🧠:<think>Okay, so I need to figure out the coefficients a, b, and c for the quadratic function f(t) = at² + bt + c. The function models the annual funding for veterans' services in Brainerd, Minnesota over the past decade. The data points given are:- In 2013, the funding was 150,000. Since t represents the number of years since 2013, that would be t = 0. So, f(0) = 150.- The funding peaked at 250,000 in 2018. 2018 is 5 years after 2013, so t = 5. Therefore, f(5) = 250.- In 2023, the funding decreased to 180,000. 2023 is 10 years after 2013, so t = 10. Thus, f(10) = 180.Alright, so I have three points: (0, 150), (5, 250), and (10, 180). Since it's a quadratic function, these three points should be sufficient to determine the coefficients a, b, and c.Let me write down the equations based on these points.First, for t = 0:f(0) = a*(0)² + b*(0) + c = c = 150. So, c = 150. That's straightforward.Next, for t = 5:f(5) = a*(5)² + b*(5) + c = 25a + 5b + c = 250.But since we already know c = 150, we can substitute that in:25a + 5b + 150 = 250.Subtract 150 from both sides:25a + 5b = 100.Let me simplify this equation by dividing both sides by 5:5a + b = 20. Let's call this Equation (1).Now, for t = 10:f(10) = a*(10)² + b*(10) + c = 100a + 10b + c = 180.Again, substitute c = 150:100a + 10b + 150 = 180.Subtract 150 from both sides:100a + 10b = 30.Divide both sides by 10:10a + b = 3. Let's call this Equation (2).Now, we have two equations:Equation (1): 5a + b = 20Equation (2): 10a + b = 3I can subtract Equation (1) from Equation (2) to eliminate b:(10a + b) - (5a + b) = 3 - 2010a + b - 5a - b = -175a = -17So, a = -17/5 = -3.4.Hmm, that seems a bit steep for a quadratic coefficient, but let's go with it.Now, substitute a = -17/5 into Equation (1):5*(-17/5) + b = 20Simplify:-17 + b = 20So, b = 20 + 17 = 37.Therefore, the coefficients are:a = -17/5, b = 37, c = 150.Let me write the function:f(t) = (-17/5)t² + 37t + 150.Wait, let me verify if this satisfies all three points.First, t = 0:f(0) = 0 + 0 + 150 = 150. Correct.t = 5:f(5) = (-17/5)*(25) + 37*5 + 150Calculate each term:(-17/5)*25 = (-17)*5 = -8537*5 = 185So, f(5) = -85 + 185 + 150 = (-85 + 185) + 150 = 100 + 150 = 250. Correct.t = 10:f(10) = (-17/5)*(100) + 37*10 + 150Calculate each term:(-17/5)*100 = (-17)*20 = -34037*10 = 370So, f(10) = -340 + 370 + 150 = (30) + 150 = 180. Correct.Alright, so the coefficients are correct.Now, moving on to the second part: predicting the year when the funding will return to the 2013 level, which is 150,000. So, we need to find t such that f(t) = 150.Given that f(t) = (-17/5)t² + 37t + 150, set this equal to 150:(-17/5)t² + 37t + 150 = 150.Subtract 150 from both sides:(-17/5)t² + 37t = 0.Factor out t:t*(-17/5 t + 37) = 0.So, the solutions are t = 0 and (-17/5 t + 37) = 0.We already know t = 0 is the starting point in 2013. The other solution is:-17/5 t + 37 = 0-17/5 t = -37Multiply both sides by (-5/17):t = (-37)*(-5/17) = (185)/17 ≈ 10.882.So, t ≈ 10.882 years after 2013.Since t = 10 is 2023, t = 10.882 would be approximately 10 years and 0.882 of a year. 0.882 of a year is roughly 0.882 * 12 ≈ 10.58 months, which is about October 2023. But since funding is annual, we might consider the next whole year, which would be 2024.But let me think: the function is quadratic, so after t ≈ 10.882, the funding would have returned to 150. So, in the year 2013 + 10.882 ≈ 2023.882, which is approximately October 2023. But since the funding is given annually, it might not exactly hit 150 in a specific year unless we consider the exact time.But the question says, "predict the year in which the funding will return to the 2013 level, assuming the quadratic trend continues."So, perhaps we can consider the exact t value and round it to the nearest whole number.t ≈ 10.882, which is approximately 11 years after 2013, so 2013 + 11 = 2024.But let's check f(10) and f(11):f(10) = 180, as given.f(11) = (-17/5)*(121) + 37*11 + 150Calculate each term:(-17/5)*121 = (-17)*24.2 = -411.437*11 = 407So, f(11) = -411.4 + 407 + 150 = (-4.4) + 150 = 145.6Hmm, that's actually less than 150. So, in 2024 (t=11), the funding is approximately 145,600, which is below the 2013 level.Wait, so the function crosses 150 at t ≈ 10.882, which is between t=10 and t=11. So, in the year 2023.882, which is approximately October 2023, the funding would be 150. But since the funding is annual, it's either 2023 or 2024.But in 2023, the funding is 180, which is higher than 150, and in 2024, it's 145.6, which is lower. So, the funding crosses 150 between 2023 and 2024.But the question is asking for the year when it returns to the 2013 level. Since it's a quadratic trend, it's symmetric around the vertex. Wait, the vertex is at t = -b/(2a). Let's calculate that.Vertex t = -b/(2a) = -37/(2*(-17/5)) = -37 / (-34/5) = (37)*(5/34) ≈ (185)/34 ≈ 5.441 years.So, the peak was at t ≈ 5.441, which is around May 2018, which makes sense because the peak was in 2018.Since the parabola is symmetric, the time it takes to go from t=0 to the vertex is about 5.441 years, and then the same time to come back down to the original level. So, the total time to return to the original level would be 2*5.441 ≈ 10.882 years, which is consistent with our earlier calculation.So, the funding returns to 150 at t ≈ 10.882, which is approximately 10 years and 10.58 months after 2013, so around October 2023. But since the funding is annual, it's not exactly a specific year. However, the question asks for the year, so we might need to consider when it would next be at 150. Since in 2023, it's 180, and in 2024, it's 145.6, which is below 150, so the funding would have returned to 150 in between 2023 and 2024. But since we can't have a fraction of a year in the answer, perhaps we can consider the next whole year after the crossing point, which would be 2024, but the funding is already below 150 in 2024. Alternatively, maybe we can say it returns to 150 in 2024, but it's actually below that. Hmm.Alternatively, perhaps the question expects us to solve for t when f(t) = 150, which is t ≈ 10.882, so 2013 + 10.882 ≈ 2023.882, which is approximately 2024. But since it's not a whole year, maybe we can say 2024, but it's actually below 150 in 2024. Alternatively, perhaps we can consider that the funding returns to 150 in 2024, but it's actually lower. Hmm, this is a bit confusing.Wait, let me think again. The function is quadratic, so after the vertex, it decreases. So, the funding was 150 in 2013, peaked in 2018, and then decreased. It passed through 150 again on the way down at t ≈ 10.882, which is 2023.882. So, in 2023, it's 180, and in 2024, it's 145.6. So, the funding would have been 150 in 2023.882, which is not a full year. So, perhaps the answer is 2024, but it's actually below 150 in that year. Alternatively, maybe the question expects us to round to the nearest whole year, which would be 2024.Alternatively, perhaps the question is considering the exact t value and converting it to a year, so 2013 + 10.882 ≈ 2023.882, which is approximately 2024. So, the answer would be 2024.But let me check the exact t value:t = 185/17 ≈ 10.88235294.So, 10.88235294 years after 2013 is 2013 + 10 + 0.88235294 years.0.88235294 years is approximately 0.88235294 * 12 ≈ 10.588 months, which is about 10 months and 18 days. So, adding that to January 2023, we get around November 2023.But since the funding is annual, it's not measured at that exact point. So, the funding in 2023 is 180, and in 2024, it's 145.6. So, the funding would have returned to 150 between 2023 and 2024, but since it's annual, it's not exactly a specific year. However, the question asks for the year, so perhaps we can consider the next whole year after the crossing point, which is 2024, even though the funding is below 150 in that year.Alternatively, maybe the question expects us to recognize that the funding returns to 150 in 2024, but it's actually lower. Hmm, this is a bit tricky.Wait, perhaps I made a mistake in calculating f(11). Let me double-check:f(11) = (-17/5)*(11)^2 + 37*11 + 15011 squared is 121.(-17/5)*121 = (-17)*24.2 = -411.437*11 = 407So, f(11) = -411.4 + 407 + 150 = (-4.4) + 150 = 145.6Yes, that's correct. So, in 2024, the funding is 145.6, which is below 150.So, the funding was 150 in 2013, went up to 250 in 2018, then came back down to 180 in 2023, and then to 145.6 in 2024. So, the funding returns to 150 between 2023 and 2024, but since it's annual, it's not exactly a year. So, perhaps the answer is 2024, but it's actually below 150 in that year. Alternatively, maybe the question expects us to consider the exact t value and say it's 2024, even though it's not exactly 150.Alternatively, perhaps the question is considering the exact t value and converting it to a year, so 2013 + 10.882 ≈ 2023.882, which is approximately 2024. So, the answer would be 2024.But let me think again: the function is quadratic, so it's symmetric around the vertex. The vertex is at t ≈ 5.441, so the time to go up to the peak is 5.441 years, and the time to come back down is another 5.441 years, totaling 10.882 years. So, the funding returns to 150 at t ≈ 10.882, which is 2023.882, so approximately 2024.Therefore, the answer is 2024.But to be precise, since the funding is 150 at t ≈ 10.882, which is 2023.882, so it's not a full year yet, but the next full year is 2024. So, perhaps the answer is 2024.Alternatively, if we consider that the funding returns to 150 in 2024, but it's actually lower, but since it's a quadratic trend, it's just a model, so maybe we can accept that.Alternatively, perhaps the question expects us to solve for t when f(t) = 150, which is t ≈ 10.882, so the year is 2013 + 10.882 ≈ 2023.882, which is approximately 2024.So, I think the answer is 2024.Final Answer1. The coefficients are ( a = -frac{17}{5} ), ( b = 37 ), and ( c = 150 ).2. The funding will return to the 2013 level in the year boxed{2024}.</think>

🧠:<think>Alright, so I have these two math problems here related to an astronaut's paintings. Let me try to tackle them one by one. I'll start with the first one about the elliptical orbit.Problem 1: The orbit is given by the equation (frac{x^2}{a^2} + frac{y^2}{b^2} = 1), which is the standard equation of an ellipse. The area of the orbit is 314 square units, and the eccentricity (e) is 0.6. I need to find the semi-major axis (a) and the semi-minor axis (b).First, I remember that the area (A) of an ellipse is given by (A = pi a b). So, if the area is 314, then:[pi a b = 314]I can write this as:[a b = frac{314}{pi}]Let me compute that. Since (pi) is approximately 3.1416, dividing 314 by 3.1416 gives:[a b approx frac{314}{3.1416} approx 100]So, (a b = 100). That's one equation.Next, the eccentricity (e) of an ellipse is given by (e = frac{c}{a}), where (c) is the distance from the center to a focus. Also, I recall that for an ellipse, (c^2 = a^2 - b^2). So, substituting (c = a e) into this equation:[(a e)^2 = a^2 - b^2]Which simplifies to:[a^2 e^2 = a^2 - b^2]Let me rearrange this to solve for (b^2):[b^2 = a^2 - a^2 e^2 = a^2 (1 - e^2)]Given that (e = 0.6), so (e^2 = 0.36). Therefore:[b^2 = a^2 (1 - 0.36) = a^2 (0.64)]So, (b = a sqrt{0.64} = a times 0.8). Therefore, (b = 0.8 a).Now, going back to the area equation:[a b = 100]Substituting (b = 0.8 a):[a times 0.8 a = 100][0.8 a^2 = 100][a^2 = frac{100}{0.8} = 125][a = sqrt{125} = 5 sqrt{5} approx 11.1803]So, (a approx 11.1803). Then, (b = 0.8 a = 0.8 times 11.1803 approx 8.9442).Let me check if this makes sense. The area should be (pi a b). Plugging in the approximate values:[pi times 11.1803 times 8.9442 approx 3.1416 times 100 = 314.16]Which is approximately 314, so that checks out. Also, the eccentricity (e = 0.6) should be correct because (c = a e approx 11.1803 times 0.6 approx 6.7082), and then (c^2 = 45), while (a^2 - b^2 = 125 - 71 = 54). Wait, that doesn't match. Hmm, maybe I made a mistake here.Wait, let me recalculate (b^2). If (b = 0.8 a), then (b^2 = 0.64 a^2). So, (a^2 - b^2 = a^2 - 0.64 a^2 = 0.36 a^2). Since (a^2 = 125), then (c^2 = 0.36 times 125 = 45), so (c = sqrt{45} approx 6.7082). Then, (e = c/a = 6.7082 / 11.1803 approx 0.6), which is correct. So, my calculations are consistent. I must have miscalculated earlier when I thought (a^2 - b^2 = 54). Actually, (a^2 = 125), (b^2 = 0.64 times 125 = 80), so (a^2 - b^2 = 125 - 80 = 45), which is correct. So, all is well.Therefore, the semi-major axis (a) is (5 sqrt{5}) units and the semi-minor axis (b) is (4 sqrt{5}) units because (0.8 a = 0.8 times 5 sqrt{5} = 4 sqrt{5}). Let me confirm:(5 sqrt{5} approx 11.1803) and (4 sqrt{5} approx 8.9442), which matches my earlier approximations. So, exact values are (a = 5 sqrt{5}) and (b = 4 sqrt{5}).Problem 2: Time dilation. The formula given is (Delta t' = Delta t sqrt{1 - frac{v^2}{c^2}}). The astronaut traveled at (v = 0.8c) for 150 days as measured on Earth. I need to find the time experienced by the astronaut, (Delta t').First, let's plug in the values. (Delta t = 150) days, (v = 0.8c), so:[Delta t' = 150 times sqrt{1 - left(frac{0.8c}{c}right)^2} = 150 times sqrt{1 - 0.64}][= 150 times sqrt{0.36} = 150 times 0.6 = 90]So, the astronaut experienced 90 days.Wait, that seems straightforward. Let me double-check. The gamma factor (gamma = frac{1}{sqrt{1 - v^2/c^2}} = frac{1}{sqrt{1 - 0.64}} = frac{1}{sqrt{0.36}} = frac{1}{0.6} approx 1.6667). So, time dilation means that the astronaut's time is shorter by a factor of gamma. So, (Delta t' = Delta t / gamma = 150 / (5/3) = 150 times (3/5) = 90). Yep, that's consistent.Alternatively, using the formula directly: (sqrt{1 - v^2/c^2} = 0.6), so 150 * 0.6 = 90. So, both ways, it's 90 days.Therefore, the astronaut experienced 90 days during the mission.Final Answer1. The semi-major axis is (boxed{5sqrt{5}}) and the semi-minor axis is (boxed{4sqrt{5}}).2. The time experienced by the astronaut is (boxed{90}) days.</think>

🧠:<think>Alright, so I have two problems here related to Dr. Alex's study on hurricanes and their effects. Let me take them one by one.Problem 1: Expected Number of HurricanesOkay, the first problem says that the expected number of hurricanes during a 6-month hurricane season is given by the integral of a sinusoidal function ( lambda(t) = A sin(Bt + C) + D ). I need to find this expected number.Hmm, I remember that in a Poisson process, the expected number of events in a given interval is the integral of the rate function over that interval. So, in this case, the expected number of hurricanes from time ( t = 0 ) to ( t = 6 ) months is the integral of ( lambda(t) ) from 0 to 6.So, mathematically, that would be:[E = int_{0}^{6} lambda(t) , dt = int_{0}^{6} left( A sin(Bt + C) + D right) dt]Alright, let's break this integral into two parts:1. The integral of ( A sin(Bt + C) ) from 0 to 6.2. The integral of ( D ) from 0 to 6.Starting with the first part:[int A sin(Bt + C) , dt]I recall that the integral of ( sin(ax + b) ) is ( -frac{1}{a} cos(ax + b) ). So, applying that here:[A int sin(Bt + C) , dt = A left( -frac{1}{B} cos(Bt + C) right) + text{constant}]So, evaluating from 0 to 6:[A left[ -frac{1}{B} cos(B cdot 6 + C) + frac{1}{B} cos(B cdot 0 + C) right]]Simplify that:[-frac{A}{B} left[ cos(6B + C) - cos(C) right]]Okay, so that's the first part.Now, the second part is integrating D from 0 to 6:[int_{0}^{6} D , dt = D cdot (6 - 0) = 6D]Putting it all together, the expected number of hurricanes is:[E = -frac{A}{B} left[ cos(6B + C) - cos(C) right] + 6D]Hmm, that seems right. Let me double-check the integral of the sine function. Yes, the integral of ( sin(Bt + C) ) is indeed ( -frac{1}{B} cos(Bt + C) ). So, the constants A and B are just scaling factors, so that part is correct.I think that's the answer for the first part. Let me write it neatly:[E = 6D - frac{A}{B} left[ cos(6B + C) - cos(C) right]]Problem 2: Expected Economic ImpactMoving on to the second problem. Dr. Alex models the economic impact ( I(x) ) of a hurricane of intensity ( x ) as ( I(x) = kx^n ), where ( k ) and ( n ) are constants. The intensity ( x ) follows a Gamma distribution with probability density function:[f(x) = frac{beta^alpha}{Gamma(alpha)} x^{alpha - 1} e^{-beta x} quad text{for } x > 0]I need to find the expected economic impact of a single hurricane. So, that would be the expected value of ( I(x) ), which is ( E[I(x)] = E[kx^n] = k E[x^n] ).So, I need to compute ( E[x^n] ) where ( x ) is Gamma distributed.I remember that for a Gamma distribution with parameters ( alpha ) and ( beta ), the expected value of ( x^k ) is given by:[E[x^k] = frac{Gamma(alpha + k)}{Gamma(alpha)} left( frac{1}{beta} right)^k]Is that right? Let me recall. The Gamma distribution has the property that if ( X sim text{Gamma}(alpha, beta) ), then ( E[X^k] = frac{Gamma(alpha + k)}{Gamma(alpha)} beta^{-k} ). Yes, that seems correct.So, in this case, ( k ) is just a constant multiplier in the economic impact function, and ( n ) is the exponent. So, substituting ( k = n ) in the formula above, we get:[E[x^n] = frac{Gamma(alpha + n)}{Gamma(alpha)} left( frac{1}{beta} right)^n]Therefore, the expected economic impact is:[E[I(x)] = k cdot frac{Gamma(alpha + n)}{Gamma(alpha)} left( frac{1}{beta} right)^n]Simplify that:[E[I(x)] = frac{k Gamma(alpha + n)}{Gamma(alpha) beta^n}]Hmm, that seems straightforward. Let me make sure I didn't mix up any parameters. The Gamma distribution is often parameterized in terms of shape ( alpha ) and rate ( beta ), so the expectation of ( x^n ) is indeed as I wrote.Alternatively, sometimes Gamma distributions are parameterized with scale parameter ( theta = 1/beta ). But in this case, the density is given as ( frac{beta^alpha}{Gamma(alpha)} x^{alpha - 1} e^{-beta x} ), which confirms that it's the rate parameterization, so ( beta ) is the rate, not the scale. So, the formula I used is correct.Therefore, the expected economic impact is:[E[I(x)] = frac{k Gamma(alpha + n)}{Gamma(alpha) beta^n}]Alternatively, this can be written using the property of the Gamma function ( Gamma(alpha + n) = (alpha + n - 1)(alpha + n - 2) cdots alpha Gamma(alpha) ), but unless specified, it's probably fine to leave it in terms of Gamma functions.So, summarizing:1. The expected number of hurricanes is ( 6D - frac{A}{B} [ cos(6B + C) - cos(C) ] ).2. The expected economic impact is ( frac{k Gamma(alpha + n)}{Gamma(alpha) beta^n} ).I think that's it. Let me just check if I interpreted the problems correctly.For the first problem, the expected number is the integral of the rate function over the period, which is straightforward. The integral of a sine function over an interval gives the difference in cosines scaled by the amplitude and frequency. The integral of the constant D is just D times the interval length, which is 6 months.For the second problem, since the economic impact is a function of the intensity, and intensity follows a Gamma distribution, the expectation is the integral of ( I(x) f(x) ) over all x, which simplifies using the known moments of the Gamma distribution.Yes, I think both answers are correct.Final Answer1. The expected number of hurricanes is boxed{6D - dfrac{A}{B} left[ cos(6B + C) - cos(C) right]}.2. The expected economic impact of a single hurricane is boxed{dfrac{k Gamma(alpha + n)}{Gamma(alpha) beta^n}}.</think>

🧠:<think>Alright, so I'm trying to help this race car driver optimize their car's performance by choosing the best combination of wheel material and tread type. There are two main factors here: rotational inertia of the wheels and the friction coefficient between the wheels and the track. Let me break this down step by step.First, I need to figure out the rotational inertia for both Material A and Material B. The formula given is ( I = frac{1}{2} m r^2 ). I know the radius ( r ) is 0.3 meters for both materials. But I don't have the mass ( m ) directly; instead, I have the densities of the materials. Material A has a density of 7800 kg/m³, and Material B is 4500 kg/m³. Both wheels have a uniform thickness of 0.05 meters.Hmm, okay, so I need to calculate the mass of each wheel. Since the wheels are cylinders, the volume ( V ) can be calculated using the formula for the volume of a cylinder: ( V = pi r^2 h ), where ( h ) is the thickness. Once I have the volume, I can find the mass by multiplying the volume by the density.Let me write that down:For Material A:1. Calculate the volume ( V_A = pi r^2 h )2. Then, mass ( m_A = rho_A times V_A )3. Finally, rotational inertia ( I_A = frac{1}{2} m_A r^2 )Same steps for Material B.Let me plug in the numbers.Radius ( r = 0.3 ) m, thickness ( h = 0.05 ) m.Calculating volume for both:( V = pi times (0.3)^2 times 0.05 )First, ( (0.3)^2 = 0.09 )Then, ( 0.09 times 0.05 = 0.0045 )So, ( V = pi times 0.0045 approx 3.1416 times 0.0045 approx 0.014137 ) m³Wait, that seems a bit small, but considering the thickness is only 0.05 meters, it might be correct.Now, for Material A:Mass ( m_A = 7800 times 0.014137 approx 7800 times 0.014137 )Calculating that:7800 * 0.01 = 787800 * 0.004137 ≈ 7800 * 0.004 = 31.2, and 7800 * 0.000137 ≈ 1.0686So total ≈ 78 + 31.2 + 1.0686 ≈ 110.2686 kgSimilarly, for Material B:( m_B = 4500 times 0.014137 approx 4500 times 0.014137 )Calculating:4500 * 0.01 = 454500 * 0.004137 ≈ 4500 * 0.004 = 18, and 4500 * 0.000137 ≈ 0.6165So total ≈ 45 + 18 + 0.6165 ≈ 63.6165 kgOkay, so Material A wheels are about 110.27 kg each, and Material B are about 63.62 kg each.Now, calculating rotational inertia:For Material A:( I_A = frac{1}{2} times 110.27 times (0.3)^2 )First, ( (0.3)^2 = 0.09 )So, ( I_A = 0.5 times 110.27 times 0.09 )Calculating 110.27 * 0.09 ≈ 9.9243Then, 0.5 * 9.9243 ≈ 4.96215 kg·m²For Material B:( I_B = frac{1}{2} times 63.6165 times 0.09 )63.6165 * 0.09 ≈ 5.7254850.5 * 5.725485 ≈ 2.86274 kg·m²So, Material A has a rotational inertia of approximately 4.96 kg·m², and Material B is about 2.86 kg·m². That means Material B has a lower rotational inertia, which is better for performance because lower inertia means easier to accelerate or decelerate, which is good for handling and speed changes.Alright, that's the first part done. Now, moving on to the friction coefficient.The friction coefficient ( mu = frac{F_f}{F_n} ). The normal force ( F_n ) is the weight of the car, which is mass times gravity. The car's mass is 1200 kg, so ( F_n = 1200 times 9.81 ) N.Calculating that:1200 * 9.81 = 11772 NSo, ( F_n = 11772 ) N.Now, for Tread X, the frictional force ( F_f ) is 4000 N, and for Tread Y, it's 4500 N.Calculating ( mu ) for both:For Tread X:( mu_X = frac{4000}{11772} approx 0.3396 )For Tread Y:( mu_Y = frac{4500}{11772} approx 0.382 )So, Tread Y has a higher friction coefficient, which means better grip on the track. That's good for traction and cornering.Putting it all together, the driver wants low rotational inertia for better performance and high friction coefficient for better grip. Therefore, the best combination would be Material B wheels (lower inertia) and Tread Y (higher friction coefficient).Wait, just to make sure I didn't make any calculation errors.Double-checking the volume:( V = pi r^2 h = pi (0.3)^2 (0.05) = pi * 0.09 * 0.05 = pi * 0.0045 ≈ 0.014137 ) m³. That seems correct.Mass for Material A: 7800 * 0.014137 ≈ 110.27 kg. Correct.Mass for Material B: 4500 * 0.014137 ≈ 63.62 kg. Correct.Inertia calculations:For A: 0.5 * 110.27 * 0.09 ≈ 4.96 kg·m². Correct.For B: 0.5 * 63.62 * 0.09 ≈ 2.86 kg·m². Correct.Friction coefficient:Normal force: 1200 * 9.81 = 11772 N. Correct.Tread X: 4000 / 11772 ≈ 0.3396. Correct.Tread Y: 4500 / 11772 ≈ 0.382. Correct.Yes, all calculations seem accurate. So, the conclusion is that Material B and Tread Y are the optimal choices.</think>