answer:<think>Alright, so I have this problem about modeling the freshness of an ingredient over time and then figuring out the optimal distance to source it from. Let me try to break this down step by step.First, part 1 is about finding the time ( t ) when the freshness of the herb drops below 50%. The freshness function is given as ( F(t) = 100e^{-kt} ), where ( k = 0.2 ). So, I need to solve for ( t ) when ( F(t) = 50 ).Let me write that equation down:( 50 = 100e^{-0.2t} )Hmm, okay. So, I can divide both sides by 100 to simplify:( 0.5 = e^{-0.2t} )Now, to solve for ( t ), I should take the natural logarithm of both sides because the natural log is the inverse of the exponential function with base ( e ). So:( ln(0.5) = ln(e^{-0.2t}) )Simplifying the right side, since ( ln(e^x) = x ):( ln(0.5) = -0.2t )Now, I can solve for ( t ) by dividing both sides by -0.2:( t = frac{ln(0.5)}{-0.2} )I know that ( ln(0.5) ) is a negative number because 0.5 is less than 1. Specifically, ( ln(0.5) approx -0.6931 ). Plugging that in:( t = frac{-0.6931}{-0.2} = frac{0.6931}{0.2} )Calculating that, 0.6931 divided by 0.2 is the same as 0.6931 multiplied by 5, which is approximately 3.4655 days.So, the freshness drops below 50% after about 3.4655 days. I can round this to maybe two decimal places, so 3.47 days.Wait, let me double-check my steps. I set ( F(t) = 50 ), substituted into the equation, took the natural log correctly, and solved for ( t ). The negative signs canceled out, and the division was correct. Yeah, that seems right.Moving on to part 2. This is about optimizing the cost function ( C = 50 + 10F(t) + 5d ). The distance ( d ) ranges from 10 to 100 miles. Also, it's given that ( t = d/10 ). So, the time since harvesting is directly proportional to the distance, which makes sense because the farther the farm, the longer it takes to get the ingredient, hence the less fresh it is.First, I need to express ( F(t) ) in terms of ( d ). Since ( t = d/10 ), substitute that into the freshness function:( F(t) = 100e^{-0.2(d/10)} )Simplifying the exponent:( 0.2(d/10) = 0.02d )So,( F(t) = 100e^{-0.02d} )Now, substitute this back into the cost function:( C = 50 + 10 times 100e^{-0.02d} + 5d )Simplify the terms:( C = 50 + 1000e^{-0.02d} + 5d )So, the cost function is ( C(d) = 50 + 1000e^{-0.02d} + 5d ). Now, I need to find the value of ( d ) between 10 and 100 that minimizes this cost.To find the minimum, I should take the derivative of ( C ) with respect to ( d ), set it equal to zero, and solve for ( d ). Then, check if it's a minimum by using the second derivative or analyzing the behavior.Let's compute the derivative ( C'(d) ):The derivative of 50 is 0.The derivative of ( 1000e^{-0.02d} ) with respect to ( d ) is ( 1000 times (-0.02)e^{-0.02d} = -20e^{-0.02d} ).The derivative of ( 5d ) is 5.So, putting it all together:( C'(d) = -20e^{-0.02d} + 5 )Set this equal to zero to find critical points:( -20e^{-0.02d} + 5 = 0 )Solving for ( d ):( -20e^{-0.02d} = -5 )Divide both sides by -5:( 4e^{-0.02d} = 1 )So,( e^{-0.02d} = 1/4 )Take the natural logarithm of both sides:( ln(e^{-0.02d}) = ln(1/4) )Simplify:( -0.02d = ln(1/4) )Again, ( ln(1/4) ) is negative. ( ln(1/4) = -ln(4) approx -1.3863 )So,( -0.02d = -1.3863 )Divide both sides by -0.02:( d = frac{-1.3863}{-0.02} = frac{1.3863}{0.02} )Calculating that, 1.3863 divided by 0.02 is the same as 1.3863 multiplied by 50, which is approximately 69.315.So, ( d approx 69.315 ) miles.Now, I need to verify if this is indeed a minimum. Let's check the second derivative.First, compute the second derivative ( C''(d) ):The derivative of ( C'(d) = -20e^{-0.02d} + 5 ) is:( C''(d) = (-20)(-0.02)e^{-0.02d} + 0 = 0.4e^{-0.02d} )Since ( e^{-0.02d} ) is always positive, ( C''(d) ) is positive. Therefore, the function is concave upward at this critical point, which means it's a local minimum.So, the optimal distance is approximately 69.315 miles. But since the distance can't be a fraction of a mile in this context, we might need to consider whether to round it to 69 or 70 miles. However, the problem doesn't specify, so I can present it as approximately 69.32 miles.But wait, let me make sure I didn't make a mistake in the derivative. The derivative of ( 1000e^{-0.02d} ) is indeed ( 1000 * (-0.02)e^{-0.02d} = -20e^{-0.02d} ). Then, the derivative of 5d is 5. So, the derivative is correct.Setting it equal to zero, solving for ( d ), that seems correct. The second derivative is positive, so it's a minimum. So, yeah, 69.315 miles is the optimal distance.But hold on, the distance is between 10 and 100 miles. 69.315 is within that range, so that's acceptable.Alternatively, maybe I should check the cost at 69 and 70 miles to see which one is lower, but since 69.315 is closer to 69, and the function is smooth, it's likely that 69 is the optimal integer value. But the question doesn't specify if ( d ) has to be an integer, so maybe it's okay to leave it as a decimal.So, summarizing:1. The time when freshness drops below 50% is approximately 3.47 days.2. The optimal distance to minimize cost is approximately 69.32 miles.I think that's it. Let me just recap the steps to make sure I didn't skip anything.For part 1, solved the exponential equation by taking natural logs, got t ≈ 3.47 days.For part 2, substituted t = d/10 into the freshness function, plugged into the cost function, took the derivative, found critical point at d ≈ 69.32, confirmed it's a minimum with the second derivative. Seems solid.Final Answer1. The freshness drops below 50% after approximately boxed{3.47} days.2. The optimal distance to minimize cost is approximately boxed{69.32} miles.</think>

answer:<think>Alright, so I've got this problem about determining the success rate of a collaboration between two companies. It involves some math, specifically figuring out constants in a formula and then optimizing another variable. Let me try to break this down step by step.First, the problem gives me a formula for the success rate ( S ):[ S = frac{k cdot M_A cdot M_B cdot C}{(M_A + M_B + C)^n} ]Here, ( k ) and ( n ) are constants that need to be determined. They've given me two data points to work with, each with different values of ( M_A ), ( M_B ), ( C ), and the resulting ( S ). My job is to find ( k ) and ( n ) using these data points.Let me write down the data points:1. First collaboration: - ( M_A = 0.3 ) - ( M_B = 0.4 ) - ( C = 0.5 ) - ( S = 0.25 )2. Second collaboration: - ( M_A = 0.5 ) - ( M_B = 0.6 ) - ( C = 0.7 ) - ( S = 0.36 )So, I have two equations with two unknowns (( k ) and ( n )). That should be solvable. Let me plug in the first data point into the formula:For the first collaboration:[ 0.25 = frac{k cdot 0.3 cdot 0.4 cdot 0.5}{(0.3 + 0.4 + 0.5)^n} ]Let me compute the numerator and denominator separately.Numerator: ( k cdot 0.3 cdot 0.4 cdot 0.5 )- 0.3 * 0.4 = 0.12- 0.12 * 0.5 = 0.06So numerator is ( 0.06k )Denominator: ( (0.3 + 0.4 + 0.5)^n = (1.2)^n )So the equation becomes:[ 0.25 = frac{0.06k}{(1.2)^n} ]Similarly, for the second collaboration:[ 0.36 = frac{k cdot 0.5 cdot 0.6 cdot 0.7}{(0.5 + 0.6 + 0.7)^n} ]Compute numerator and denominator:Numerator: ( k cdot 0.5 cdot 0.6 cdot 0.7 )- 0.5 * 0.6 = 0.3- 0.3 * 0.7 = 0.21So numerator is ( 0.21k )Denominator: ( (0.5 + 0.6 + 0.7)^n = (1.8)^n )So the equation becomes:[ 0.36 = frac{0.21k}{(1.8)^n} ]Now, I have two equations:1. ( 0.25 = frac{0.06k}{(1.2)^n} )2. ( 0.36 = frac{0.21k}{(1.8)^n} )I need to solve for ( k ) and ( n ). Let me denote equation 1 as Eq(1) and equation 2 as Eq(2).Let me solve Eq(1) for ( k ):[ 0.25 = frac{0.06k}{(1.2)^n} ]Multiply both sides by ( (1.2)^n ):[ 0.25 cdot (1.2)^n = 0.06k ]Then, divide both sides by 0.06:[ k = frac{0.25 cdot (1.2)^n}{0.06} ]Simplify:[ k = frac{0.25}{0.06} cdot (1.2)^n ][ k = frac{25}{6} cdot (1.2)^n ][ k approx 4.1667 cdot (1.2)^n ]Similarly, solve Eq(2) for ( k ):[ 0.36 = frac{0.21k}{(1.8)^n} ]Multiply both sides by ( (1.8)^n ):[ 0.36 cdot (1.8)^n = 0.21k ]Divide both sides by 0.21:[ k = frac{0.36 cdot (1.8)^n}{0.21} ]Simplify:[ k = frac{0.36}{0.21} cdot (1.8)^n ][ k approx 1.7143 cdot (1.8)^n ]Now, I have two expressions for ( k ):1. ( k approx 4.1667 cdot (1.2)^n )2. ( k approx 1.7143 cdot (1.8)^n )Since both equal ( k ), I can set them equal to each other:[ 4.1667 cdot (1.2)^n = 1.7143 cdot (1.8)^n ]Now, I need to solve for ( n ). Let me write this equation as:[ frac{(1.2)^n}{(1.8)^n} = frac{1.7143}{4.1667} ]Simplify the left side:[ left( frac{1.2}{1.8} right)^n = frac{1.7143}{4.1667} ]Compute ( frac{1.2}{1.8} ):[ frac{1.2}{1.8} = frac{2}{3} approx 0.6667 ]Compute the right side:[ frac{1.7143}{4.1667} approx 0.4114 ]So, the equation becomes:[ (0.6667)^n = 0.4114 ]To solve for ( n ), take the natural logarithm of both sides:[ lnleft( (0.6667)^n right) = ln(0.4114) ][ n cdot ln(0.6667) = ln(0.4114) ]Compute the logarithms:- ( ln(0.6667) approx -0.4055 )- ( ln(0.4114) approx -0.8909 )So,[ n = frac{-0.8909}{-0.4055} approx 2.196 ]So, ( n approx 2.196 ). Let me round this to two decimal places for simplicity, so ( n approx 2.20 ).Now, plug this value of ( n ) back into one of the expressions for ( k ). Let's use the first one:[ k approx 4.1667 cdot (1.2)^{2.20} ]Compute ( (1.2)^{2.20} ). Let me calculate this step by step.First, ( 1.2^2 = 1.44 ). Then, ( 1.2^{0.20} ). To compute ( 1.2^{0.20} ), I can use logarithms or a calculator. Let me approximate it.Alternatively, I can use the formula ( a^b = e^{b ln a} ).Compute ( ln(1.2) approx 0.1823 ).So, ( 1.2^{0.20} = e^{0.20 cdot 0.1823} = e^{0.03646} approx 1.0371 ).Therefore, ( 1.2^{2.20} = 1.2^2 cdot 1.2^{0.20} approx 1.44 cdot 1.0371 approx 1.493 ).So, ( k approx 4.1667 cdot 1.493 approx 6.23 ).Let me check this with the second expression for ( k ):[ k approx 1.7143 cdot (1.8)^{2.20} ]Compute ( (1.8)^{2.20} ).Again, ( 1.8^2 = 3.24 ). Then, ( 1.8^{0.20} ).Compute ( ln(1.8) approx 0.5878 ).So, ( 1.8^{0.20} = e^{0.20 cdot 0.5878} = e^{0.11756} approx 1.1247 ).Therefore, ( 1.8^{2.20} = 1.8^2 cdot 1.8^{0.20} approx 3.24 cdot 1.1247 approx 3.643 ).So, ( k approx 1.7143 cdot 3.643 approx 6.24 ).Hmm, that's consistent with the previous calculation, approximately 6.23 or 6.24. So, ( k approx 6.23 ).Let me note that the slight difference is due to rounding errors in the intermediate steps. So, I can take ( k approx 6.23 ) and ( n approx 2.20 ).But to be more precise, maybe I should carry more decimal places.Wait, let me see. Maybe I can solve for ( n ) more accurately.Going back to the equation:[ (0.6667)^n = 0.4114 ]Taking natural logs:[ n = frac{ln(0.4114)}{ln(0.6667)} ]Compute ( ln(0.4114) approx -0.8909 ) and ( ln(0.6667) approx -0.4055 ).So,[ n = frac{-0.8909}{-0.4055} approx 2.196 ]So, ( n approx 2.196 ). Let me use more precise value, say ( n = 2.196 ).Then, compute ( (1.2)^{2.196} ).Again, using natural logs:( ln(1.2) approx 0.1823 )So,( ln(1.2^{2.196}) = 2.196 times 0.1823 approx 0.3996 )Thus,( 1.2^{2.196} = e^{0.3996} approx 1.490 )Similarly, compute ( (1.8)^{2.196} ):( ln(1.8) approx 0.5878 )So,( ln(1.8^{2.196}) = 2.196 times 0.5878 approx 1.291 )Thus,( 1.8^{2.196} = e^{1.291} approx 3.633 )Now, compute ( k ) using the first equation:( k = 4.1667 times 1.490 approx 6.233 )Using the second equation:( k = 1.7143 times 3.633 approx 6.227 )So, both give approximately 6.23. Therefore, ( k approx 6.23 ) and ( n approx 2.196 ).To make it more precise, perhaps we can carry more decimal places, but for practical purposes, maybe we can take ( k = 6.23 ) and ( n = 2.20 ).Wait, but let me check if these values actually satisfy the original equations.First, for the first data point:Compute ( S = frac{6.23 cdot 0.3 cdot 0.4 cdot 0.5}{(0.3 + 0.4 + 0.5)^{2.20}} )Compute numerator:6.23 * 0.3 = 1.8691.869 * 0.4 = 0.74760.7476 * 0.5 = 0.3738Denominator:0.3 + 0.4 + 0.5 = 1.21.2^{2.20} ≈ 1.490So, S ≈ 0.3738 / 1.490 ≈ 0.2509, which is approximately 0.25. That's good.For the second data point:Compute ( S = frac{6.23 cdot 0.5 cdot 0.6 cdot 0.7}{(0.5 + 0.6 + 0.7)^{2.20}} )Compute numerator:6.23 * 0.5 = 3.1153.115 * 0.6 = 1.8691.869 * 0.7 = 1.3083Denominator:0.5 + 0.6 + 0.7 = 1.81.8^{2.20} ≈ 3.633So, S ≈ 1.3083 / 3.633 ≈ 0.360, which is exactly 0.36. Perfect.So, these values of ( k ) and ( n ) satisfy both data points.Therefore, ( k approx 6.23 ) and ( n approx 2.20 ).But let me see if I can express ( n ) as a fraction or something. 2.196 is approximately 2.2, which is 11/5. Let me check if 11/5 is 2.2, yes. So, maybe ( n = 11/5 ) or 2.2 exactly.Wait, let me see. If I take ( n = 11/5 = 2.2 ), then let's compute ( (0.6667)^{2.2} ).Wait, 0.6667 is 2/3, so (2/3)^{2.2}.But 2.2 is 11/5, so (2/3)^{11/5}.Alternatively, maybe it's better to just keep ( n ) as approximately 2.2.Alternatively, perhaps the exact value is 2.2, given that 2.196 is very close to 2.2.So, for simplicity, I can take ( n = 2.2 ) and ( k approx 6.23 ).But let me see if I can compute ( k ) more precisely.From the first equation:( k = frac{0.25 cdot (1.2)^n}{0.06} )With ( n = 2.196 ), ( (1.2)^{2.196} approx 1.490 )So,( k = 0.25 / 0.06 * 1.490 ≈ (4.1667) * 1.490 ≈ 6.233 )Similarly, from the second equation:( k = frac{0.36 cdot (1.8)^n}{0.21} )With ( n = 2.196 ), ( (1.8)^{2.196} ≈ 3.633 )So,( k = 0.36 / 0.21 * 3.633 ≈ (1.7143) * 3.633 ≈ 6.227 )So, ( k ) is approximately 6.23.Therefore, I can conclude that ( k approx 6.23 ) and ( n approx 2.20 ).Now, moving on to part 2.Once ( k ) and ( n ) are determined, the strategist wants to maximize the success rate ( S ) for a potential collaboration where ( M_A = 0.45 ) and ( M_B = 0.55 ). What value of ( C ) will maximize ( S )?So, we need to find the value of ( C ) that maximizes ( S ) given ( M_A = 0.45 ), ( M_B = 0.55 ), ( k approx 6.23 ), and ( n approx 2.20 ).So, the function to maximize is:[ S(C) = frac{6.23 cdot 0.45 cdot 0.55 cdot C}{(0.45 + 0.55 + C)^{2.20}} ]Simplify the constants:First, compute ( 6.23 cdot 0.45 cdot 0.55 ).Compute 0.45 * 0.55:0.45 * 0.55 = 0.2475Then, 6.23 * 0.2475 ≈ 6.23 * 0.2475Compute 6 * 0.2475 = 1.4850.23 * 0.2475 ≈ 0.057So total ≈ 1.485 + 0.057 ≈ 1.542So, numerator is approximately 1.542 * CDenominator is ( (0.45 + 0.55 + C)^{2.20} = (1.0 + C)^{2.20} )So, the function simplifies to:[ S(C) = frac{1.542 C}{(1.0 + C)^{2.20}} ]We need to find the value of ( C ) that maximizes ( S(C) ).To find the maximum, we can take the derivative of ( S(C) ) with respect to ( C ), set it equal to zero, and solve for ( C ).Let me denote ( S(C) = frac{1.542 C}{(1 + C)^{2.20}} )Let me compute the derivative ( S'(C) ).Using the quotient rule:If ( S(C) = frac{u}{v} ), then ( S'(C) = frac{u'v - uv'}{v^2} )Here, ( u = 1.542 C ), so ( u' = 1.542 )( v = (1 + C)^{2.20} ), so ( v' = 2.20 (1 + C)^{1.20} )So,[ S'(C) = frac{1.542 cdot (1 + C)^{2.20} - 1.542 C cdot 2.20 (1 + C)^{1.20}}{(1 + C)^{4.40}} ]Simplify numerator:Factor out ( 1.542 (1 + C)^{1.20} ):Numerator = ( 1.542 (1 + C)^{1.20} [ (1 + C) - 2.20 C ] )Simplify inside the brackets:( (1 + C) - 2.20 C = 1 + C - 2.20 C = 1 - 1.20 C )So, numerator becomes:( 1.542 (1 + C)^{1.20} (1 - 1.20 C) )Therefore, the derivative is:[ S'(C) = frac{1.542 (1 + C)^{1.20} (1 - 1.20 C)}{(1 + C)^{4.40}} ]Simplify the denominator:( (1 + C)^{4.40} = (1 + C)^{1.20} cdot (1 + C)^{3.20} )So, cancel out ( (1 + C)^{1.20} ):[ S'(C) = frac{1.542 (1 - 1.20 C)}{(1 + C)^{3.20}} ]Set ( S'(C) = 0 ):The denominator is always positive for ( C > 0 ), so set numerator equal to zero:[ 1.542 (1 - 1.20 C) = 0 ]So,[ 1 - 1.20 C = 0 ][ 1.20 C = 1 ][ C = frac{1}{1.20} ][ C = frac{10}{12} = frac{5}{6} approx 0.8333 ]So, the critical point is at ( C approx 0.8333 ).To ensure this is a maximum, we can check the second derivative or analyze the behavior of the first derivative around this point.Alternatively, since the function ( S(C) ) tends to zero as ( C ) approaches zero and as ( C ) approaches infinity, and there is only one critical point, this critical point must be a maximum.Therefore, the value of ( C ) that maximizes ( S ) is ( C = frac{5}{6} ) or approximately 0.8333.But let me verify this calculation.We had:[ S'(C) = frac{1.542 (1 - 1.20 C)}{(1 + C)^{3.20}} ]Setting numerator to zero:( 1 - 1.20 C = 0 )So,( C = 1 / 1.20 = 5/6 ≈ 0.8333 )Yes, that seems correct.Alternatively, let me think about the general form of the function.The function ( S(C) = frac{k M_A M_B C}{(M_A + M_B + C)^n} )This is a function of the form ( S(C) = frac{a C}{(b + C)^n} ), where ( a = k M_A M_B ) and ( b = M_A + M_B ).To find the maximum, we can take the derivative and set it to zero.In general, for ( S(C) = frac{a C}{(b + C)^n} ), the derivative is:[ S'(C) = frac{a (b + C)^n - a C n (b + C)^{n - 1}}{(b + C)^{2n}}} ]Wait, no, better to use the quotient rule as before.But regardless, the critical point occurs when the derivative is zero, which leads to:( (b + C)^n - C n (b + C)^{n - 1} = 0 )Factor out ( (b + C)^{n - 1} ):( (b + C)^{n - 1} [ (b + C) - n C ] = 0 )So,( (b + C) - n C = 0 )( b + C - n C = 0 )( b = (n - 1) C )Thus,( C = frac{b}{n - 1} )In our case, ( b = M_A + M_B = 0.45 + 0.55 = 1.0 ), and ( n = 2.20 )So,( C = frac{1.0}{2.20 - 1} = frac{1.0}{1.20} = frac{5}{6} ≈ 0.8333 )Yes, that's consistent with our earlier result.Therefore, the value of ( C ) that maximizes ( S ) is ( frac{5}{6} ) or approximately 0.8333.So, to summarize:1. The constants are ( k approx 6.23 ) and ( n approx 2.20 ).2. The value of ( C ) that maximizes ( S ) when ( M_A = 0.45 ) and ( M_B = 0.55 ) is ( C = frac{5}{6} ) or approximately 0.8333.I think that's it. Let me just double-check if I made any calculation errors.In part 1, solving for ( k ) and ( n ), the process involved setting up two equations and solving for the unknowns. I used logarithms to solve for ( n ) and then back-calculated ( k ). The values checked out with both data points, so that seems solid.In part 2, using calculus to find the maximum of ( S(C) ), I derived the critical point and confirmed it's a maximum. The result is ( C = 5/6 ), which seems reasonable given the parameters.Therefore, I'm confident in these results.Final Answer1. The constants are ( k approx boxed{6.23} ) and ( n approx boxed{2.20} ).2. The value of ( C ) that maximizes ( S ) is ( boxed{dfrac{5}{6}} ).</think>

answer:<think>To determine the total area of the Essex Company Mills complex, I'll start by calculating the area of Building A and then move on to Building B.Building A:Building A is a rectangle with a length of 120 meters and a width of 80 meters. The area of a rectangle is calculated by multiplying its length by its width. So, multiplying 120 meters by 80 meters gives an area of 9,600 square meters.Building B:Building B is an irregular pentagon with vertices at the coordinates (0,0), (30,0), (50,40), (20,60), and (0,40). To find the area of an irregular polygon, I'll use the Shoelace Theorem. This involves listing the coordinates of the vertices in order, multiplying them in a specific pattern, and then taking half the absolute difference of the sums.First, I'll list the coordinates and repeat the first vertex at the end to complete the cycle:(0,0), (30,0), (50,40), (20,60), (0,40), (0,0).Next, I'll calculate the sum of the products of the coordinates going one way:(0*0) + (30*40) + (50*60) + (20*40) + (0*0) = 0 + 1,200 + 3,000 + 800 + 0 = 5,000.Then, I'll calculate the sum of the products going the other way:(0*30) + (0*50) + (40*20) + (60*0) + (40*0) = 0 + 0 + 800 + 0 + 0 = 800.Subtracting the second sum from the first gives 5,000 - 800 = 4,200. Taking half of this absolute value gives the area: 4,200 / 2 = 2,100 square meters.Total Area:Finally, I'll add the areas of Building A and Building B to find the total area to be restored: 9,600 + 2,100 = 11,700 square meters.</think>

answer:<think>Alright, so I have this problem about voter apathy in Louisiana. A local resident is trying to study this by surveying 1000 people. They came up with this Voter Apathy Index (VAI) which is a score between 0 and 1, where 0 is completely engaged and 1 is totally apathetic. The VAI scores are modeled by a continuous random variable X with a probability density function f(x) = k(1 - x²) for 0 ≤ x ≤ 1. First, I need to find the normalization constant k. Hmm, okay. I remember that for a probability density function, the total area under the curve from 0 to 1 must equal 1. So, the integral of f(x) from 0 to 1 should be 1. That means I need to compute the integral of k(1 - x²) dx from 0 to 1 and set it equal to 1, then solve for k.Let me write that down:∫₀¹ k(1 - x²) dx = 1So, factoring out the constant k, it becomes:k ∫₀¹ (1 - x²) dx = 1Now, I need to compute the integral of (1 - x²) from 0 to 1. Let's do that step by step.The integral of 1 dx is just x, and the integral of x² dx is (x³)/3. So, putting it together:∫₀¹ (1 - x²) dx = [x - (x³)/3] from 0 to 1Evaluating at 1: 1 - (1³)/3 = 1 - 1/3 = 2/3Evaluating at 0: 0 - 0 = 0So, subtracting, we get 2/3 - 0 = 2/3Therefore, the integral is 2/3. Plugging back into the equation:k * (2/3) = 1So, solving for k:k = 1 / (2/3) = 3/2So, k is 3/2. That seems straightforward.Now, moving on to the second part. The voter turnout rate T is inversely proportional to the VAI score and is modeled as T = c(1 - X), where c is a constant. They observed that the average voter turnout is 55%, so we need to find the expected value of T, which should be 0.55.Wait, hold on. The expected value of T is the average voter turnout, which is 55%, so E[T] = 0.55. Since T = c(1 - X), then E[T] = c * E[1 - X] = c * (1 - E[X])So, I need to compute E[X], the expected value of X, and then solve for c.First, let's find E[X]. The expected value of a continuous random variable is given by:E[X] = ∫₀¹ x * f(x) dxWe already know f(x) = (3/2)(1 - x²), so:E[X] = ∫₀¹ x * (3/2)(1 - x²) dxFactor out the constant 3/2:E[X] = (3/2) ∫₀¹ x(1 - x²) dxLet me expand the integrand:x(1 - x²) = x - x³So, the integral becomes:(3/2) ∫₀¹ (x - x³) dxCompute each integral separately:∫ x dx = (x²)/2∫ x³ dx = (x⁴)/4So, putting it together:(3/2) [ (x²)/2 - (x⁴)/4 ] from 0 to 1Evaluate at 1:(1²)/2 - (1⁴)/4 = 1/2 - 1/4 = 1/4Evaluate at 0:0 - 0 = 0So, subtracting, we get 1/4 - 0 = 1/4Multiply by 3/2:(3/2) * (1/4) = 3/8Therefore, E[X] = 3/8So, going back to E[T] = c * (1 - E[X]) = c * (1 - 3/8) = c * (5/8)We know that E[T] is 0.55, so:c * (5/8) = 0.55Solving for c:c = 0.55 * (8/5) = (0.55 * 8)/5Calculate 0.55 * 8:0.55 * 8 = 4.4Then, 4.4 / 5 = 0.88So, c is 0.88Wait, but let me double-check that calculation.0.55 * 8 = 4.4, yes. Then 4.4 divided by 5 is 0.88, correct.So, c is 0.88.But hold on, is that all? Let me recap.We had E[T] = c * (1 - E[X]) = 0.55We found E[X] = 3/8, so 1 - 3/8 = 5/8.Thus, c = 0.55 / (5/8) = 0.55 * (8/5) = (0.55 * 8)/5 = 4.4 / 5 = 0.88Yes, that seems correct.So, the expected value of T is 0.55, which is 55%, as given.Wait, but the question says "calculate the expected value of T." But we were told that the average voter turnout is 55%, so E[T] is 0.55. So, maybe the question is just asking to confirm that, given the model, E[T] is indeed 0.55, which it is, because we set it up that way.But maybe I misread. Let me check."Assuming that the voter turnout rate T is inversely proportional to the VAI score and can be modeled as T = c(1 - X) where c is a constant. If the average voter turnout in Louisiana is observed to be 55%, calculate the expected value of T."Wait, so maybe they just want us to compute E[T], which is 0.55, but perhaps they want it expressed in terms of c or something else? Hmm.But no, since we were told the average is 55%, so E[T] is 0.55. So, maybe the question is just confirming that, but in the process, we had to find c.But in the problem statement, it says "calculate the expected value of T." Since E[T] is given as 55%, which is 0.55, perhaps the answer is 0.55. But maybe they want it in terms of the model, so we have to compute it through the expectation.Wait, but in the problem statement, part 2 says: "If the average voter turnout in Louisiana is observed to be 55%, calculate the expected value of T."So, perhaps it's just telling us that E[T] is 0.55, so we don't need to compute it, but maybe they want us to find c? Hmm.Wait, let me read again."Assuming that the voter turnout rate T is inversely proportional to the VAI score and can be modeled as T = c(1 - X) where c is a constant. If the average voter turnout in Louisiana is observed to be 55%, calculate the expected value of T."Hmm, maybe it's just telling us that E[T] is 0.55, so the expected value is 0.55, which is 55%. So, perhaps the answer is 0.55.But in the process, we had to find c, which is 0.88, but the question is asking for E[T], which is given as 55%. So, maybe the answer is 0.55.But let me think again. Maybe they want us to compute E[T] using the model, given that T = c(1 - X), and we have to find E[T] which is 0.55, but in the process, we had to find c. So, perhaps the answer is 0.55, but in the process, we found c = 0.88.Wait, but the question is phrased as: "calculate the expected value of T." So, perhaps they just want us to state that it's 0.55, but given that we have to model it, we have to compute it through the expectation, which we did, and it comes out to 0.55.So, maybe the answer is 0.55, but we had to go through the process of finding c to confirm that.Alternatively, maybe the question is expecting us to compute E[T] without using the given average, but that doesn't make sense because the average is given as 55%.Wait, perhaps I misread the question. Let me check again."Assuming that the voter turnout rate T is inversely proportional to the VAI score and can be modeled as T = c(1 - X) where c is a constant. If the average voter turnout in Louisiana is observed to be 55%, calculate the expected value of T."Hmm, so maybe they just want us to compute E[T] given the model, which is 0.55, but in the process, we had to find c. So, the expected value is 0.55.But perhaps, to be thorough, we should present both the value of c and the expected value of T.Wait, but the question specifically says "calculate the expected value of T," so maybe just 0.55 is the answer. But since we had to compute c to get there, perhaps we need to include that.But in the problem statement, part 2 is separate from part 1, so maybe they just want the expected value, which is 0.55, but expressed as a decimal or percentage.Wait, 55% is 0.55, so that's the expected value.But let me think again. If we model T = c(1 - X), and we know E[T] = 0.55, then we can write E[T] = c * E[1 - X] = c * (1 - E[X]) = 0.55We found E[X] = 3/8, so 1 - 3/8 = 5/8, so c = 0.55 / (5/8) = 0.88So, c is 0.88, but the expected value of T is 0.55.So, the answer is 0.55.But maybe they want it in percentage, so 55%.But the question says "calculate the expected value of T," and since T is a rate, it's a proportion, so 0.55 is fine.Alternatively, maybe they want it expressed as a fraction, 11/20, but 0.55 is probably acceptable.So, to sum up:1. The normalization constant k is 3/2.2. The expected value of T is 0.55.But let me just make sure I didn't make any mistakes in the calculations.For part 1:∫₀¹ k(1 - x²) dx = 1Computed integral as 2/3, so k = 3/2. That seems correct.For part 2:E[X] = ∫₀¹ x * (3/2)(1 - x²) dx = 3/8Then, E[T] = c * (1 - 3/8) = c * 5/8 = 0.55So, c = 0.55 * 8/5 = 0.88Thus, E[T] is 0.55, which is 55%.Yes, that all checks out.So, final answers:1. k = 3/22. E[T] = 0.55But let me write them in the required format.For part 1, the normalization constant k is 3/2, which is 1.5.For part 2, the expected value of T is 0.55, which is 55%.But the question says "calculate the expected value of T," so 0.55 is the answer.So, I think that's it.</think>

answer:<think>Alright, so I have this problem about Samantha and her energy levels. It seems like a recursive relation problem where her energy each day depends on the previous day's energy, the time she spends on tai chi, and the quality of her sleep. Let me try to break this down step by step.First, let's look at Sub-problem 1. The recursive relation given is:[ E_{n+1} = alpha E_n + beta t_n + gamma S_n ]We are given the constants: α = 0.7, β = 2, γ = 1.5, and E₀ = 50. Also, the sleep hours Sₙ follow a pattern:[ S_n = 6 + 0.5 cosleft(frac{pi n}{7}right) ]So, we need to find the general form of Eₙ in terms of n, tₙ, and the given constants.Hmm, recursive relations can sometimes be solved by finding the homogeneous solution and a particular solution. Let me recall that method.The general form of a linear recurrence relation is:[ E_{n+1} = a E_n + b f(n) ]Which in this case, a is 0.7, and f(n) is β tₙ + γ Sₙ.So, the homogeneous part is:[ E_{n+1} = 0.7 E_n ]The solution to the homogeneous equation is:[ E_n^{(h)} = C (0.7)^n ]Where C is a constant determined by initial conditions.Now, for the particular solution, since the nonhomogeneous term is β tₙ + γ Sₙ, which is a function of n, we need to find a particular solution Eₙ^{(p)} such that:[ E_{n+1}^{(p)} = 0.7 E_n^{(p)} + 2 t_n + 1.5 S_n ]But since tₙ and Sₙ are given functions of n, we might need to express Eₙ^{(p)} in terms of a summation.Wait, actually, since the recurrence is linear and nonhomogeneous, the general solution is the sum of the homogeneous solution and the particular solution. The particular solution can be found using the method of summation.So, the general solution is:[ E_n = E_n^{(h)} + E_n^{(p)} ]Where:[ E_n^{(p)} = sum_{k=0}^{n-1} 0.7^{n-1 - k} (2 t_k + 1.5 S_k) ]Therefore, combining both:[ E_n = C (0.7)^n + sum_{k=0}^{n-1} 0.7^{n-1 - k} (2 t_k + 1.5 S_k) ]Now, applying the initial condition E₀ = 50. Let's plug n = 0 into the general solution:When n = 0,[ E_0 = C (0.7)^0 + sum_{k=0}^{-1} ... ]But the summation from k=0 to -1 is zero, so:[ 50 = C (1) implies C = 50 ]Therefore, the general form is:[ E_n = 50 (0.7)^n + sum_{k=0}^{n-1} 0.7^{n-1 - k} (2 t_k + 1.5 S_k) ]Alternatively, we can write the summation as:[ E_n = 50 (0.7)^n + sum_{m=0}^{n-1} 0.7^{m} (2 t_{n-1 - m} + 1.5 S_{n-1 - m}) ]But perhaps it's clearer to leave it as is.So, for Sub-problem 1, the general form is:[ E_n = 50 (0.7)^n + sum_{k=0}^{n-1} 0.7^{n-1 - k} (2 t_k + 1.5 S_k) ]Now, moving on to Sub-problem 2. We are told that tₙ = 1 + 0.1n. So, we need to compute Eₙ after 14 days.Given that, let's substitute tₙ into the general form.First, let's note that Sₙ is given by:[ S_n = 6 + 0.5 cosleft(frac{pi n}{7}right) ]So, for each day n, we can compute Sₙ.So, our general formula is:[ E_n = 50 (0.7)^n + sum_{k=0}^{n-1} 0.7^{n-1 - k} (2 (1 + 0.1k) + 1.5 (6 + 0.5 cosleft(frac{pi k}{7}right))) ]Simplify the terms inside the summation:First, compute 2(1 + 0.1k):2*(1 + 0.1k) = 2 + 0.2kThen, compute 1.5*(6 + 0.5 cos(πk/7)):1.5*6 = 91.5*0.5 = 0.75So, 1.5*(6 + 0.5 cos(πk/7)) = 9 + 0.75 cos(πk/7)Therefore, adding both parts:2 + 0.2k + 9 + 0.75 cos(πk/7) = 11 + 0.2k + 0.75 cos(πk/7)So, the summation becomes:Sum from k=0 to n-1 of 0.7^{n-1 -k}*(11 + 0.2k + 0.75 cos(πk/7))Therefore, Eₙ is:50*(0.7)^n + Sum_{k=0}^{n-1} 0.7^{n-1 -k}*(11 + 0.2k + 0.75 cos(πk/7))This seems a bit complicated, but perhaps we can separate the summation into three separate sums:Sum1 = Sum_{k=0}^{n-1} 0.7^{n-1 -k}*11Sum2 = Sum_{k=0}^{n-1} 0.7^{n-1 -k}*0.2kSum3 = Sum_{k=0}^{n-1} 0.7^{n-1 -k}*0.75 cos(πk/7)So, Eₙ = 50*(0.7)^n + Sum1 + Sum2 + Sum3Let me compute each sum separately.First, Sum1:Sum1 = 11 * Sum_{k=0}^{n-1} 0.7^{n-1 -k}Let’s change variable: let m = n-1 -k, so when k=0, m = n-1; when k = n-1, m=0.So, Sum1 = 11 * Sum_{m=0}^{n-1} 0.7^{m} = 11 * (1 - 0.7^n)/(1 - 0.7) = 11*(1 - 0.7^n)/0.3 ≈ 11*(1 - 0.7^n)/0.3Compute 11 / 0.3 ≈ 36.6667So, Sum1 ≈ 36.6667*(1 - 0.7^n)Second, Sum2:Sum2 = 0.2 * Sum_{k=0}^{n-1} k * 0.7^{n-1 -k}Again, let m = n-1 -k, so k = n-1 -mWhen k=0, m = n-1; when k = n-1, m=0.So, Sum2 = 0.2 * Sum_{m=0}^{n-1} (n-1 - m) * 0.7^{m}= 0.2 * [ (n-1) Sum_{m=0}^{n-1} 0.7^{m} - Sum_{m=0}^{n-1} m*0.7^{m} ]We can compute these sums.First, Sum_{m=0}^{n-1} 0.7^{m} = (1 - 0.7^n)/0.3Second, Sum_{m=0}^{n-1} m*0.7^{m} is a standard sum. The formula for Sum_{m=0}^{N} m r^m is r(1 - (N+1) r^N + N r^{N+1}) / (1 - r)^2So, here, r = 0.7, N = n-1.Therefore,Sum_{m=0}^{n-1} m*0.7^{m} = 0.7*(1 - n*0.7^{n-1} + (n-1)*0.7^n ) / (1 - 0.7)^2Compute denominator: (1 - 0.7)^2 = 0.3^2 = 0.09So,Sum = 0.7*(1 - n*0.7^{n-1} + (n-1)*0.7^n ) / 0.09Simplify numerator:0.7*(1 - n*0.7^{n-1} + (n-1)*0.7^n ) = 0.7 - 0.7n*0.7^{n-1} + 0.7(n-1)*0.7^n= 0.7 - n*0.7^{n} + (n-1)*0.7^{n+1}So, Sum = [0.7 - n*0.7^{n} + (n-1)*0.7^{n+1}] / 0.09Therefore, going back to Sum2:Sum2 = 0.2 * [ (n-1)*(1 - 0.7^n)/0.3 - [0.7 - n*0.7^{n} + (n-1)*0.7^{n+1}]/0.09 ]Let me compute each part step by step.First term inside the brackets: (n-1)*(1 - 0.7^n)/0.3Second term: [0.7 - n*0.7^{n} + (n-1)*0.7^{n+1}]/0.09So, let's compute:Sum2 = 0.2 * [ (n-1)*(1 - 0.7^n)/0.3 - (0.7 - n*0.7^{n} + (n-1)*0.7^{n+1}) / 0.09 ]Let me factor out 1/0.09 from both terms:Note that 1/0.3 = 10/3 ≈ 3.3333, and 1/0.09 ≈ 11.1111.But perhaps better to write in fractions:0.3 = 3/10, so 1/0.3 = 10/30.09 = 9/100, so 1/0.09 = 100/9So,Sum2 = 0.2 * [ (n-1)*(1 - 0.7^n)*(10/3) - (0.7 - n*0.7^{n} + (n-1)*0.7^{n+1})*(100/9) ]Compute each part:First part: (n-1)*(1 - 0.7^n)*(10/3)Second part: (0.7 - n*0.7^{n} + (n-1)*0.7^{n+1})*(100/9)So, putting it all together:Sum2 = 0.2 * [ (10/3)(n - 1)(1 - 0.7^n) - (100/9)(0.7 - n*0.7^n + (n - 1)*0.7^{n+1}) ]This is getting quite involved. Maybe it's better to compute this numerically for n=14.But before that, let me see if I can find a pattern or simplify.Alternatively, perhaps it's better to compute Eₙ step by step for each day up to n=14, since n=14 is manageable.Given that, maybe we can compute E₁, E₂, ..., E₁₄ step by step.Given that, let's see:We have E₀ = 50For each day n from 0 to 13, compute E_{n+1} = 0.7 E_n + 2 t_n + 1.5 S_nGiven t_n = 1 + 0.1nS_n = 6 + 0.5 cos(π n /7)So, let's compute each term step by step.Let me create a table for n from 0 to 13:Compute E₁ to E₁₄.Let me start:n=0:E₀ = 50t₀ = 1 + 0.1*0 = 1S₀ = 6 + 0.5 cos(0) = 6 + 0.5*1 = 6.5E₁ = 0.7*50 + 2*1 + 1.5*6.5Compute:0.7*50 = 352*1 = 21.5*6.5 = 9.75So, E₁ = 35 + 2 + 9.75 = 46.75n=1:t₁ = 1 + 0.1*1 = 1.1S₁ = 6 + 0.5 cos(π*1/7) ≈ 6 + 0.5 cos(π/7) ≈ 6 + 0.5*0.9009688679 ≈ 6 + 0.450484434 ≈ 6.450484434E₂ = 0.7*46.75 + 2*1.1 + 1.5*6.450484434Compute:0.7*46.75 ≈ 32.7252*1.1 = 2.21.5*6.450484434 ≈ 9.675726651Sum: 32.725 + 2.2 + 9.675726651 ≈ 44.60072665n=2:t₂ = 1 + 0.1*2 = 1.2S₂ = 6 + 0.5 cos(2π/7) ≈ 6 + 0.5*0.623489802 ≈ 6 + 0.311744901 ≈ 6.311744901E₃ = 0.7*44.60072665 + 2*1.2 + 1.5*6.311744901Compute:0.7*44.60072665 ≈ 31.220508662*1.2 = 2.41.5*6.311744901 ≈ 9.467617352Sum: 31.22050866 + 2.4 + 9.467617352 ≈ 43.08812601n=3:t₃ = 1 + 0.1*3 = 1.3S₃ = 6 + 0.5 cos(3π/7) ≈ 6 + 0.5*0.222520934 ≈ 6 + 0.111260467 ≈ 6.111260467E₄ = 0.7*43.08812601 + 2*1.3 + 1.5*6.111260467Compute:0.7*43.08812601 ≈ 30.161688212*1.3 = 2.61.5*6.111260467 ≈ 9.166890701Sum: 30.16168821 + 2.6 + 9.166890701 ≈ 41.92857891n=4:t₄ = 1 + 0.1*4 = 1.4S₄ = 6 + 0.5 cos(4π/7) ≈ 6 + 0.5*(-0.222520934) ≈ 6 - 0.111260467 ≈ 5.888739533E₅ = 0.7*41.92857891 + 2*1.4 + 1.5*5.888739533Compute:0.7*41.92857891 ≈ 29.350005242*1.4 = 2.81.5*5.888739533 ≈ 8.8331093Sum: 29.35000524 + 2.8 + 8.8331093 ≈ 40.98311454n=5:t₅ = 1 + 0.1*5 = 1.5S₅ = 6 + 0.5 cos(5π/7) ≈ 6 + 0.5*(-0.623489803) ≈ 6 - 0.311744901 ≈ 5.688255099E₆ = 0.7*40.98311454 + 2*1.5 + 1.5*5.688255099Compute:0.7*40.98311454 ≈ 28.688180182*1.5 = 31.5*5.688255099 ≈ 8.532382649Sum: 28.68818018 + 3 + 8.532382649 ≈ 39.22056283n=6:t₆ = 1 + 0.1*6 = 1.6S₆ = 6 + 0.5 cos(6π/7) ≈ 6 + 0.5*(-0.9009688679) ≈ 6 - 0.450484434 ≈ 5.549515566E₇ = 0.7*39.22056283 + 2*1.6 + 1.5*5.549515566Compute:0.7*39.22056283 ≈ 27.4543942*1.6 = 3.21.5*5.549515566 ≈ 8.324273349Sum: 27.454394 + 3.2 + 8.324273349 ≈ 38.97866735n=7:t₇ = 1 + 0.1*7 = 1.7S₇ = 6 + 0.5 cos(7π/7) = 6 + 0.5 cos(π) = 6 + 0.5*(-1) = 6 - 0.5 = 5.5E₈ = 0.7*38.97866735 + 2*1.7 + 1.5*5.5Compute:0.7*38.97866735 ≈ 27.285067152*1.7 = 3.41.5*5.5 = 8.25Sum: 27.28506715 + 3.4 + 8.25 ≈ 38.93506715n=8:t₈ = 1 + 0.1*8 = 1.8S₈ = 6 + 0.5 cos(8π/7) ≈ 6 + 0.5 cos(π + π/7) = 6 + 0.5*(-cos(π/7)) ≈ 6 - 0.5*0.9009688679 ≈ 6 - 0.450484434 ≈ 5.549515566E₉ = 0.7*38.93506715 + 2*1.8 + 1.5*5.549515566Compute:0.7*38.93506715 ≈ 27.254547012*1.8 = 3.61.5*5.549515566 ≈ 8.324273349Sum: 27.25454701 + 3.6 + 8.324273349 ≈ 39.17882036n=9:t₉ = 1 + 0.1*9 = 1.9S₉ = 6 + 0.5 cos(9π/7) ≈ 6 + 0.5 cos(π + 2π/7) ≈ 6 - 0.5*cos(2π/7) ≈ 6 - 0.5*0.623489803 ≈ 6 - 0.311744901 ≈ 5.688255099E₁₀ = 0.7*39.17882036 + 2*1.9 + 1.5*5.688255099Compute:0.7*39.17882036 ≈ 27.425174252*1.9 = 3.81.5*5.688255099 ≈ 8.532382649Sum: 27.42517425 + 3.8 + 8.532382649 ≈ 39.7575569n=10:t₁₀ = 1 + 0.1*10 = 2.0S₁₀ = 6 + 0.5 cos(10π/7) ≈ 6 + 0.5 cos(π + 3π/7) ≈ 6 - 0.5*cos(3π/7) ≈ 6 - 0.5*0.222520934 ≈ 6 - 0.111260467 ≈ 5.888739533E₁₁ = 0.7*39.7575569 + 2*2.0 + 1.5*5.888739533Compute:0.7*39.7575569 ≈ 27.830289832*2.0 = 41.5*5.888739533 ≈ 8.8331093Sum: 27.83028983 + 4 + 8.8331093 ≈ 40.66339913n=11:t₁₁ = 1 + 0.1*11 = 2.1S₁₁ = 6 + 0.5 cos(11π/7) ≈ 6 + 0.5 cos(π + 4π/7) ≈ 6 - 0.5*cos(4π/7) ≈ 6 - 0.5*(-0.222520934) ≈ 6 + 0.111260467 ≈ 6.111260467E₁₂ = 0.7*40.66339913 + 2*2.1 + 1.5*6.111260467Compute:0.7*40.66339913 ≈ 28.464379392*2.1 = 4.21.5*6.111260467 ≈ 9.166890701Sum: 28.46437939 + 4.2 + 9.166890701 ≈ 41.83127009n=12:t₁₂ = 1 + 0.1*12 = 2.2S₁₂ = 6 + 0.5 cos(12π/7) ≈ 6 + 0.5 cos(2π - 2π/7) ≈ 6 + 0.5*cos(2π/7) ≈ 6 + 0.5*0.623489803 ≈ 6 + 0.311744901 ≈ 6.311744901E₁₃ = 0.7*41.83127009 + 2*2.2 + 1.5*6.311744901Compute:0.7*41.83127009 ≈ 29.281889062*2.2 = 4.41.5*6.311744901 ≈ 9.467617352Sum: 29.28188906 + 4.4 + 9.467617352 ≈ 43.14950641n=13:t₁₃ = 1 + 0.1*13 = 2.3S₁₃ = 6 + 0.5 cos(13π/7) ≈ 6 + 0.5 cos(2π - π/7) ≈ 6 + 0.5*cos(π/7) ≈ 6 + 0.5*0.9009688679 ≈ 6 + 0.450484434 ≈ 6.450484434E₁₄ = 0.7*43.14950641 + 2*2.3 + 1.5*6.450484434Compute:0.7*43.14950641 ≈ 30.204654492*2.3 = 4.61.5*6.450484434 ≈ 9.675726651Sum: 30.20465449 + 4.6 + 9.675726651 ≈ 44.48038114So, after computing each day step by step, E₁₄ ≈ 44.48038114Rounding to a reasonable decimal place, say two decimal places: 44.48But let me check my calculations to ensure I didn't make any errors.Looking back:At n=0: E₁=46.75n=1: E₂≈44.6007n=2: E₃≈43.0881n=3: E₄≈41.9286n=4: E₅≈40.9831n=5: E₆≈39.2206n=6: E₇≈38.9787n=7: E₈≈38.9351n=8: E₉≈39.1788n=9: E₁₀≈39.7576n=10: E₁₁≈40.6634n=11: E₁₂≈41.8313n=12: E₁₃≈43.1495n=13: E₁₄≈44.4804Yes, seems consistent. The energy level fluctuates but overall seems to be increasing after a certain point, which makes sense because as tₙ increases, the contribution from tai chi becomes more significant.So, after 14 days, E₁₄ ≈ 44.48But let me check if I made any calculation errors in the last step:E₁₄ = 0.7*43.14950641 + 2*2.3 + 1.5*6.450484434Compute each term:0.7*43.14950641 = 30.204654492*2.3 = 4.61.5*6.450484434 = 9.675726651Sum: 30.20465449 + 4.6 = 34.80465449 + 9.675726651 = 44.48038114Yes, correct.So, the energy level after 14 days is approximately 44.48.But let me see if I can represent this more accurately. Since all the computations were approximate, but perhaps we can carry more decimal places.But given the context, two decimal places should suffice.Therefore, the final answer is approximately 44.48.Final AnswerSub-problem 1: The general form of ( E_n ) is ( boxed{50 (0.7)^n + sum_{k=0}^{n-1} 0.7^{n-1 - k} (2 t_k + 1.5 S_k)} ).Sub-problem 2: Samantha's energy level after 14 days is ( boxed{44.48} ).</think>

answer:<think>Okay, so I have this problem about a cognitive development index (CDI) that follows a logistic growth model. The mother is evaluating an early childhood education program, and she wants to know if it's effective and safe. There are two sub-problems here, so I'll tackle them one by one.Starting with Sub-problem 1: I need to calculate the CDI at t = 3 months and t = 12 months. The formula given is:[ I(t) = frac{L}{1 + e^{-k(t - t_0)}} ]They provided the values: L is 100, k is 0.5, and t0 is 6 months. So, plugging these into the equation, it becomes:[ I(t) = frac{100}{1 + e^{-0.5(t - 6)}} ]Alright, so for t = 3 months, let's compute that. First, compute the exponent part: -0.5*(3 - 6). That would be -0.5*(-3) which is 1.5. So, e^1.5. Hmm, e is approximately 2.71828, so e^1.5 is about 4.4817. Then, the denominator becomes 1 + 4.4817 = 5.4817. So, I(3) is 100 divided by 5.4817. Let me calculate that: 100 / 5.4817 ≈ 18.24. So, approximately 18.24.Wait, that seems low. Let me double-check. The exponent is -0.5*(3 - 6) = -0.5*(-3) = 1.5. So, e^1.5 is indeed about 4.4817. Then, 1 + 4.4817 is 5.4817. 100 divided by that is approximately 18.24. Yeah, that seems correct. So, at 3 months, the CDI is about 18.24.Now, for t = 12 months. Let's compute the exponent: -0.5*(12 - 6) = -0.5*6 = -3. So, e^-3. e^-3 is approximately 0.0498. Then, the denominator is 1 + 0.0498 = 1.0498. So, I(12) is 100 / 1.0498 ≈ 95.26. Hmm, that seems high, but considering it's after 12 months, which is 6 months after the inflection point, so it's on the steeper part of the logistic curve. So, yeah, 95.26 makes sense.Wait, actually, the logistic curve is symmetric around the inflection point. So, at t0 = 6, the CDI is L/2, which is 50. Then, as t increases beyond 6, it grows rapidly towards 100. So, at t = 12, which is 6 months after the inflection point, it's almost at the maximum. So, 95.26 is reasonable.So, Sub-problem 1 seems done. Now, moving on to Sub-problem 2.The mother is concerned about risks. The study says that for every 10 point increase in CDI above 50, the risk of an adverse effect increases by 5%. The base risk is 10% when CDI is 50. So, I need to formulate an expression for R(I), the risk as a function of CDI I.Let me parse this. When I = 50, R = 10%. For every 10 points above 50, the risk increases by 5%. So, if CDI is 60, the risk is 10% + 5% = 15%. If it's 70, it's 10% + 10% = 20%, and so on.So, the risk increases by 5% for each 10 points above 50. So, the increase in risk is 5% per 10 points. So, the formula would be:R(I) = 10% + (I - 50)/10 * 5%But let me write that in mathematical terms. Let's convert percentages to decimals for calculation purposes. 10% is 0.10, 5% is 0.05.So,R(I) = 0.10 + ((I - 50)/10) * 0.05Simplify that:R(I) = 0.10 + (I - 50) * 0.005Alternatively, we can write it as:R(I) = 0.10 + 0.005*(I - 50)So, that's the expression. Let me test it with I = 50. Then, R = 0.10 + 0.005*(0) = 0.10, which is 10%. Good. For I = 60, R = 0.10 + 0.005*(10) = 0.10 + 0.05 = 0.15, which is 15%. Correct. So, the formula seems correct.Now, the question is to calculate the risk when CDI is 70. So, plug I = 70 into the formula.R(70) = 0.10 + 0.005*(70 - 50) = 0.10 + 0.005*20 = 0.10 + 0.10 = 0.20, which is 20%.So, the risk is 20% when CDI is 70.Wait, let me make sure. So, each 10 points above 50 adds 5%. So, 70 is 20 points above 50, which is 2 increments of 10 points. So, 2*5% = 10%, added to the base 10%, gives 20%. Yep, that's correct.So, summarizing:Sub-problem 1:At t = 3 months, CDI ≈ 18.24At t = 12 months, CDI ≈ 95.26Sub-problem 2:Risk function R(I) = 0.10 + 0.005*(I - 50)At I = 70, R = 20%I think that's all.Final AnswerSub-problem 1: The CDI at 3 months is boxed{18.24} and at 12 months is boxed{95.26}.Sub-problem 2: The risk when the CDI is 70 is boxed{20%}.</think>