_A(t) ) goes from 4750 to 5000, while ( P_B(t) ) goes from ~4975 to ~4985. So, ( P_A(t) ) crosses ( P_B(t) ) again somewhere between t=25 and t=26.6667.Therefore, there are two times when ( P_A(t) = P_B(t) ): one between t=0 and t=10, and another between t=25 and t=26.6667.But the question says "determine the time ( t ) when both villages will have the same population." It doesn't specify which one, but usually, such problems expect the first intersection. But let me check the problem statement again."1. Determine the time ( t ) (in years) when both villages will have the same population."It doesn't specify the first time or any particular time, but given the context, it's likely asking for the first time they meet. Because after that, Village B's population overtakes Village A's, and then Village A's population overtakes again as it continues to grow linearly beyond the carrying capacity of Village B.But let me confirm by solving the equation numerically.So, the equation is:( 150t + 1000 = frac{5000}{1 + 9e^{-0.3t}} )Let me rearrange it:( 1 + 9e^{-0.3t} = frac{5000}{150t + 1000} )So,( 9e^{-0.3t} = frac{5000}{150t + 1000} - 1 )( e^{-0.3t} = frac{1}{9} left( frac{5000}{150t + 1000} - 1 right) )Let me define a function ( f(t) = e^{-0.3t} - frac{1}{9} left( frac{5000}{150t + 1000} - 1 right) ). We need to find ( t ) such that ( f(t) = 0 ).Let me compute ( f(t) ) at different points.At t=0:( f(0) = e^{0} - (1/9)(5000/1000 -1 ) = 1 - (1/9)(5 -1 ) = 1 - (4/9) ≈ 0.5556 >0At t=5:Compute ( e^{-1.5} ≈0.2231 )Compute ( 5000/(150*5 +1000)=5000/(750 +1000)=5000/1750≈2.8571So, ( (2.8571 -1)/9 ≈1.8571/9≈0.2063Thus, ( f(5)=0.2231 -0.2063≈0.0168 >0At t=6:( e^{-1.8}≈0.1653( 5000/(900 +1000)=5000/1900≈2.6316( (2.6316 -1)/9≈1.6316/9≈0.1813( f(6)=0.1653 -0.1813≈-0.016 <0So, between t=5 and t=6, f(t) crosses zero.Using linear approximation:At t=5, f=0.0168At t=6, f=-0.016So, the root is approximately at t=5 + (0 -0.0168)/( -0.016 -0.0168 )*(1)Wait, let me use linear interpolation.The change in t is 1, from 5 to 6.The change in f is -0.016 -0.0168 = -0.0328We need to find t where f(t)=0.From t=5 to t=6, f decreases by 0.0328 over 1 year.At t=5, f=0.0168So, the fraction needed to reach 0 is 0.0168 / 0.0328 ≈0.512So, t≈5 + 0.512 ≈5.512 years.Let me compute f(5.5):t=5.5Compute ( e^{-0.3*5.5}=e^{-1.65}≈0.1922Compute ( 5000/(150*5.5 +1000)=5000/(825 +1000)=5000/1825≈2.7397Compute ( (2.7397 -1)/9≈1.7397/9≈0.1933Thus, f(5.5)=0.1922 -0.1933≈-0.0011Almost zero. So, t≈5.5 is close.Compute f(5.4):t=5.4( e^{-0.3*5.4}=e^{-1.62}≈0.2019( 5000/(150*5.4 +1000)=5000/(810 +1000)=5000/1810≈2.7624( (2.7624 -1)/9≈1.7624/9≈0.1958f(5.4)=0.2019 -0.1958≈0.0061So, f(5.4)=0.0061, f(5.5)=-0.0011So, the root is between 5.4 and 5.5.Using linear approximation:Between t=5.4 and t=5.5, f(t) goes from 0.0061 to -0.0011, a change of -0.0072 over 0.1 years.We need to find t where f(t)=0.The fraction is 0.0061 / 0.0072 ≈0.847So, t≈5.4 + (0 -0.0061)/(-0.0072)*0.1 ≈5.4 + (0.0061/0.0072)*0.1≈5.4 +0.0847≈5.4847So, approximately t≈5.485 years.Let me check t=5.485:Compute ( e^{-0.3*5.485}=e^{-1.6455}≈0.1947Compute ( 5000/(150*5.485 +1000)=5000/(822.75 +1000)=5000/1822.75≈2.743Compute ( (2.743 -1)/9≈1.743/9≈0.1937Thus, f(5.485)=0.1947 -0.1937≈0.001Almost zero. So, t≈5.485 is a good approximation.But let's do one more iteration.Compute f(5.485)=0.001Compute f(5.49):t=5.49( e^{-0.3*5.49}=e^{-1.647}≈0.1942( 5000/(150*5.49 +1000)=5000/(823.5 +1000)=5000/1823.5≈2.742( (2.742 -1)/9≈1.742/9≈0.1936f(5.49)=0.1942 -0.1936≈0.0006Still positive.t=5.495:( e^{-0.3*5.495}=e^{-1.6485}≈0.1940( 5000/(150*5.495 +1000)=5000/(824.25 +1000)=5000/1824.25≈2.741( (2.741 -1)/9≈1.741/9≈0.1934f(5.495)=0.1940 -0.1934≈0.0006Wait, similar to t=5.49.Wait, maybe my approximations are getting too close. Alternatively, perhaps using Newton-Raphson would be better.Let me denote ( f(t) = e^{-0.3t} - frac{1}{9} left( frac{5000}{150t + 1000} - 1 right) )We need to find t where f(t)=0.Compute f(t) and f’(t):f(t) = e^{-0.3t} - (1/9)(5000/(150t +1000) -1 )f’(t) = -0.3e^{-0.3t} - (1/9)( -5000*150/(150t +1000)^2 )Simplify f’(t):f’(t) = -0.3e^{-0.3t} + (1/9)(75000)/(150t +1000)^2Let me compute f(5.485)=0.001f’(5.485):Compute -0.3e^{-1.6455}= -0.3*0.1947≈-0.0584Compute (1/9)(75000)/(150*5.485 +1000)^2First, denominator: 150*5.485≈822.75, so 822.75 +1000=1822.75So, denominator squared≈(1822.75)^2≈3,322,500Thus, (1/9)(75000)/3,322,500≈(8333.33)/3,322,500≈0.00251So, f’(5.485)= -0.0584 +0.00251≈-0.0559Using Newton-Raphson:t1 = t0 - f(t0)/f’(t0)t0=5.485f(t0)=0.001f’(t0)= -0.0559t1=5.485 - (0.001)/(-0.0559)≈5.485 +0.0179≈5.5029Compute f(5.5029):t=5.5029Compute ( e^{-0.3*5.5029}=e^{-1.65087}≈0.1922Compute ( 5000/(150*5.5029 +1000)=5000/(825.435 +1000)=5000/1825.435≈2.739Compute ( (2.739 -1)/9≈1.739/9≈0.1932Thus, f(5.5029)=0.1922 -0.1932≈-0.001So, f(t1)= -0.001Compute f’(t1):f’(5.5029)= -0.3e^{-1.65087} + (1/9)(75000)/(150*5.5029 +1000)^2Compute -0.3*0.1922≈-0.0577Compute denominator:150*5.5029≈825.435, so 825.435 +1000=1825.435Denominator squared≈(1825.435)^2≈3,332,000Thus, (1/9)(75000)/3,332,000≈(8333.33)/3,332,000≈0.0025So, f’(5.5029)= -0.0577 +0.0025≈-0.0552Now, t2 = t1 - f(t1)/f’(t1)=5.5029 - (-0.001)/(-0.0552)=5.5029 -0.0181≈5.4848Wait, that's oscillating between 5.4848 and 5.5029. Maybe I need to average or use a better method.Alternatively, since f(5.485)=0.001 and f(5.5029)=-0.001, the root is approximately halfway between 5.485 and 5.5029, which is around 5.494.Let me compute f(5.494):t=5.494( e^{-0.3*5.494}=e^{-1.6482}≈0.1940( 5000/(150*5.494 +1000)=5000/(824.1 +1000)=5000/1824.1≈2.741( (2.741 -1)/9≈1.741/9≈0.1934f(5.494)=0.1940 -0.1934≈0.0006Still positive.t=5.497:( e^{-0.3*5.497}=e^{-1.6491}≈0.1938( 5000/(150*5.497 +1000)=5000/(824.55 +1000)=5000/1824.55≈2.741( (2.741 -1)/9≈0.1934f(5.497)=0.1938 -0.1934≈0.0004Still positive.t=5.499:( e^{-0.3*5.499}=e^{-1.6497}≈0.1937( 5000/(150*5.499 +1000)=5000/(824.85 +1000)=5000/1824.85≈2.741( (2.741 -1)/9≈0.1934f(5.499)=0.1937 -0.1934≈0.0003Still positive.t=5.5:f(t)= -0.001 as before.So, it seems the root is around t≈5.498.Given the oscillation, perhaps the root is approximately t≈5.5 years.But to be more precise, let's use the secant method between t=5.485 (f=0.001) and t=5.5 (f=-0.001)The secant method formula:t_new = t1 - f(t1)*(t1 - t0)/(f(t1) - f(t0))Here, t0=5.485, f(t0)=0.001t1=5.5, f(t1)=-0.001So,t_new =5.5 - (-0.001)*(5.5 -5.485)/( -0.001 -0.001 )=5.5 - (-0.001)*(0.015)/(-0.002)=5.5 - (0.000015)/(-0.002)=5.5 - (-0.0075)=5.5 +0.0075=5.5075But f(5.5075) would be more negative, so maybe not helpful.Alternatively, average t=5.485 and t=5.5: (5.485 +5.5)/2=5.4925Compute f(5.4925):t=5.4925( e^{-0.3*5.4925}=e^{-1.64775}≈0.1942( 5000/(150*5.4925 +1000)=5000/(823.875 +1000)=5000/1823.875≈2.742( (2.742 -1)/9≈1.742/9≈0.1935f(5.4925)=0.1942 -0.1935≈0.0007Still positive.t=5.495:f(t)=0.0006t=5.4975:f(t)=?t=5.4975( e^{-0.3*5.4975}=e^{-1.64925}≈0.1938( 5000/(150*5.4975 +1000)=5000/(824.625 +1000)=5000/1824.625≈2.741( (2.741 -1)/9≈0.1934f(t)=0.1938 -0.1934≈0.0004Still positive.t=5.499:f(t)=0.0003t=5.5:f(t)=-0.001So, the root is between 5.499 and5.5.Using linear approximation:From t=5.499 to t=5.5, f(t) goes from 0.0003 to -0.001, a change of -0.0013 over 0.001 years.We need to find t where f(t)=0.Fraction=0.0003 /0.0013≈0.2308So, t≈5.499 +0.2308*0.001≈5.499 +0.00023≈5.4992So, approximately t≈5.4992 years.Thus, the first intersection is approximately at t≈5.5 years.But let's check the second intersection.Earlier, we saw that at t=25, ( P_A=4750 ), ( P_B≈4975 )At t=26.6667, ( P_A=5000 ), ( P_B≈4985 )So, between t=25 and t=26.6667, ( P_A(t) ) crosses ( P_B(t) ) again.Let me compute ( f(t) ) at t=26:( P_A=150*26 +1000=3900 +1000=4900( P_B=5000/(1 +9e^{-7.8}) )Compute e^{-7.8}≈0.000386Denominator≈1 +9*0.000386≈1 +0.003474≈1.003474Thus, ( P_B≈5000 /1.003474≈4982.64 )So, ( P_A=4900 < P_B≈4982.64 )At t=26.5:( P_A=150*26.5 +1000=3975 +1000=4975( P_B=5000/(1 +9e^{-8.0}) )e^{-8}≈0.000335Denominator≈1 +9*0.000335≈1.003015( P_B≈5000 /1.003015≈4985.06 )So, ( P_A=4975 < P_B≈4985.06 )At t=26.6667:( P_A=5000 )( P_B≈4985 )So, ( P_A=5000 > P_B≈4985 )Thus, between t=26.5 and t=26.6667, ( P_A(t) ) crosses ( P_B(t) ).Let me compute f(t) at t=26.6:( P_A=150*26.6 +1000=3990 +1000=4990( P_B=5000/(1 +9e^{-7.98}) )e^{-7.98}≈e^{-8 +0.02}=e^{-8}*e^{0.02}≈0.000335*1.0202≈0.000342Denominator≈1 +9*0.000342≈1 +0.003078≈1.003078( P_B≈5000 /1.003078≈4985.56 )So, ( P_A=4990 > P_B≈4985.56 )Thus, between t=26.5 and t=26.6, ( P_A(t) ) crosses ( P_B(t) ).Let me compute f(t) at t=26.55:( P_A=150*26.55 +1000=3982.5 +1000=4982.5( P_B=5000/(1 +9e^{-7.965}) )e^{-7.965}=e^{-8 +0.035}=e^{-8}*e^{0.035}≈0.000335*1.0356≈0.000347Denominator≈1 +9*0.000347≈1 +0.003123≈1.003123( P_B≈5000 /1.003123≈4985.14 )So, ( P_A=4982.5 < P_B≈4985.14 )At t=26.575:( P_A=150*26.575 +1000=3986.25 +1000=4986.25( P_B=5000/(1 +9e^{-7.9725}) )e^{-7.9725}=e^{-8 +0.0275}=e^{-8}*e^{0.0275}≈0.000335*1.0279≈0.000344Denominator≈1 +9*0.000344≈1 +0.003096≈1.003096( P_B≈5000 /1.003096≈4985.38 )So, ( P_A=4986.25 > P_B≈4985.38 )Thus, between t=26.55 and t=26.575, ( P_A(t) ) crosses ( P_B(t) ).Using linear approximation:At t=26.55, ( P_A=4982.5 < P_B≈4985.14 )At t=26.575, ( P_A=4986.25 > P_B≈4985.38 )The difference in t is 0.025 years.The difference in ( P_A ) is 4986.25 -4982.5=3.75The difference in ( P_B ) is 4985.38 -4985.14=0.24We need to find t where ( P_A(t) = P_B(t) ). Let me denote the crossing point as t=26.55 + Δt.At t=26.55, ( P_A=4982.5 ), ( P_B≈4985.14 ). So, ( P_A < P_B ) by 2.64.At t=26.575, ( P_A=4986.25 ), ( P_B≈4985.38 ). So, ( P_A > P_B ) by 0.87.The total change needed is 2.64 to overcome.The rate of change of ( P_A ) is 150 per year, and ( P_B ) is decreasing as t increases.But perhaps it's easier to compute the exact point.Let me set up the equation:150t +1000 = 5000/(1 +9e^{-0.3t})Let me denote t=26.55 + ΔtBut this might be too time-consuming. Alternatively, since the crossing is very close to t=26.56, perhaps we can approximate it as t≈26.56 years.But given the problem statement, it's likely that they only expect the first crossing time, which is approximately 5.5 years.However, the problem didn't specify which crossing, so perhaps both times are valid. But usually, such problems expect the first time.But let me check the problem statement again:"1. Determine the time ( t ) (in years) when both villages will have the same population."It says "the time", implying a single time. But given the functions, there are two times. So, perhaps the answer expects both times.But in the context, maybe the first time is more relevant, as after that, Village B's population overtakes Village A's, but Village A's population continues to grow and eventually surpasses Village B's again.But the problem doesn't specify, so perhaps both times are needed.But given the logistic model, the population of Village B approaches 5000, while Village A's population grows beyond that. So, the second crossing is when Village A's population surpasses Village B's carrying capacity.But the problem is about when both villages have the same population, which can happen twice.But perhaps the answer expects both times.But in the first part, the problem is to determine the time ( t ). So, maybe both times are required.But let me see, in the second part, the carrying capacity increases by 10%, so the new carrying capacity is 5500. Then, the new logistic function is ( P_B(t) = 5500/(1 +9e^{-0.3t}) ). Then, we need to find the new time when ( P_A(t) = P_B(t) ).But in the first part, since the problem didn't specify, maybe it's just the first time.But to be thorough, I think both times are valid, but perhaps the problem expects the first time.Given that, I think the answer is approximately 5.5 years.But let me check with t=5.5:( P_A=150*5.5 +1000=825 +1000=1825( P_B=5000/(1 +9e^{-1.65})≈5000/(1 +9*0.1922)≈5000/(1 +1.7298)≈5000/2.7298≈1832So, ( P_A=1825 ), ( P_B≈1832 ). Close, but not exact.Wait, earlier calculations showed that at t≈5.499, ( P_A≈150*5.499 +1000≈824.85 +1000=1824.85 )( P_B≈5000/(1 +9e^{-1.6497})≈5000/(1 +9*0.1937)≈5000/(1 +1.7433)≈5000/2.7433≈1822.5So, at t≈5.499, ( P_A≈1824.85 ), ( P_B≈1822.5 ). So, they are very close, but not exactly equal.Thus, the exact solution requires more precise calculation, but for the purposes of this problem, t≈5.5 years is a reasonable approximation.Therefore, the answer to part 1 is approximately 5.5 years.For part 2, the carrying capacity of Village B increases by 10%, so the new carrying capacity is 5000*1.1=5500. The logistic function becomes ( P_B(t) = frac{5500}{1 + 9e^{-0.3t}} ). We need to find the new time ( t ) when ( P_A(t) = P_B(t) ).So, the equation is:( 150t + 1000 = frac{5500}{1 + 9e^{-0.3t}} )Similarly, we can rearrange:( 1 + 9e^{-0.3t} = frac{5500}{150t + 1000} )( 9e^{-0.3t} = frac{5500}{150t + 1000} -1 )( e^{-0.3t} = frac{1}{9} left( frac{5500}{150t + 1000} -1 right) )Again, this is a transcendental equation, so we'll need to solve it numerically.Let me define ( f(t) = e^{-0.3t} - frac{1}{9} left( frac{5500}{150t + 1000} -1 right) ). We need to find t where f(t)=0.Let me compute f(t) at different points.First, let's estimate where the crossing might occur.At t=0:( P_A=1000 ), ( P_B=5500/(1 +9)=5500/10=550 ). So, ( P_A > P_B )At t=10:( P_A=2500 ), ( P_B=5500/(1 +9e^{-3})≈5500/(1 +9*0.0498)≈5500/1.448≈3793 ). So, ( P_A < P_B )Thus, the first crossing is between t=0 and t=10.But let's check at t=5:( P_A=150*5 +1000=1750( P_B=5500/(1 +9e^{-1.5})≈5500/(1 +9*0.2231)≈5500/(1 +2.0079)≈5500/3.0079≈1829So, ( P_A=1750 < P_B≈1829 )At t=4:( P_A=150*4 +1000=600 +1000=1600( P_B=5500/(1 +9e^{-1.2})≈5500/(1 +9*0.3012)≈5500/(1 +2.7108)≈5500/3.7108≈1482So, ( P_A=1600 > P_B≈1482 )Thus, the crossing is between t=4 and t=5.Let me compute f(t) at t=4.5:( P_A=150*4.5 +1000=675 +1000=1675( P_B=5500/(1 +9e^{-1.35})≈5500/(1 +9*0.2592)≈5500/(1 +2.3328)≈5500/3.3328≈1650So, ( P_A=1675 > P_B≈1650 )At t=4.75:( P_A=150*4.75 +1000=712.5 +1000=1712.5( P_B=5500/(1 +9e^{-1.425})≈5500/(1 +9*0.2395)≈5500/(1 +2.1555)≈5500/3.1555≈1743So, ( P_A=1712.5 < P_B≈1743 )Thus, the crossing is between t=4.5 and t=4.75.Let me compute f(t) at t=4.6:( P_A=150*4.6 +1000=690 +1000=1690( P_B=5500/(1 +9e^{-1.38})≈5500/(1 +9*0.2513)≈5500/(1 +2.2617)≈5500/3.2617≈1685So, ( P_A=1690 > P_B≈1685 )At t=4.65:( P_A=150*4.65 +1000=697.5 +1000=1697.5( P_B=5500/(1 +9e^{-1.395})≈5500/(1 +9*0.2473)≈5500/(1 +2.2257)≈5500/3.2257≈1705So, ( P_A=1697.5 < P_B≈1705 )Thus, the crossing is between t=4.6 and t=4.65.Let me compute f(t) at t=4.625:( P_A=150*4.625 +1000=693.75 +1000=1693.75( P_B=5500/(1 +9e^{-1.3875})≈5500/(1 +9*0.2503)≈5500/(1 +2.2527)≈5500/3.2527≈1689So, ( P_A=1693.75 > P_B≈1689 )At t=4.6375:( P_A=150*4.6375 +1000=695.625 +1000=1695.625( P_B=5500/(1 +9e^{-1.39125})≈5500/(1 +9*0.2495)≈5500/(1 +2.2455)≈5500/3.2455≈1695So, ( P_A=1695.625 > P_B≈1695 )At t=4.64:( P_A=150*4.64 +1000=696 +1000=1696( P_B=5500/(1 +9e^{-1.392})≈5500/(1 +9*0.2493)≈5500/(1 +2.2437)≈5500/3.2437≈1695.5So, ( P_A=1696 > P_B≈1695.5 )At t=4.645:( P_A=150*4.645 +1000=696.75 +1000=1696.75( P_B=5500/(1 +9e^{-1.3935})≈5500/(1 +9*0.2491)≈5500/(1 +2.2419)≈5500/3.2419≈1696.4So, ( P_A=1696.75 > P_B≈1696.4 )At t=4.6475:( P_A=150*4.6475 +1000=697.125 +1000=1697.125( P_B=5500/(1 +9e^{-1.39425})≈5500/(1 +9*0.2490)≈5500/(1 +2.241)≈5500/3.241≈1696.8So, ( P_A=1697.125 > P_B≈1696.8 )At t=4.649:( P_A=150*4.649 +1000=697.35 +1000=1697.35( P_B=5500/(1 +9e^{-1.3947})≈5500/(1 +9*0.2489)≈5500/(1 +2.2401)≈5500/3.2401≈1697.1So, ( P_A=1697.35 > P_B≈1697.1 )At t=4.6495:( P_A=150*4.6495 +1000=697.425 +1000=1697.425( P_B=5500/(1 +9e^{-1.39485})≈5500/(1 +9*0.2489)≈5500/(1 +2.2401)≈5500/3.2401≈1697.1So, ( P_A=1697.425 > P_B≈1697.1 )At t=4.6499:( P_A≈150*4.6499 +1000≈697.485 +1000≈1697.485( P_B≈5500/(1 +9e^{-1.39497})≈5500/(1 +9*0.2489)≈5500/(1 +2.2401)≈5500/3.2401≈1697.1So, ( P_A≈1697.485 > P_B≈1697.1 )Thus, the crossing is very close to t≈4.65 years.But let me check at t=4.65:( P_A=150*4.65 +1000=697.5 +1000=1697.5( P_B=5500/(1 +9e^{-1.395})≈5500/(1 +9*0.2489)≈5500/(1 +2.2401)≈5500/3.2401≈1697.1So, ( P_A=1697.5 > P_B≈1697.1 )Thus, the crossing is just above t=4.65.Using linear approximation between t=4.65 and t=4.6501:At t=4.65, ( P_A=1697.5 ), ( P_B≈1697.1 ). So, ( P_A > P_B ) by 0.4.At t=4.6501, ( P_A=150*4.6501 +1000≈697.515 +1000=1697.515( P_B≈5500/(1 +9e^{-1.39503})≈5500/(1 +9*0.2489)≈5500/3.2401≈1697.1So, ( P_A=1697.515 > P_B≈1697.1 )Thus, the crossing is very close to t=4.65.Therefore, the new time when both villages have the same population is approximately t≈4.65 years.But let me check if this makes sense. Since the carrying capacity increased, Village B's population grows faster initially, so the crossing time should be earlier than the original 5.5 years. Indeed, 4.65 years is earlier than 5.5 years.Thus, the answer to part 2 is approximately 4.65 years.