answer:<think>Alright, so I have this problem where Dr. Smith needs to allocate research funds between pure and applied mathematics. The total fund is F dollars. I need to figure out how to split this between x (pure) and y (applied) to maximize the combined satisfaction, which is given by S_p(x) = ln(x + 1) and S_a(y) = sqrt(y). First, I should set up the problem as an optimization problem. The goal is to maximize S_p(x) + S_a(y) subject to the constraint x + y = F. That makes sense because all the funds have to be allocated, so x plus y can't exceed F, and since they have to use all the funds, it's exactly F.So, the function to maximize is ln(x + 1) + sqrt(y), with the constraint x + y = F. Since we have a constraint, I think I can use substitution to reduce it to a single variable problem. Let me express y in terms of x: y = F - x. Then, substitute that into the satisfaction function.So, the combined satisfaction becomes S(x) = ln(x + 1) + sqrt(F - x). Now, I need to find the value of x that maximizes S(x). To do this, I should take the derivative of S with respect to x, set it equal to zero, and solve for x.Let's compute the derivative S'(x). The derivative of ln(x + 1) is 1/(x + 1), and the derivative of sqrt(F - x) is (1/(2*sqrt(F - x)))*(-1). So, putting it together:S'(x) = 1/(x + 1) - 1/(2*sqrt(F - x)).Set this equal to zero for maximization:1/(x + 1) = 1/(2*sqrt(F - x)).Now, I need to solve for x. Let me write that equation again:1/(x + 1) = 1/(2*sqrt(F - x)).Cross-multiplying gives:2*sqrt(F - x) = x + 1.Now, let's square both sides to eliminate the square root:[2*sqrt(F - x)]^2 = (x + 1)^24*(F - x) = x^2 + 2x + 1.Expanding the left side:4F - 4x = x^2 + 2x + 1.Bring all terms to one side:x^2 + 2x + 1 + 4x - 4F = 0x^2 + 6x + 1 - 4F = 0.Wait, that seems a bit off. Let me double-check my expansion:Left side after squaring: 4F - 4x.Right side: (x + 1)^2 = x^2 + 2x + 1.So, bringing everything to the left:4F - 4x - x^2 - 2x - 1 = 0- x^2 - 6x + 4F - 1 = 0.Multiplying both sides by -1 to make it more standard:x^2 + 6x - 4F + 1 = 0.So, the quadratic equation is x^2 + 6x + (1 - 4F) = 0.Now, solving for x using the quadratic formula:x = [-6 ± sqrt(36 - 4*1*(1 - 4F))]/2x = [-6 ± sqrt(36 - 4 + 16F)]/2x = [-6 ± sqrt(32 + 16F)]/2x = [-6 ± sqrt(16*(2 + F))]/2x = [-6 ± 4*sqrt(2 + F)]/2x = (-6/2) ± (4/2)*sqrt(2 + F)x = -3 ± 2*sqrt(2 + F).Since x represents the amount of funds allocated, it must be non-negative. Therefore, we discard the negative solution:x = -3 + 2*sqrt(2 + F).But wait, we need to ensure that x is positive. Let's check:Since sqrt(2 + F) is always positive, and 2*sqrt(2 + F) is greater than 3 for F > ( (3/2)^2 - 2 ) = (9/4 - 2) = 1/4. So, for F > 1/4, x is positive. If F is less than or equal to 1/4, x might be negative, which doesn't make sense. Hmm, but F is the total fund, so it's likely a positive number, but we should consider the domain.But in the context, F is a total research fund, so it's a positive number, but we need to ensure that x is non-negative. So, let's see:x = -3 + 2*sqrt(2 + F) >= 0So, 2*sqrt(2 + F) >= 3sqrt(2 + F) >= 3/2Square both sides:2 + F >= 9/4F >= 9/4 - 2 = 1/4.So, if F >= 1/4, x is non-negative. If F < 1/4, then x would be negative, which isn't feasible, so in that case, the optimal x would be 0, and y would be F.But since the problem doesn't specify the value of F, I think we can proceed with the solution x = -3 + 2*sqrt(2 + F), and y = F - x = F - (-3 + 2*sqrt(2 + F)) = F + 3 - 2*sqrt(2 + F).Wait, let me compute y:y = F - x = F - (-3 + 2*sqrt(2 + F)) = F + 3 - 2*sqrt(2 + F).But we need to ensure that y is also non-negative. Let's check:y = F + 3 - 2*sqrt(2 + F) >= 0Let me see if this holds. Let's denote sqrt(2 + F) as t, so t = sqrt(2 + F), which implies t^2 = 2 + F, so F = t^2 - 2.Substituting into y:y = (t^2 - 2) + 3 - 2t = t^2 + 1 - 2t = (t - 1)^2.Since (t - 1)^2 is always non-negative, y is always non-negative. So, that's good.Therefore, the optimal allocation is:x = -3 + 2*sqrt(2 + F)y = (sqrt(2 + F) - 1)^2.Wait, let me verify that:From y = (t - 1)^2 where t = sqrt(2 + F), so y = (sqrt(2 + F) - 1)^2.Yes, that's correct.So, that's the solution for part 1.Now, moving on to part 2. The university administration imposes a condition that the funding for applied mathematics research should be at least twice the funding for pure mathematics research. So, y >= 2x.We need to reformulate the optimization problem with this additional constraint and find the new optimal x and y.So, the constraints now are:x + y = Fy >= 2xand x >= 0, y >= 0.So, we need to maximize S(x) = ln(x + 1) + sqrt(F - x) subject to y >= 2x, which translates to F - x >= 2x, so F >= 3x, or x <= F/3.So, x is now bounded above by F/3.Therefore, the feasible region for x is 0 <= x <= F/3.So, we need to check if the previous optimal x (from part 1) is within this feasible region. If it is, then that's still the optimal solution. If not, then the maximum occurs at the boundary, which is x = F/3.So, let's compute x from part 1: x = -3 + 2*sqrt(2 + F).We need to check if x <= F/3.So, set up the inequality:-3 + 2*sqrt(2 + F) <= F/3.Let me solve for F:2*sqrt(2 + F) <= F/3 + 3Divide both sides by 2:sqrt(2 + F) <= (F/3 + 3)/2Let me denote t = sqrt(2 + F), so t >= sqrt(2).Then, t <= ( (t^2 - 2)/3 + 3 ) / 2Simplify the right side:(t^2 - 2)/3 + 3 = (t^2 - 2 + 9)/3 = (t^2 + 7)/3So, t <= (t^2 + 7)/(6)Multiply both sides by 6:6t <= t^2 + 7Bring all terms to one side:t^2 - 6t + 7 >= 0Solve the quadratic inequality t^2 - 6t + 7 >= 0.The discriminant is 36 - 28 = 8, so roots are t = [6 ± sqrt(8)]/2 = [6 ± 2*sqrt(2)]/2 = 3 ± sqrt(2).So, the quadratic is positive outside the interval (3 - sqrt(2), 3 + sqrt(2)).Since t = sqrt(2 + F) >= sqrt(2) ≈ 1.414, and 3 - sqrt(2) ≈ 1.586, which is greater than sqrt(2). So, for t >= 3 - sqrt(2), the inequality t^2 - 6t + 7 >= 0 holds when t >= 3 + sqrt(2) or t <= 3 - sqrt(2). But since t >= sqrt(2) ≈ 1.414, and 3 - sqrt(2) ≈ 1.586, which is just slightly larger than sqrt(2), so for t >= 3 - sqrt(2), the inequality holds when t >= 3 + sqrt(2).Therefore, the inequality sqrt(2 + F) <= (F/3 + 3)/2 holds only when sqrt(2 + F) >= 3 + sqrt(2), which is not possible because sqrt(2 + F) is increasing with F, but 3 + sqrt(2) is a constant. Wait, this seems a bit confusing.Alternatively, perhaps it's better to consider specific values. Let's test for F.Suppose F is such that x from part 1 is less than or equal to F/3. Then, the optimal x remains the same. Otherwise, we have to set x = F/3.So, let's find the threshold F where x = F/3.Set x = -3 + 2*sqrt(2 + F) = F/3.Solve for F:-3 + 2*sqrt(2 + F) = F/3Multiply both sides by 3:-9 + 6*sqrt(2 + F) = FBring -9 to the right:6*sqrt(2 + F) = F + 9Square both sides:36*(2 + F) = (F + 9)^272 + 36F = F^2 + 18F + 81Bring all terms to one side:F^2 + 18F + 81 - 72 - 36F = 0F^2 - 18F + 9 = 0Solve using quadratic formula:F = [18 ± sqrt(324 - 36)]/2 = [18 ± sqrt(288)]/2 = [18 ± 12*sqrt(2)]/2 = 9 ± 6*sqrt(2).Since F must be positive, we take the positive root:F = 9 + 6*sqrt(2).So, when F = 9 + 6*sqrt(2), x from part 1 equals F/3.Therefore, for F <= 9 + 6*sqrt(2), the optimal x from part 1 is greater than F/3, which violates the constraint y >= 2x. Therefore, in this case, the optimal x must be set to F/3, and y = 2F/3.Wait, no, because y = F - x, so if x = F/3, then y = 2F/3, which satisfies y = 2x.But wait, actually, if x = F/3, then y = F - F/3 = 2F/3, which is exactly twice x, so y = 2x.Therefore, for F <= 9 + 6*sqrt(2), the optimal x from part 1 would exceed F/3, which is not allowed, so we have to set x = F/3.But wait, let me think again. The threshold F is 9 + 6*sqrt(2). So, if F is less than this value, then x from part 1 would be greater than F/3, which is not allowed, so we have to set x = F/3.If F is greater than or equal to 9 + 6*sqrt(2), then x from part 1 is less than or equal to F/3, so it's feasible, and we can use that solution.Therefore, the optimal solution depends on the value of F.So, summarizing:If F >= 9 + 6*sqrt(2), then the optimal x is -3 + 2*sqrt(2 + F), and y = F - x.If F < 9 + 6*sqrt(2), then the optimal x is F/3, and y = 2F/3.But let me verify this with an example.Suppose F = 10, which is less than 9 + 6*sqrt(2) ≈ 9 + 8.485 ≈ 17.485.So, for F = 10, the optimal x from part 1 would be:x = -3 + 2*sqrt(2 + 10) = -3 + 2*sqrt(12) ≈ -3 + 2*3.464 ≈ -3 + 6.928 ≈ 3.928.But F/3 ≈ 3.333. So, 3.928 > 3.333, which violates the constraint y >= 2x. Therefore, we have to set x = F/3 ≈ 3.333, y = 6.666.But wait, let's compute the satisfaction for both cases.Case 1: x = 3.928, y = 6.072.S = ln(3.928 + 1) + sqrt(6.072) ≈ ln(4.928) + sqrt(6.072) ≈ 1.6 + 2.464 ≈ 4.064.Case 2: x = 3.333, y = 6.666.S = ln(4.333) + sqrt(6.666) ≈ 1.466 + 2.582 ≈ 4.048.So, in this case, the satisfaction is slightly higher when x = 3.928, but it violates the constraint. Therefore, we have to choose x = 3.333, even though it gives a slightly lower satisfaction.Wait, but that seems counterintuitive. If the constraint reduces the satisfaction, but it's a requirement, so we have to comply.Alternatively, perhaps I made a mistake in the threshold calculation.Wait, when F = 9 + 6*sqrt(2), x from part 1 equals F/3.Let me compute x when F = 9 + 6*sqrt(2):x = -3 + 2*sqrt(2 + 9 + 6*sqrt(2)) = -3 + 2*sqrt(11 + 6*sqrt(2)).Let me compute sqrt(11 + 6*sqrt(2)).Note that (sqrt(2) + 3)^2 = 2 + 6*sqrt(2) + 9 = 11 + 6*sqrt(2). So, sqrt(11 + 6*sqrt(2)) = sqrt(2) + 3.Therefore, x = -3 + 2*(sqrt(2) + 3) = -3 + 2*sqrt(2) + 6 = 3 + 2*sqrt(2).Now, F = 9 + 6*sqrt(2), so F/3 = 3 + 2*sqrt(2), which matches x. So, indeed, when F = 9 + 6*sqrt(2), x from part 1 equals F/3.Therefore, for F >= 9 + 6*sqrt(2), x from part 1 is feasible, so we can use it.For F < 9 + 6*sqrt(2), x from part 1 would exceed F/3, so we have to set x = F/3.Therefore, the optimal solution is:If F >= 9 + 6*sqrt(2):x = -3 + 2*sqrt(2 + F)y = F - x = 3 + 2*sqrt(2 + F) - 2*sqrt(2 + F) + 2*sqrt(2 + F) ?Wait, no, let me compute y correctly.Wait, y = F - x = F - (-3 + 2*sqrt(2 + F)) = F + 3 - 2*sqrt(2 + F).Alternatively, since x = -3 + 2*sqrt(2 + F), then y = F - x = F + 3 - 2*sqrt(2 + F).But earlier, I expressed y as (sqrt(2 + F) - 1)^2, which is correct because:(sqrt(2 + F) - 1)^2 = (2 + F) - 2*sqrt(2 + F) + 1 = F + 3 - 2*sqrt(2 + F), which matches y.So, y = (sqrt(2 + F) - 1)^2.Therefore, the optimal solution is:If F >= 9 + 6*sqrt(2):x = -3 + 2*sqrt(2 + F)y = (sqrt(2 + F) - 1)^2Else:x = F/3y = 2F/3.So, that's the solution for part 2.</think>

answer:<think>Okay, so I have this problem where my co-worker is analyzing workplace safety incidents over time using a differential equation. The first part is to find the general solution for the differential equation given. Let me write it down again:[frac{dI(t)}{dt} + kI(t) = C sin(omega t)]with the initial condition ( I(0) = I_0 ). Hmm, this looks like a linear first-order differential equation. I remember that for such equations, we can use an integrating factor to solve them. The standard form is:[frac{dy}{dt} + P(t)y = Q(t)]So, in this case, ( P(t) = k ) and ( Q(t) = C sin(omega t) ). The integrating factor ( mu(t) ) is given by:[mu(t) = e^{int P(t) dt} = e^{int k dt} = e^{kt}]Multiplying both sides of the differential equation by the integrating factor:[e^{kt} frac{dI(t)}{dt} + k e^{kt} I(t) = C e^{kt} sin(omega t)]The left side of the equation is the derivative of ( e^{kt} I(t) ) with respect to t. So, we can write:[frac{d}{dt} [e^{kt} I(t)] = C e^{kt} sin(omega t)]Now, to find ( I(t) ), we need to integrate both sides with respect to t:[e^{kt} I(t) = int C e^{kt} sin(omega t) dt + D]Where D is the constant of integration. So, the integral on the right side is the tricky part. I need to compute:[int e^{kt} sin(omega t) dt]I recall that integrals involving products of exponentials and sine or cosine functions can be solved using integration by parts twice and then solving for the integral. Alternatively, I can use a formula for such integrals. Let me try the integration by parts method.Let me set:Let ( u = sin(omega t) ), so ( du = omega cos(omega t) dt )Let ( dv = e^{kt} dt ), so ( v = frac{1}{k} e^{kt} )Then, integration by parts gives:[int e^{kt} sin(omega t) dt = frac{e^{kt}}{k} sin(omega t) - frac{omega}{k} int e^{kt} cos(omega t) dt]Now, I need to compute the integral ( int e^{kt} cos(omega t) dt ). Let me use integration by parts again.Let ( u = cos(omega t) ), so ( du = -omega sin(omega t) dt )Let ( dv = e^{kt} dt ), so ( v = frac{1}{k} e^{kt} )Thus,[int e^{kt} cos(omega t) dt = frac{e^{kt}}{k} cos(omega t) + frac{omega}{k} int e^{kt} sin(omega t) dt]Now, substitute this back into the previous equation:[int e^{kt} sin(omega t) dt = frac{e^{kt}}{k} sin(omega t) - frac{omega}{k} left( frac{e^{kt}}{k} cos(omega t) + frac{omega}{k} int e^{kt} sin(omega t) dt right )]Simplify the right-hand side:[int e^{kt} sin(omega t) dt = frac{e^{kt}}{k} sin(omega t) - frac{omega e^{kt}}{k^2} cos(omega t) - frac{omega^2}{k^2} int e^{kt} sin(omega t) dt]Now, let me collect the integral terms on the left side:[int e^{kt} sin(omega t) dt + frac{omega^2}{k^2} int e^{kt} sin(omega t) dt = frac{e^{kt}}{k} sin(omega t) - frac{omega e^{kt}}{k^2} cos(omega t)]Factor out the integral:[left( 1 + frac{omega^2}{k^2} right ) int e^{kt} sin(omega t) dt = frac{e^{kt}}{k} sin(omega t) - frac{omega e^{kt}}{k^2} cos(omega t)]Simplify the left side:[frac{k^2 + omega^2}{k^2} int e^{kt} sin(omega t) dt = frac{e^{kt}}{k} sin(omega t) - frac{omega e^{kt}}{k^2} cos(omega t)]Multiply both sides by ( frac{k^2}{k^2 + omega^2} ):[int e^{kt} sin(omega t) dt = frac{k e^{kt}}{k^2 + omega^2} sin(omega t) - frac{omega e^{kt}}{k^2 + omega^2} cos(omega t) + D]Wait, actually, since we already accounted for the constant of integration earlier, I think we can write it as:[int e^{kt} sin(omega t) dt = frac{e^{kt}}{k^2 + omega^2} (k sin(omega t) - omega cos(omega t)) + D]So, going back to our equation:[e^{kt} I(t) = C cdot frac{e^{kt}}{k^2 + omega^2} (k sin(omega t) - omega cos(omega t)) + D]We can divide both sides by ( e^{kt} ):[I(t) = frac{C}{k^2 + omega^2} (k sin(omega t) - omega cos(omega t)) + D e^{-kt}]Now, we need to apply the initial condition ( I(0) = I_0 ). Let's plug in t = 0:[I(0) = frac{C}{k^2 + omega^2} (0 - omega cdot 1) + D e^{0} = I_0]Simplify:[- frac{C omega}{k^2 + omega^2} + D = I_0]Solving for D:[D = I_0 + frac{C omega}{k^2 + omega^2}]Therefore, the general solution is:[I(t) = frac{C}{k^2 + omega^2} (k sin(omega t) - omega cos(omega t)) + left( I_0 + frac{C omega}{k^2 + omega^2} right ) e^{-kt}]Hmm, let me check if this makes sense. As t approaches infinity, the term with ( e^{-kt} ) will go to zero, assuming k is positive, which it should be because it's a decay constant. So, the long-term behavior is dominated by the steady-state solution, which is the sinusoidal part.So, the general solution is a transient exponential decay plus a steady-state oscillation.Alright, that seems correct.Now, moving on to part 2. The co-worker modifies the differential equation to:[frac{dI(t)}{dt} + (k + alpha) I(t) = C sin(omega t)]where ( alpha ) is a positive constant. We need to analyze how increasing ( alpha ) affects the long-term behavior of ( I(t) ). Specifically, derive the steady-state solution and discuss the impact of increasing ( alpha ).So, similar to part 1, this is a linear first-order differential equation. The only difference is that the coefficient of I(t) is now ( k + alpha ) instead of k. So, let's denote ( k' = k + alpha ), so the equation becomes:[frac{dI(t)}{dt} + k' I(t) = C sin(omega t)]Following the same steps as in part 1, the integrating factor is ( e^{k' t} ). Multiplying through:[e^{k' t} frac{dI(t)}{dt} + k' e^{k' t} I(t) = C e^{k' t} sin(omega t)]Which is the derivative of ( e^{k' t} I(t) ):[frac{d}{dt} [e^{k' t} I(t)] = C e^{k' t} sin(omega t)]Integrate both sides:[e^{k' t} I(t) = int C e^{k' t} sin(omega t) dt + D]Again, the integral is similar to before, but with ( k' ) instead of k. So, following the same integration by parts method, the integral becomes:[int e^{k' t} sin(omega t) dt = frac{e^{k' t}}{(k')^2 + omega^2} (k' sin(omega t) - omega cos(omega t)) + D]Therefore, plugging back into the equation:[e^{k' t} I(t) = C cdot frac{e^{k' t}}{(k')^2 + omega^2} (k' sin(omega t) - omega cos(omega t)) + D]Divide both sides by ( e^{k' t} ):[I(t) = frac{C}{(k')^2 + omega^2} (k' sin(omega t) - omega cos(omega t)) + D e^{-k' t}]Applying the initial condition ( I(0) = I_0 ):[I(0) = frac{C}{(k')^2 + omega^2} (0 - omega cdot 1) + D = I_0]Simplify:[- frac{C omega}{(k')^2 + omega^2} + D = I_0]Solving for D:[D = I_0 + frac{C omega}{(k')^2 + omega^2}]So, the general solution is:[I(t) = frac{C}{(k')^2 + omega^2} (k' sin(omega t) - omega cos(omega t)) + left( I_0 + frac{C omega}{(k')^2 + omega^2} right ) e^{-k' t}]Again, as t approaches infinity, the exponential term ( e^{-k' t} ) goes to zero, so the steady-state solution is:[I_{ss}(t) = frac{C}{(k')^2 + omega^2} (k' sin(omega t) - omega cos(omega t))]Alternatively, this can be written in amplitude-phase form. Let me compute the amplitude:The amplitude of the steady-state solution is:[A = frac{C}{sqrt{(k')^2 + omega^2}}]So, the amplitude is inversely proportional to ( sqrt{(k')^2 + omega^2} ). Since ( k' = k + alpha ), increasing ( alpha ) increases ( k' ), which in turn increases the denominator, thus decreasing the amplitude A.Therefore, increasing ( alpha ) reduces the amplitude of the steady-state oscillations, meaning that the number of incidents will oscillate with a smaller magnitude around the equilibrium. This implies that the new safety protocol is effective in reducing the number of incidents over time, especially in the long term.Additionally, the exponential term ( e^{-k' t} ) decays faster when ( k' ) is larger (i.e., when ( alpha ) is larger), meaning the transient part of the solution diminishes more quickly, leading to a quicker convergence to the steady-state solution.So, in summary, increasing ( alpha ) leads to a smaller amplitude in the steady-state solution and a faster decay of the transient solution, both of which result in a lower number of incidents in the long term.Final Answer1. The general solution is: [ boxed{I(t) = frac{C}{k^2 + omega^2}(k sin(omega t) - omega cos(omega t)) + left(I_0 + frac{C omega}{k^2 + omega^2}right)e^{-kt}} ]2. The steady-state solution is: [ boxed{I_{ss}(t) = frac{C}{(k + alpha)^2 + omega^2}((k + alpha) sin(omega t) - omega cos(omega t))} ] Increasing ( alpha ) decreases the amplitude of the steady-state solution, thereby reducing the number of incidents over time.</think>

answer:<think>Alright, so I have this problem about Alex, a basketball player who's struggling with self-confidence and fear of failure. His coach suggested working on free throw accuracy, and Alex wants to model his performance using a binomial distribution. There are two main parts to this problem: calculating the probability of making exactly 8 out of 10 free throws, and then figuring out the expected number and variance if he attempts an infinite number of free throws. Let me break this down step by step.First, let's tackle part a: calculating the probability of making exactly 8 successful free throws out of 10 attempts. I remember that the binomial distribution is used when there are a fixed number of independent trials, each with two possible outcomes (success or failure), and the probability of success is constant. In this case, each free throw is a trial, making a basket is a success, and missing is a failure. The probability of success, p, is given as 60%, or 0.6. The number of trials, n, is 10, and we want exactly k=8 successes.The formula for the binomial probability is:P(k) = C(n, k) * p^k * (1-p)^(n-k)Where C(n, k) is the combination of n things taken k at a time. I think this is also written as "n choose k." So, I need to compute this combination first.Calculating C(10, 8): This is the number of ways to choose 8 successes out of 10 attempts. The formula for combinations is:C(n, k) = n! / (k!(n - k)!)So plugging in the numbers:C(10, 8) = 10! / (8! * (10 - 8)!) = 10! / (8! * 2!) I know that 10! is 10 factorial, which is 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1, but since 8! is in the denominator, a lot of terms will cancel out.Let me compute it step by step:10! = 10 × 9 × 8!So, 10! / 8! = 10 × 9 = 90Then, 90 / (2!) = 90 / 2 = 45So, C(10, 8) is 45.Now, moving on to the rest of the formula:p^k = (0.6)^8(1 - p)^(n - k) = (0.4)^(10 - 8) = (0.4)^2So, putting it all together:P(8) = 45 * (0.6)^8 * (0.4)^2I need to compute (0.6)^8 and (0.4)^2.Let me calculate (0.6)^8 first. I can do this step by step:0.6^1 = 0.60.6^2 = 0.360.6^3 = 0.2160.6^4 = 0.12960.6^5 = 0.077760.6^6 = 0.0466560.6^7 = 0.02799360.6^8 = 0.01679616So, (0.6)^8 is approximately 0.01679616.Next, (0.4)^2 is straightforward:0.4 * 0.4 = 0.16So, now multiply all the parts together:45 * 0.01679616 * 0.16First, multiply 45 and 0.01679616:45 * 0.01679616 = ?Let me compute 45 * 0.01679616.0.01679616 * 45:First, 0.01 * 45 = 0.450.00679616 * 45:Compute 0.006 * 45 = 0.270.00079616 * 45 ≈ 0.0358272Adding them together: 0.45 + 0.27 = 0.72; 0.72 + 0.0358272 ≈ 0.7558272So, approximately 0.7558272.Now, multiply this by 0.16:0.7558272 * 0.16Let me compute 0.7558272 * 0.16:First, 0.7 * 0.16 = 0.1120.0558272 * 0.16 ≈ 0.008932352Adding them together: 0.112 + 0.008932352 ≈ 0.120932352So, approximately 0.120932352.Therefore, the probability of making exactly 8 out of 10 free throws is approximately 0.1209, or 12.09%.Wait, let me verify that calculation because 0.7558272 * 0.16 seems a bit off. Let me compute it more accurately.0.7558272 * 0.16:Break it down:0.7558272 * 0.1 = 0.075582720.7558272 * 0.06 = 0.045349632Adding them together: 0.07558272 + 0.045349632 = 0.120932352Yes, that's correct. So, approximately 0.120932352, which is about 12.09%.So, rounding to four decimal places, that's approximately 0.1209.Alternatively, if I use a calculator, 45 * (0.6)^8 * (0.4)^2.Let me compute (0.6)^8 again:0.6^2 = 0.360.6^4 = (0.36)^2 = 0.12960.6^8 = (0.1296)^2 ≈ 0.01679616(0.4)^2 = 0.16So, 45 * 0.01679616 = 0.75582720.7558272 * 0.16 = 0.120932352So, yes, 0.120932352, which is approximately 12.09%.So, part a is approximately 12.09%.Now, moving on to part b: determining the expected number and variance of successful free throws if Alex continues to attempt free throws indefinitely.Hmm, so if Alex attempts an infinite number of free throws, what is the expected number and variance?Wait, the binomial distribution is for a finite number of trials. If we have an infinite number of trials, we might be moving into the realm of the Poisson distribution or something else, but actually, in probability theory, as the number of trials n approaches infinity, and the probability p approaches zero in such a way that np remains constant, we get the Poisson distribution. However, in this case, Alex is keeping p constant at 0.6, and n is going to infinity. So, what happens to the expected value and variance?Wait, for a binomial distribution, the expected value is n*p, and the variance is n*p*(1-p). So, if n approaches infinity, then the expected value would also approach infinity, and the variance would approach infinity as well, since both are linear in n.But that seems a bit too straightforward. Maybe the question is referring to the long-term behavior in terms of the law of large numbers? Or perhaps it's expecting the expected value and variance per free throw, but that doesn't make much sense because as n increases, both expectation and variance increase.Wait, let me read the question again: "If Alex attempts an infinite number of free throws, what is the expected number of successful free throws and the variance of the number of successful free throws?"So, it's still a binomial distribution with n approaching infinity. So, E[X] = n*p, which would be infinity, and Var(X) = n*p*(1-p), which would also be infinity.But that seems trivial. Maybe the question is misworded, or perhaps it's expecting the limit as n approaches infinity, but in that case, both expectation and variance tend to infinity.Alternatively, maybe it's referring to the probability distribution in the limit, but for a binomial distribution with n approaching infinity and p fixed, it doesn't converge to any particular distribution except in the case where p approaches zero appropriately.Wait, another thought: perhaps it's referring to the expected value and variance per attempt, but that would just be p and p*(1-p), respectively, which are 0.6 and 0.24. But the question says "the expected number of successful free throws and the variance of the number of successful free throws" when attempting an infinite number. So, if n is infinite, then the expected number would be infinite, and the variance would also be infinite.But that seems a bit too straightforward, so maybe I'm missing something.Alternatively, perhaps the question is referring to the proportion of successful free throws, rather than the number. In that case, as n approaches infinity, the proportion would converge to p, which is 0.6, and the variance of the proportion would be p*(1-p)/n, which would approach zero. But the question specifically says "the number of successful free throws," not the proportion.So, given that, I think the answer is that the expected number is infinity, and the variance is also infinity.But maybe the question is expecting the answer in terms of the limit of the binomial distribution as n approaches infinity, but since p is fixed, it doesn't converge to any particular distribution, unless p approaches zero as n increases.Wait, another angle: perhaps it's referring to the Poisson binomial distribution, but no, that's when each trial has a different probability.Alternatively, maybe it's a typo, and they meant to say "a large number" instead of "infinite number," but since it's specified as infinite, I think we have to go with that.Alternatively, perhaps they are referring to the expected value and variance per free throw, but that would be 0.6 and 0.24, but the question says "the number of successful free throws," which is a count, not a proportion.So, to clarify, if n is finite, say n=10, then E[X] = 10*0.6=6, Var(X)=10*0.6*0.4=2.4.If n is infinite, then E[X] = infinity, Var(X)=infinity.Alternatively, if we consider the limit as n approaches infinity, the distribution doesn't converge to a proper distribution, but rather, the mean and variance both go to infinity.Therefore, I think the answer is that the expected number is infinity, and the variance is also infinity.But let me think again: is there another interpretation? Maybe they are asking about the expected value and variance per free throw, but that would be 0.6 and 0.24, but the wording is about the number of successful free throws when attempting infinitely many, so it's about the total count, not per attempt.Therefore, I think the answer is that both the expected number and variance are infinite.Alternatively, if we consider that as n approaches infinity, the distribution becomes approximately normal with mean mu = np and variance sigma^2 = np(1-p). But even then, as n approaches infinity, mu and sigma^2 both approach infinity.So, in conclusion, for part b, the expected number is infinity, and the variance is also infinity.But let me check if there's another way to interpret it. Maybe they are asking about the expected value and variance of the proportion, but the question specifically says "the number of successful free throws," so it's about the count, not the proportion.Therefore, I think the answer is that the expected number is infinity, and the variance is also infinity.Alternatively, maybe they are expecting the answer in terms of the limit of the binomial coefficients or something else, but I don't think so.So, to summarize:a. The probability of making exactly 8 out of 10 free throws is approximately 12.09%.b. The expected number of successful free throws is infinity, and the variance is also infinity.But wait, let me double-check part a. Maybe I made a calculation error.Let me compute 45 * (0.6)^8 * (0.4)^2 again.First, (0.6)^8: 0.6^2=0.36, 0.6^4=0.1296, 0.6^8=0.01679616.(0.4)^2=0.16.So, 45 * 0.01679616 = 0.7558272.0.7558272 * 0.16 = 0.120932352.Yes, that's correct. So, approximately 0.1209, or 12.09%.Alternatively, using a calculator, 45 * 0.01679616 * 0.16 = 0.120932352.So, that's correct.Therefore, the answers are:a. Approximately 12.09%b. Expected number: infinity, Variance: infinityBut wait, maybe the question is referring to the limit as n approaches infinity, but in terms of the distribution, it doesn't converge, so the mean and variance both go to infinity.Alternatively, if we consider the Central Limit Theorem, as n increases, the distribution becomes approximately normal with mean mu and variance sigma^2, but as n approaches infinity, mu and sigma^2 both approach infinity.Therefore, I think the answer is as above.Final Answera. The probability is boxed{0.1209}.b. The expected number is boxed{infty} and the variance is boxed{infty}.</think>

answer:<think>First, I need to determine the amount of each type of flour required to make 1 kilogram of the gluten-free blend. The blend consists of rice flour, almond flour, and tapioca starch in the proportions of 40%, 35%, and 25%, respectively.For the first part, I'll calculate each component:- Rice flour: 40% of 1000 grams is 400 grams.- Almond flour: 35% of 1000 grams is 350 grams.- Tapioca starch: 25% of 1000 grams is 250 grams.Next, for the second part, the recipe requires 500 grams of the gluten-free flour blend. The seeds mixture should be 10% of this amount, which is 50 grams. The seeds are mixed in a 3:2 ratio of chia seeds to flax seeds.To find the exact weights:- The total parts of the ratio are 3 + 2 = 5 parts.- Each part is 50 grams divided by 5, which is 10 grams per part.- Chia seeds: 3 parts × 10 grams = 30 grams.- Flax seeds: 2 parts × 10 grams = 20 grams.This ensures the correct proportions of both the flour blend and the seeds mixture for the recipe.</think>

answer:<think>Okay, so I have this problem about optimizing a marketing campaign for a luxury condominium in Singapore. Let me try to break it down step by step. First, the problem is divided into two sub-problems. The first one is calculating the total potential revenue from selling all 150 units. The second one is figuring out the minimum marketing budget needed to achieve a 90% probability of selling each unit and then calculating the expected revenue from that.Starting with the first sub-problem: total potential revenue. I know the average price per square foot is S4,500, and each unit is 1,200 square feet. So, to find the price per unit, I should multiply the price per square foot by the size of each unit. That makes sense because if each square foot is S4,500, then 1,200 square feet would be 4,500 multiplied by 1,200. Let me write that down: 4,500 * 1,200. Hmm, 4,500 times 1,000 is 4,500,000, and 4,500 times 200 is 900,000. So adding those together, that's 4,500,000 + 900,000 = 5,400,000. So each unit is priced at S5,400,000. Now, since there are 150 units, the total potential revenue would be 150 multiplied by 5,400,000. Let me calculate that. 150 * 5,400,000. Well, 100 * 5,400,000 is 540,000,000, and 50 * 5,400,000 is 270,000,000. Adding those together gives 540,000,000 + 270,000,000 = 810,000,000. So the total potential revenue is S810,000,000. That seems straightforward.Moving on to the second sub-problem: determining the minimum marketing budget required to achieve a 90% probability of selling each unit. The data says that for every S10,000 spent on marketing, the probability of selling a unit increases by 0.5%, up to a maximum of 95%. Wait, so the probability starts at some base rate, and each S10,000 increases it by 0.5%. But the problem doesn't specify what the base probability is. Hmm, that's a bit confusing. It just says that for every S10,000, the probability increases by 0.5%, up to 95%. So I think we can assume that without any marketing, the probability is 0%, but that might not make sense. Alternatively, maybe the base probability is 0%, and each S10,000 adds 0.5%, but you can't go beyond 95%. But actually, the problem says "for every S10,000 spent on marketing, the probability of selling a unit increases by 0.5%, up to a maximum probability of 95%." So it doesn't specify the starting point. Maybe we can assume that without any marketing, the probability is 0%, and each S10,000 adds 0.5% until it reaches 95%. So to get to 90%, how much do we need to spend? Let's see. Each S10,000 gives 0.5%, so how many increments of 0.5% do we need to reach 90%? Let me calculate: 90% divided by 0.5% per S10,000. So 90 / 0.5 = 180. So 180 increments of S10,000. Therefore, the total marketing budget would be 180 * 10,000 = S1,800,000. But wait, the maximum probability is 95%, so 90% is within the limit. So that should be the minimum budget required. Now, calculating the expected revenue. Expected revenue is the probability of selling each unit multiplied by the revenue per unit, summed over all units. Since each unit has a 90% chance of being sold, and there are 150 units, the expected number of units sold is 150 * 0.9 = 135 units. Each unit is sold at S5,400,000, so the expected revenue is 135 * 5,400,000. Let me compute that. 100 * 5,400,000 is 540,000,000, and 35 * 5,400,000 is... 35 times 5,400,000. 30 * 5,400,000 is 162,000,000, and 5 * 5,400,000 is 27,000,000. Adding those together gives 162,000,000 + 27,000,000 = 189,000,000. So total expected revenue is 540,000,000 + 189,000,000 = 729,000,000. Wait, but is that correct? Because the expected revenue per unit is 0.9 * 5,400,000, so per unit it's 4,860,000. Then, 150 units would be 150 * 4,860,000. Let me compute that: 100 * 4,860,000 = 486,000,000, and 50 * 4,860,000 = 243,000,000. Adding those gives 486,000,000 + 243,000,000 = 729,000,000. Yes, that's the same result. So that seems consistent.But wait, the marketing budget is S1,800,000. Should we subtract that from the expected revenue to get the net expected revenue? The problem says "calculate the expected revenue from the sales if this marketing budget is utilized." It doesn't specify whether to subtract the marketing cost or not. It just asks for the expected revenue, so I think we just need to report the expected sales revenue, not net profit. So S729,000,000 is the expected revenue.Let me just recap to make sure I didn't miss anything. For the first part, total potential revenue is 150 units * (4,500 * 1,200) = 150 * 5,400,000 = 810,000,000. That seems right.For the second part, to get 90% probability, since each S10,000 gives 0.5%, we need 90 / 0.5 = 180 increments, so 180 * 10,000 = 1,800,000. Then, expected units sold is 150 * 0.9 = 135, so revenue is 135 * 5,400,000 = 729,000,000. I think that's all. I don't see any mistakes in the calculations. The key was understanding how the marketing budget translates into probability and then applying that to find the expected revenue.</think>

answer:<think>Okay, so I have this problem where I'm an Irish aerospace engineer working for NASA, and I need to design a trajectory for a spacecraft that does a gravity assist around Earth before heading to Mars. There are two parts to this problem: one involving orbital mechanics and another dealing with relativistic effects. Let me try to tackle each part step by step.Starting with the first sub-problem about orbital mechanics. The spacecraft is in a circular low Earth orbit (LEO) at an altitude of 300 km. I need to calculate the velocity change (∆v) required to transition from this LEO to a hyperbolic escape trajectory. The given parameters are Earth's gravitational parameter (μ = 3.986 x 10^14 m³/s²) and Earth's radius (R_E = 6371 km). Hmm, okay. So, I remember that for a spacecraft in a circular orbit, the velocity can be calculated using the formula v = sqrt(μ / r), where r is the radius of the orbit. Since the spacecraft is at an altitude of 300 km, I need to add that to Earth's radius to get the total orbital radius. Let me convert everything into meters for consistency. Earth's radius is 6371 km, which is 6,371,000 meters. The altitude is 300 km, which is 300,000 meters. So, the total orbital radius r is 6,371,000 + 300,000 = 6,671,000 meters.Now, calculating the velocity in LEO: v = sqrt(μ / r). Plugging in the numbers, that's sqrt(3.986e14 / 6.671e6). Let me compute that. First, 3.986e14 divided by 6.671e6 is approximately 5.976e7. Taking the square root of that gives me about 7,730 m/s. Wait, that seems a bit high. Let me double-check. Oh, no, wait, 6.671e6 is 6,671,000 meters. So, 3.986e14 divided by 6.671e6 is indeed approximately 5.976e7. The square root of 5.976e7 is sqrt(59,760,000). Let me compute that. The square root of 59,760,000 is approximately 7,730 m/s. Yeah, that's correct because typical LEO velocities are around 7.8 km/s, so 7.73 km/s is close enough.Now, to transition to a hyperbolic escape trajectory, the spacecraft needs to achieve the escape velocity from Earth. The escape velocity from Earth is given by v_escape = sqrt(2 * μ / r). So, let's compute that. Using the same r of 6,671,000 meters, 2 * μ is 7.972e14. Divided by 6.671e6 gives approximately 1.195e8. Taking the square root of that gives about 10,930 m/s. Wait, that can't be right because the escape velocity from Earth is known to be around 11.2 km/s at the surface, but since we're at an altitude, it should be slightly less. Hmm, maybe I made a calculation error.Wait, let's recalculate. 2 * μ is 2 * 3.986e14 = 7.972e14. Divided by 6.671e6 is 7.972e14 / 6.671e6 ≈ 1.195e8. The square root of 1.195e8 is sqrt(119,500,000) ≈ 10,930 m/s. But wait, escape velocity at Earth's surface is about 11.2 km/s, so at 300 km altitude, it should be a bit less. Maybe my formula is correct, but perhaps I should consider that the escape velocity is sqrt(2) times the circular orbit velocity. Since the circular velocity was 7.73 km/s, then the escape velocity should be about 10.93 km/s, which is consistent. So, the required velocity for escape is approximately 10.93 km/s.But wait, the spacecraft is already moving at 7.73 km/s in LEO. So, the velocity change needed is the difference between the escape velocity and the current velocity. So, ∆v = v_escape - v_circular = 10.93 km/s - 7.73 km/s = 3.2 km/s. Is that right? Wait, no, because when transitioning to a hyperbolic trajectory, the required velocity is actually the escape velocity, but the spacecraft is already moving at the circular velocity. So, the ∆v needed is the difference. So, yes, 3.2 km/s. But wait, I think I might be missing something here. Because in orbital mechanics, the ∆v required to escape is actually the difference between the current velocity and the escape velocity, but depending on the direction of the burn. If the burn is prograde, then you add the velocity, so the ∆v is the difference. So, yes, 3.2 km/s.Wait, let me confirm. The formula for ∆v to escape is indeed v_escape - v_circular. So, 10.93 - 7.73 = 3.2 km/s. That seems correct.Okay, so that's the first part. Now, moving on to the second sub-problem about relativistic effects. Once the spacecraft reaches a velocity close to 0.1c (which is 0.1 times the speed of light, approximately 3 x 10^8 m/s), we need to calculate the time dilation experienced by the onboard clock relative to an Earth-based clock over a period of 1 year (Earth time). We need to use the Lorentz factor (γ) for this.The Lorentz factor is given by γ = 1 / sqrt(1 - v²/c²). So, first, let's compute v. 0.1c is 0.1 * 3e8 m/s = 3e7 m/s. So, v = 3e7 m/s, c = 3e8 m/s. So, v² = (3e7)^2 = 9e14 m²/s². c² = (3e8)^2 = 9e16 m²/s². So, v²/c² = 9e14 / 9e16 = 1e-2 = 0.01. Therefore, 1 - v²/c² = 0.99. So, sqrt(0.99) is approximately 0.994987. Therefore, γ = 1 / 0.994987 ≈ 1.005038.So, the Lorentz factor γ is approximately 1.005. This means that time on the spacecraft will pass slower than on Earth by a factor of γ. So, for every second on Earth, the spacecraft clock will have elapsed 1/γ seconds. Wait, no, actually, it's the other way around. From Earth's perspective, the spacecraft clock is running slower. So, if a year passes on Earth, the spacecraft clock will have experienced less time.Wait, let me clarify. The Lorentz factor γ is the factor by which time dilates. So, if γ > 1, then the moving clock (spacecraft) runs slower. So, the time experienced by the spacecraft (t') is t = t_earth / γ. So, over 1 year (Earth time), the spacecraft clock will have experienced t' = 1 year / γ.But wait, actually, the formula is t' = γ t. Wait, no, I think I'm getting confused. Let me recall. The Lorentz transformation for time is t' = γ (t - (v x)/c²). But in the case where the spacecraft is moving at velocity v relative to Earth, and we're considering the time experienced by the spacecraft, it's t' = t / γ. So, if Earth time is t, then spacecraft time is t' = t / γ.Wait, no, actually, it's the other way around. The moving clock runs slower. So, for a given Earth time t, the spacecraft clock will have elapsed t' = t / γ. So, if γ is 1.005, then t' = t / 1.005 ≈ 0.995 t. So, over 1 year Earth time, the spacecraft clock would have experienced approximately 0.995 years, which is about 0.995 * 365 days ≈ 363 days. So, the time dilation is about 2 days less over a year.Wait, but let me compute it more accurately. γ is approximately 1.005038. So, t' = t / γ = 1 year / 1.005038 ≈ 0.995038 years. So, the difference is 1 - 0.995038 = 0.004962 years. Converting that to days: 0.004962 * 365 ≈ 1.81 days. So, approximately 1.81 days less over a year.But wait, let me make sure I'm using the correct formula. The time experienced by the spacecraft is t' = t * sqrt(1 - v²/c²). Since γ = 1 / sqrt(1 - v²/c²), then t' = t / γ. So, yes, t' = t * sqrt(1 - v²/c²). So, sqrt(1 - 0.01) = sqrt(0.99) ≈ 0.994987. So, t' ≈ 0.994987 * t. So, over 1 year, t' ≈ 0.994987 years, which is about 0.994987 * 365 ≈ 362.17 days. So, the difference is about 2.83 days. Wait, that's conflicting with my earlier calculation. Hmm.Wait, let's compute it step by step. γ = 1.005038. So, t' = t / γ = 1 / 1.005038 ≈ 0.995038 years. So, 0.995038 * 365 ≈ 362.18 days. So, the spacecraft clock would have experienced approximately 362.18 days, while Earth has experienced 365 days. So, the difference is about 2.82 days. So, the spacecraft clock is behind by approximately 2.82 days after 1 Earth year.Alternatively, using the other formula: t' = t * sqrt(1 - v²/c²) = 1 * sqrt(0.99) ≈ 0.994987 years, which is the same as above. So, yes, approximately 2.82 days less.Wait, but let me check the exact value. sqrt(0.99) is approximately 0.994987437. So, 0.994987437 * 365 ≈ 362.17 days. So, the difference is 365 - 362.17 = 2.83 days. So, approximately 2.83 days of time dilation over 1 year.But wait, I think I might have made a mistake in the calculation of γ. Let me recalculate γ. v = 0.1c, so v²/c² = 0.01. So, 1 - v²/c² = 0.99. sqrt(0.99) ≈ 0.994987. So, γ = 1 / 0.994987 ≈ 1.005038. So, that's correct.Therefore, the time experienced by the spacecraft is t' = t / γ = 1 / 1.005038 ≈ 0.995038 years. So, 0.995038 * 365 ≈ 362.18 days. So, the spacecraft clock is behind by about 2.82 days after 1 year.Alternatively, using the approximation for small velocities, the time dilation can be approximated as Δt ≈ (v²)/(2c²) * t. So, v = 0.1c, so v² = 0.01c². Therefore, Δt ≈ (0.01c²)/(2c²) * t = 0.005 * t. So, over 1 year, Δt ≈ 0.005 * 365 ≈ 1.825 days. But this is an approximation and gives a smaller value. The exact calculation gives about 2.83 days, so the approximation is less accurate.Therefore, the exact time dilation experienced is approximately 2.83 days over 1 year. So, the spacecraft clock would be behind by about 2.83 days compared to Earth-based clocks.Wait, but let me make sure I'm interpreting this correctly. The spacecraft is moving at 0.1c relative to Earth. From Earth's frame, the spacecraft's clock is running slower. So, over 1 year on Earth, the spacecraft's clock would have advanced less. So, the time experienced by the spacecraft is t' = t * sqrt(1 - v²/c²) ≈ 0.994987 * 365 ≈ 362.17 days. So, the difference is about 2.83 days.Alternatively, if we consider the spacecraft's frame, they would see Earth's clocks as running slower, but that's a different perspective. However, the problem asks for the time dilation experienced by the onboard clock relative to Earth-based clocks, so we're considering Earth's frame, where the spacecraft is moving, so the spacecraft's clock is slower.So, to sum up, the velocity change required for the escape trajectory is approximately 3.2 km/s, and the time dilation over 1 year is about 2.83 days.Wait, but let me double-check the first part again because I'm a bit unsure. The ∆v required to escape from Earth from a circular orbit is indeed the difference between the escape velocity and the circular velocity. So, v_escape = sqrt(2) * v_circular. Therefore, ∆v = v_escape - v_circular = v_circular (sqrt(2) - 1). So, v_circular is 7.73 km/s, so ∆v = 7.73 * (1.4142 - 1) ≈ 7.73 * 0.4142 ≈ 3.2 km/s. Yes, that's correct.Alternatively, using the formula for ∆v to escape from a circular orbit: ∆v = sqrt(μ / r) * (sqrt(2) - 1). So, sqrt(μ / r) is the circular velocity, which we calculated as 7.73 km/s. So, ∆v = 7.73 * (1.4142 - 1) ≈ 3.2 km/s. So, that's consistent.Therefore, the answers are:1. ∆v ≈ 3.2 km/s2. Time dilation ≈ 2.83 days over 1 year.But let me express these more precisely.For the first part, let's compute it more accurately.Given:μ = 3.986e14 m³/s²R_E = 6371 km = 6,371,000 mAltitude = 300 km = 300,000 mSo, r = 6,371,000 + 300,000 = 6,671,000 mv_circular = sqrt(μ / r) = sqrt(3.986e14 / 6.671e6) = sqrt(5.976e7) ≈ 7,730 m/sv_escape = sqrt(2 * μ / r) = sqrt(2 * 3.986e14 / 6.671e6) = sqrt(1.195e8) ≈ 10,930 m/s∆v = v_escape - v_circular = 10,930 - 7,730 = 3,200 m/s = 3.2 km/sFor the second part:v = 0.1c = 3e7 m/sc = 3e8 m/sv²/c² = (3e7)^2 / (3e8)^2 = 9e14 / 9e16 = 1e-2 = 0.01γ = 1 / sqrt(1 - 0.01) = 1 / sqrt(0.99) ≈ 1.005038Time experienced by spacecraft: t' = t / γ = 1 year / 1.005038 ≈ 0.995038 yearsConvert 0.995038 years to days: 0.995038 * 365 ≈ 362.17 daysDifference: 365 - 362.17 ≈ 2.83 daysSo, the spacecraft clock would be behind by approximately 2.83 days after 1 year.Alternatively, using the exact value of sqrt(0.99):sqrt(0.99) ≈ 0.994987437t' = 1 * 0.994987437 ≈ 0.994987437 years0.994987437 * 365 ≈ 362.17 daysDifference ≈ 2.83 daysSo, that's consistent.Therefore, the final answers are:1. ∆v ≈ 3.2 km/s2. Time dilation ≈ 2.83 days over 1 year.</think>