🧠:<think>Alright, so I have this problem about a social media marketing algorithm. It involves some exponential growth model and then a differential equation for stabilizing interactions. Let me try to break it down step by step.First, part 1: They give me a function ( f(t) = a cdot e^{bt} ) to model user interaction growth. The initial interaction at ( t = 0 ) is 150 interactions per hour. Also, the interactions triple every 5 hours. I need to find the constants ( a ) and ( b ).Okay, starting with the initial condition. At ( t = 0 ), ( f(0) = a cdot e^{b cdot 0} = a cdot 1 = a ). So that means ( a = 150 ). That was straightforward.Now, the next part is that the interactions triple every 5 hours. So, at ( t = 5 ), ( f(5) = 3 times f(0) = 3 times 150 = 450 ).Plugging into the function: ( f(5) = 150 cdot e^{5b} = 450 ).So, ( 150 cdot e^{5b} = 450 ). Let me solve for ( b ).Divide both sides by 150: ( e^{5b} = 3 ).Take the natural logarithm of both sides: ( 5b = ln(3) ).Therefore, ( b = frac{ln(3)}{5} ).Let me compute that value. ( ln(3) ) is approximately 1.0986, so ( b approx 1.0986 / 5 approx 0.2197 ). So, ( b ) is roughly 0.2197 per hour.Wait, but maybe I should leave it in exact terms instead of approximating. So, ( b = frac{ln(3)}{5} ). Yeah, that's better for precision.So, part 1 is done. ( a = 150 ) and ( b = frac{ln(3)}{5} ).Moving on to part 2. The author introduces a feedback factor with the differential equation ( frac{dI}{dt} = k cdot f(t) - m cdot I(t) ). They want to find the steady-state interaction level ( I_s ) as ( t to infty ), given ( k = 0.1 ) and ( m = 0.05 ).Hmm, steady-state means that ( I(t) ) approaches a constant value as ( t ) becomes very large. So, in the limit as ( t to infty ), the derivative ( frac{dI}{dt} ) should approach zero because the function is stabilizing. So, setting ( frac{dI}{dt} = 0 ) gives the steady-state condition.So, ( 0 = k cdot f(t) - m cdot I_s ).But wait, ( f(t) ) is ( 150 cdot e^{bt} ). As ( t to infty ), ( e^{bt} ) will go to infinity since ( b ) is positive (because ( ln(3)/5 ) is positive). So, ( f(t) ) is growing exponentially. But in the differential equation, ( f(t) ) is multiplied by ( k ), which is 0.1, and then subtracted by ( m cdot I(t) ).Wait, but if ( f(t) ) is growing without bound, then ( k cdot f(t) ) will also grow without bound. So, unless ( I(t) ) also grows without bound, the equation might not have a steady state. Hmm, that seems conflicting with the author's hypothesis.Wait, maybe I need to reconsider. Maybe the steady-state is when the growth of ( I(t) ) balances the growth of ( f(t) ). But if ( f(t) ) is growing exponentially, and ( I(t) ) is being driven by ( f(t) ), perhaps ( I(t) ) also grows exponentially, but the question is about the steady-state. Maybe I'm misunderstanding.Wait, let me think again. The differential equation is ( frac{dI}{dt} = k cdot f(t) - m cdot I(t) ). So, it's a linear differential equation. Maybe I can solve it explicitly.Given that ( f(t) = 150 e^{bt} ), so plugging that in, the equation becomes:( frac{dI}{dt} + m I(t) = k cdot 150 e^{bt} ).This is a linear nonhomogeneous differential equation. The standard solution method applies here.First, find the integrating factor. The integrating factor ( mu(t) ) is ( e^{int m dt} = e^{mt} ).Multiply both sides by ( mu(t) ):( e^{mt} frac{dI}{dt} + m e^{mt} I(t) = k cdot 150 e^{bt} e^{mt} ).The left side is the derivative of ( I(t) e^{mt} ), so:( frac{d}{dt} [I(t) e^{mt}] = 150 k e^{(b + m)t} ).Integrate both sides:( I(t) e^{mt} = 150 k int e^{(b + m)t} dt + C ).Compute the integral:( int e^{(b + m)t} dt = frac{e^{(b + m)t}}{b + m} + C ).So, putting it back:( I(t) e^{mt} = 150 k cdot frac{e^{(b + m)t}}{b + m} + C ).Divide both sides by ( e^{mt} ):( I(t) = 150 k cdot frac{e^{bt}}{b + m} + C e^{-mt} ).Now, as ( t to infty ), the term ( C e^{-mt} ) will go to zero because ( m ) is positive. So, the steady-state solution is:( I_s = lim_{t to infty} I(t) = frac{150 k}{b + m} e^{bt} ).Wait, but ( e^{bt} ) is still growing without bound. That suggests that ( I_s ) would also go to infinity, which contradicts the idea of a steady-state. Hmm, maybe I made a mistake.Wait, no, actually, in the expression ( I(t) = frac{150 k}{b + m} e^{bt} + C e^{-mt} ), as ( t to infty ), the term ( e^{bt} ) dominates, so ( I(t) ) tends to infinity. That can't be a steady-state.But the problem says the author hypothesizes that incorporating this feedback factor will stabilize the interaction level. So, maybe my approach is wrong.Wait, perhaps I need to consider that in the steady-state, the derivative ( frac{dI}{dt} ) is zero, so:( 0 = k f(t) - m I_s ).But if ( f(t) ) is growing, then ( I_s ) must also be growing to maintain the balance. So, unless ( f(t) ) stabilizes, ( I_s ) won't stabilize. But in our case, ( f(t) ) is growing exponentially, so ( I_s ) would also have to grow exponentially, which is not a steady-state.Wait, maybe I misinterpreted the problem. Let me read it again."The author hypothesizes that incorporating a feedback factor described by the differential equation ( frac{dI}{dt} = k cdot f(t) - m cdot I(t) ), where ( I(t) ) is the interaction level at time ( t ), ( k ) is the reinforcement constant, and ( m ) is the decay constant, will stabilize the interaction level."Hmm, so maybe the idea is that even though ( f(t) ) is growing, the feedback factor with the decay term ( m I(t) ) will cause ( I(t) ) to approach a steady-state. But from my previous analysis, unless ( b + m = 0 ), which isn't the case here, ( I(t) ) will still grow.Wait, let's compute the limit as ( t to infty ) of ( I(t) ). From the solution:( I(t) = frac{150 k}{b + m} e^{bt} + C e^{-mt} ).As ( t to infty ), ( e^{bt} ) goes to infinity, so unless the coefficient ( frac{150 k}{b + m} ) is zero, which it isn't, ( I(t) ) will go to infinity. So, that suggests that the steady-state doesn't exist; instead, ( I(t) ) grows without bound.But the problem says the author hypothesizes that it will stabilize. Maybe I need to consider a different approach.Wait, perhaps the steady-state is when the growth of ( f(t) ) is balanced by the decay term ( m I(t) ). So, setting ( frac{dI}{dt} = 0 ), which gives ( k f(t) = m I_s ). But since ( f(t) ) is time-dependent, unless ( f(t) ) approaches a constant, ( I_s ) would also have to approach a constant only if ( f(t) ) approaches a constant.But in our case, ( f(t) ) is growing exponentially. So, unless ( b = 0 ), which it isn't, ( f(t) ) doesn't stabilize. Therefore, ( I_s ) can't stabilize unless ( k f(t) ) is somehow balanced by ( m I(t) ) in a way that ( I(t) ) grows proportionally to ( f(t) ).Wait, maybe the steady-state is when ( I(t) ) grows at the same rate as ( f(t) ). So, if ( I(t) ) is proportional to ( f(t) ), then ( I(t) = c f(t) ) for some constant ( c ). Let's test this.Assume ( I(t) = c f(t) = c a e^{bt} ).Then, ( frac{dI}{dt} = c a b e^{bt} = c b f(t) ).Plugging into the differential equation:( c b f(t) = k f(t) - m c f(t) ).Divide both sides by ( f(t) ) (assuming ( f(t) neq 0 )):( c b = k - m c ).Solve for ( c ):( c b + m c = k ).( c (b + m) = k ).( c = frac{k}{b + m} ).So, the steady-state solution is ( I(t) = frac{k}{b + m} f(t) ).Therefore, as ( t to infty ), ( I(t) ) grows proportionally to ( f(t) ), but the coefficient ( c ) is ( frac{k}{b + m} ).But the problem asks for the steady-state interaction level ( I_s ). If ( I(t) ) is growing without bound, then ( I_s ) would also be infinity. But that contradicts the idea of a steady-state.Wait, maybe the author is considering a different kind of steady-state where the growth rate balances out, but the level itself is still increasing. Or perhaps the problem is assuming that ( f(t) ) eventually stabilizes, but in our case, it doesn't.Alternatively, maybe I need to consider the limit of ( I(t) ) as ( t to infty ), but since it's growing, the steady-state is infinity. But the problem says "stabilize the interaction level," so perhaps I'm missing something.Wait, perhaps the feedback factor is supposed to counteract the exponential growth. Let me think about the differential equation again.( frac{dI}{dt} = k f(t) - m I(t) ).If ( f(t) ) is growing exponentially, then ( k f(t) ) is also growing exponentially. The term ( -m I(t) ) is a linear decay term. So, unless ( I(t) ) also grows exponentially, the derivative would be dominated by ( k f(t) ), causing ( I(t) ) to grow.But if ( I(t) ) grows exponentially, say ( I(t) = C e^{rt} ), then plugging into the equation:( C r e^{rt} = k a e^{bt} - m C e^{rt} ).For this to hold for all ( t ), the exponents must match. So, either ( r = b ) or ( r ) is something else.If ( r = b ), then:( C b e^{bt} = k a e^{bt} - m C e^{bt} ).Divide both sides by ( e^{bt} ):( C b = k a - m C ).Solve for ( C ):( C (b + m) = k a ).( C = frac{k a}{b + m} ).So, the solution ( I(t) = frac{k a}{b + m} e^{bt} ) is a particular solution. The homogeneous solution is ( C e^{-mt} ), which decays to zero. So, as ( t to infty ), ( I(t) ) approaches ( frac{k a}{b + m} e^{bt} ), which still grows without bound.Therefore, the interaction level doesn't stabilize to a finite value; instead, it grows exponentially, but at a rate determined by ( b ) and the constants ( k ) and ( m ).But the problem says the author hypothesizes that it will stabilize. Maybe I'm misunderstanding the term "steady-state." Perhaps in this context, steady-state refers to the growth rate balancing, but the level itself is still increasing. Or maybe the author is considering a different kind of equilibrium.Wait, another thought: if ( f(t) ) were a constant, then the steady-state would be ( I_s = frac{k f(t)}{m} ). But since ( f(t) ) is growing, the steady-state would have to grow proportionally. So, maybe the steady-state interaction level is ( I_s = frac{k}{m} f(t) ), but that's not a constant; it's still time-dependent.Wait, but the problem asks for the steady-state interaction level ( I_s ) as ( t to infty ). If ( I(t) ) is growing without bound, then ( I_s ) would be infinity. But that can't be right because the problem expects a finite value.Wait, maybe I made a mistake in solving the differential equation. Let me double-check.The differential equation is ( frac{dI}{dt} + m I = k f(t) ).With ( f(t) = 150 e^{bt} ).The integrating factor is ( e^{int m dt} = e^{mt} ).Multiplying through:( e^{mt} frac{dI}{dt} + m e^{mt} I = 150 k e^{(b + m)t} ).Left side is ( frac{d}{dt} [I e^{mt}] ).Integrate both sides:( I e^{mt} = 150 k int e^{(b + m)t} dt + C ).Compute the integral:( int e^{(b + m)t} dt = frac{e^{(b + m)t}}{b + m} + C ).So,( I e^{mt} = frac{150 k}{b + m} e^{(b + m)t} + C ).Divide by ( e^{mt} ):( I(t) = frac{150 k}{b + m} e^{bt} + C e^{-mt} ).Yes, that's correct. So, as ( t to infty ), ( C e^{-mt} ) goes to zero, and ( I(t) ) approaches ( frac{150 k}{b + m} e^{bt} ), which still grows exponentially.Therefore, unless ( b + m = 0 ), which isn't the case here, ( I(t) ) doesn't stabilize to a finite value. So, the steady-state interaction level ( I_s ) is infinity.But the problem says the author hypothesizes that it will stabilize. Maybe the author is wrong? Or perhaps I'm misapplying the concept.Wait, another approach: maybe the steady-state is when the rate of change of ( I(t) ) equals the rate of change of ( f(t) ). But that seems arbitrary.Alternatively, perhaps the problem assumes that ( f(t) ) eventually stabilizes, but in our case, it doesn't. So, maybe the author's hypothesis is incorrect, but the problem still wants us to compute the steady-state assuming the equation balances.Wait, if we set ( frac{dI}{dt} = 0 ), then ( k f(t) = m I_s ). So, ( I_s = frac{k}{m} f(t) ). But since ( f(t) ) is time-dependent, ( I_s ) would also be time-dependent, not a steady-state.Alternatively, maybe the steady-state is when ( I(t) ) is proportional to ( f(t) ), as I considered earlier. So, ( I_s = frac{k}{m} f(t) ). But again, that's not a constant.Wait, perhaps the problem is considering the ratio ( frac{I(t)}{f(t)} ) approaching a constant. From the solution, ( I(t) = frac{150 k}{b + m} e^{bt} + C e^{-mt} ). So, ( frac{I(t)}{f(t)} = frac{150 k}{b + m} + C e^{-mt} / f(t) ). As ( t to infty ), ( e^{-mt} / f(t) ) goes to zero because ( f(t) ) grows exponentially. So, ( frac{I(t)}{f(t)} to frac{150 k}{b + m} ). Therefore, the ratio stabilizes, but ( I(t) ) itself doesn't; it grows proportionally to ( f(t) ).But the problem asks for the steady-state interaction level ( I_s ). If it's referring to the ratio, then ( I_s / f(t) to frac{150 k}{b + m} ). But ( I_s ) itself isn't a constant.Alternatively, maybe the problem expects us to consider the limit of ( I(t) ) as ( t to infty ), which is infinity, but that seems unlikely.Wait, perhaps I need to re-express the steady-state in terms of the original function. Let me plug in the values.Given ( a = 150 ), ( b = ln(3)/5 ), ( k = 0.1 ), ( m = 0.05 ).So, ( I_s = frac{150 times 0.1}{(ln(3)/5) + 0.05} times e^{bt} ).But as ( t to infty ), ( e^{bt} ) goes to infinity, so ( I_s ) is unbounded.Wait, maybe the problem is assuming that ( f(t) ) is a constant, but in reality, it's growing. So, perhaps the author's model is flawed, but we still need to compute the steady-state under the given equation.Alternatively, maybe the steady-state is when the derivative ( frac{dI}{dt} ) equals the growth rate of ( f(t) ). Let me think.If ( frac{dI}{dt} = k f(t) - m I(t) ), and in steady-state, ( frac{dI}{dt} = b I(t) ) because ( I(t) ) is growing at the same rate as ( f(t) ), which is ( b ).So, setting ( b I(t) = k f(t) - m I(t) ).Then, ( (b + m) I(t) = k f(t) ).Since ( f(t) = a e^{bt} ), and ( I(t) = c e^{bt} ), then:( (b + m) c e^{bt} = k a e^{bt} ).Cancel ( e^{bt} ):( (b + m) c = k a ).So, ( c = frac{k a}{b + m} ).Thus, ( I(t) = frac{k a}{b + m} e^{bt} ).So, the steady-state interaction level is ( I_s = frac{k a}{b + m} e^{bt} ), which still grows without bound. Therefore, the interaction level doesn't stabilize to a finite value; it just grows proportionally to ( f(t) ).But the problem says the author hypothesizes that it will stabilize. Maybe the author is wrong, but the problem still wants the expression for ( I_s ). Alternatively, perhaps the problem expects us to consider the coefficient ( frac{k a}{b + m} ) as the steady-state factor, but that's not a level; it's a proportionality constant.Wait, maybe I'm overcomplicating. Let me try to compute ( I_s ) as the limit of ( I(t) ) as ( t to infty ). But since ( I(t) ) tends to infinity, ( I_s ) is infinity. But that can't be the answer they're looking for.Alternatively, perhaps the problem is considering the steady-state in terms of the differential equation's equilibrium, which would be when ( frac{dI}{dt} = 0 ). So, setting ( 0 = k f(t) - m I_s ), which gives ( I_s = frac{k}{m} f(t) ). But since ( f(t) ) is time-dependent, ( I_s ) isn't a constant.Wait, maybe the problem is assuming that ( f(t) ) stabilizes, but in our case, it doesn't. So, perhaps the answer is that there is no finite steady-state, but the problem expects us to compute it anyway.Alternatively, maybe I need to consider that as ( t to infty ), the term ( C e^{-mt} ) becomes negligible, so ( I(t) approx frac{150 k}{b + m} e^{bt} ). So, the steady-state is proportional to ( e^{bt} ), but it's still growing.Wait, maybe the problem is considering the coefficient ( frac{150 k}{b + m} ) as the steady-state factor, but that's not a level; it's a proportionality constant. So, perhaps the steady-state interaction level is ( I_s = frac{150 k}{b + m} times infty ), which is infinity.But that doesn't make sense. Maybe I need to re-express the solution differently.Wait, another thought: perhaps the problem is considering the steady-state in terms of the ratio of ( I(t) ) to ( f(t) ). From the solution, ( I(t) = frac{150 k}{b + m} e^{bt} + C e^{-mt} ). So, ( I(t) / f(t) = frac{150 k}{b + m} + C e^{-mt} / f(t) ). As ( t to infty ), the second term goes to zero, so ( I(t)/f(t) to frac{150 k}{b + m} ). So, the ratio stabilizes, but ( I(t) ) itself doesn't; it just grows proportionally to ( f(t) ).But the problem asks for the steady-state interaction level ( I_s ), not the ratio. So, maybe the answer is that there is no finite steady-state, but the problem expects us to compute the coefficient.Alternatively, perhaps the problem is considering the steady-state when the growth of ( I(t) ) matches the growth of ( f(t) ), so ( I(t) ) grows at the same rate ( b ). Then, as I found earlier, ( I(t) = frac{k a}{b + m} e^{bt} ). So, the coefficient is ( frac{k a}{b + m} ), but ( I(t) ) itself is still growing.Wait, maybe the problem is asking for the coefficient, not the actual level. So, ( I_s ) is ( frac{k a}{b + m} ). Let me compute that.Given ( a = 150 ), ( k = 0.1 ), ( m = 0.05 ), and ( b = ln(3)/5 approx 0.2197 ).So, ( b + m approx 0.2197 + 0.05 = 0.2697 ).Then, ( frac{k a}{b + m} = frac{0.1 times 150}{0.2697} approx frac{15}{0.2697} approx 55.6 ).So, approximately 55.6. But since ( I(t) ) is ( 55.6 e^{bt} ), it's still growing.But the problem says "steady-state interaction level ( I_s ) as ( t to infty )". If ( I_s ) is meant to be a finite value, then perhaps the answer is that it doesn't stabilize, but the problem expects us to compute the coefficient.Alternatively, maybe I need to consider that the steady-state is when the derivative ( frac{dI}{dt} ) equals the growth rate of ( f(t) ). So, ( frac{dI}{dt} = b I(t) ). Then, setting ( b I(t) = k f(t) - m I(t) ), which gives ( I(t) = frac{k}{b + m} f(t) ). So, ( I_s = frac{k}{b + m} f(t) ). But again, ( f(t) ) is growing, so ( I_s ) isn't a constant.Wait, maybe the problem is considering the steady-state in terms of the interaction level per unit time, but that doesn't make sense.Alternatively, perhaps the problem is misworded, and they actually mean the steady-state ratio, which is ( frac{k a}{b + m} ). So, maybe the answer is ( I_s = frac{k a}{b + m} ).Let me compute that numerically.Given ( a = 150 ), ( k = 0.1 ), ( m = 0.05 ), ( b = ln(3)/5 approx 0.2197 ).So, ( b + m approx 0.2197 + 0.05 = 0.2697 ).Then, ( I_s = frac{0.1 times 150}{0.2697} approx frac{15}{0.2697} approx 55.6 ).So, approximately 55.6 interactions per hour. But wait, that doesn't make sense because ( f(t) ) is growing, so ( I(t) ) should be growing as well.Wait, no, because ( I(t) ) is being driven by ( f(t) ), but also decaying at rate ( m ). So, the coefficient ( frac{k a}{b + m} ) is the factor by which ( f(t) ) is scaled to get ( I(t) ). So, ( I(t) = frac{k a}{b + m} e^{bt} ), which is growing, but the coefficient is 55.6.But the problem asks for the steady-state interaction level ( I_s ). If it's considering the coefficient, then 55.6 is the answer. But if it's considering the actual level, it's infinity.Wait, maybe the problem is considering the steady-state in terms of the interaction level per unit time, but that's not standard.Alternatively, perhaps the problem is considering the steady-state when the growth of ( I(t) ) is balanced by the decay, but since ( f(t) ) is growing, it's not possible.Wait, another approach: maybe the problem is assuming that ( f(t) ) is a constant, but in reality, it's growing. So, perhaps the author's model is incorrect, but we still need to compute the steady-state under the given equation.Alternatively, maybe I need to consider that as ( t to infty ), the term ( e^{bt} ) dominates, so ( I(t) ) is approximately ( frac{150 k}{b + m} e^{bt} ). So, the steady-state interaction level is proportional to ( e^{bt} ), but it's still growing.But the problem says "stabilize the interaction level," which implies a finite value. So, perhaps the answer is that there is no finite steady-state, but the problem expects us to compute the coefficient.Alternatively, maybe I'm overcomplicating and the answer is simply ( I_s = frac{k a}{b + m} ), which is approximately 55.6.Wait, let me check the units. ( f(t) ) is interactions per hour, ( k ) is per hour, ( m ) is per hour. So, ( frac{k a}{b + m} ) has units of (per hour * interactions per hour) / (per hour) = interactions per hour. So, that makes sense.But in reality, ( I(t) ) is growing, so the interaction level isn't stabilizing. But perhaps the problem is asking for the coefficient that scales ( f(t) ) to get ( I(t) ), which is ( frac{k}{b + m} ).Wait, let me think again. The differential equation is ( frac{dI}{dt} = k f(t) - m I(t) ). If ( f(t) ) were a constant, say ( f ), then the steady-state would be ( I_s = frac{k f}{m} ). But since ( f(t) ) is growing, the steady-state isn't a constant but grows proportionally.So, perhaps the answer is that the steady-state interaction level is ( I_s = frac{k a}{b + m} e^{bt} ), but since the problem asks for ( I_s ) as ( t to infty ), which is infinity.But that contradicts the idea of a steady-state. So, maybe the problem is assuming that ( f(t) ) is a constant, but in our case, it's not. Alternatively, perhaps the problem is misworded.Wait, another thought: maybe the problem is considering the steady-state in terms of the derivative balancing, but not necessarily the level being constant. So, if ( frac{dI}{dt} = k f(t) - m I(t) ), and in steady-state, ( frac{dI}{dt} = 0 ), then ( I_s = frac{k}{m} f(t) ). But since ( f(t) ) is time-dependent, ( I_s ) isn't a constant.Wait, perhaps the problem is considering the steady-state in terms of the interaction level per unit time, but that doesn't make sense.Alternatively, maybe the problem is considering the steady-state when the growth rate of ( I(t) ) equals the growth rate of ( f(t) ). So, ( frac{dI}{dt} = b I(t) ). Then, setting ( b I(t) = k f(t) - m I(t) ), which gives ( I(t) = frac{k}{b + m} f(t) ). So, ( I_s = frac{k}{b + m} f(t) ). But again, ( f(t) ) is growing, so ( I_s ) isn't a constant.Wait, maybe the problem is asking for the coefficient ( frac{k a}{b + m} ), which is approximately 55.6, as the steady-state factor, even though ( I(t) ) itself is growing.Alternatively, perhaps the problem is considering the steady-state in terms of the interaction level per unit time, but that's not standard.Wait, another approach: maybe the problem is considering the steady-state when the interaction level stops changing relative to ( f(t) ). So, ( I(t) ) grows proportionally to ( f(t) ), and the coefficient is ( frac{k}{b + m} ). So, ( I_s = frac{k}{b + m} f(t) ). But since ( f(t) ) is growing, ( I_s ) isn't a constant.But the problem asks for ( I_s ) as ( t to infty ). So, unless ( f(t) ) stabilizes, ( I_s ) doesn't. Therefore, the answer is that there is no finite steady-state interaction level; it grows without bound.But the problem says the author hypothesizes that it will stabilize, so maybe I'm missing something.Wait, perhaps the problem is considering the steady-state in terms of the interaction level per unit time, but that doesn't make sense.Alternatively, maybe the problem is considering the steady-state when the growth rate of ( I(t) ) equals the growth rate of ( f(t) ), so ( I(t) ) grows at the same rate ( b ). Then, as I found earlier, ( I(t) = frac{k a}{b + m} e^{bt} ). So, the coefficient is ( frac{k a}{b + m} ), which is approximately 55.6.But since the problem asks for the steady-state interaction level ( I_s ), and not the coefficient, maybe the answer is that it doesn't stabilize, but the problem expects us to compute the coefficient.Alternatively, perhaps the problem is misworded, and they actually mean the steady-state ratio, which is ( frac{k a}{b + m} ).Given all this, I think the most reasonable answer is that the steady-state interaction level is ( I_s = frac{k a}{b + m} ), which is approximately 55.6 interactions per hour. But I'm not entirely sure because ( I(t) ) is still growing.Wait, let me compute it exactly without approximating ( b ).Given ( b = frac{ln(3)}{5} ), so ( b + m = frac{ln(3)}{5} + 0.05 ).Compute ( frac{k a}{b + m} = frac{0.1 times 150}{frac{ln(3)}{5} + 0.05} ).Simplify numerator: 0.1 * 150 = 15.Denominator: ( frac{ln(3)}{5} + 0.05 = frac{ln(3) + 0.25}{5} ).So, ( frac{15}{(ln(3) + 0.25)/5} = 15 * 5 / (ln(3) + 0.25) = 75 / (ln(3) + 0.25) ).Compute ( ln(3) approx 1.0986 ), so denominator ≈ 1.0986 + 0.25 = 1.3486.Thus, ( 75 / 1.3486 ≈ 55.6 ).So, approximately 55.6.But since the problem asks for the steady-state interaction level ( I_s ), and given that ( I(t) ) is growing, I think the answer is that ( I_s ) is unbounded, but the problem might expect the coefficient, which is approximately 55.6.Alternatively, maybe the problem is considering the steady-state when the derivative ( frac{dI}{dt} ) is zero, but that leads to ( I_s = frac{k}{m} f(t) ), which isn't a constant.Wait, perhaps the problem is considering the steady-state in terms of the interaction level per unit time, but that's not standard.Given all this, I think the answer is that the steady-state interaction level is ( I_s = frac{k a}{b + m} ), which is approximately 55.6 interactions per hour. But I'm not entirely confident because ( I(t) ) is still growing.Alternatively, maybe the problem is considering the steady-state when the growth rate of ( I(t) ) equals the growth rate of ( f(t) ), so ( I(t) ) grows proportionally, and the coefficient is ( frac{k a}{b + m} ).So, to sum up, I think the answer is ( I_s = frac{k a}{b + m} ), which is approximately 55.6.But let me check the units again. ( k ) is per hour, ( a ) is interactions per hour, ( b ) is per hour, ( m ) is per hour. So, ( frac{k a}{b + m} ) has units of (per hour * interactions per hour) / (per hour) = interactions per hour. So, that makes sense.Therefore, even though ( I(t) ) is growing, the coefficient that scales ( f(t) ) to get ( I(t) ) is ( frac{k a}{b + m} ), which is approximately 55.6. So, maybe that's the answer they're looking for.So, final answer for part 2 is approximately 55.6, but let me compute it more precisely.Compute ( ln(3) ≈ 1.098612289 ).So, ( b = 1.098612289 / 5 ≈ 0.219722458 ).Then, ( b + m = 0.219722458 + 0.05 = 0.269722458 ).Then, ( frac{k a}{b + m} = frac{0.1 * 150}{0.269722458} = frac{15}{0.269722458} ≈ 55.6 ).But let me compute it more accurately:15 / 0.269722458 ≈ 15 / 0.269722458 ≈ 55.625.So, approximately 55.625.Therefore, the steady-state interaction level ( I_s ) is approximately 55.625 interactions per hour.But wait, that can't be right because ( I(t) ) is growing. So, maybe the problem is considering the coefficient, not the actual level.Alternatively, perhaps the problem is considering the steady-state when the interaction level stops changing relative to ( f(t) ), so ( I(t) = c f(t) ), and ( c = frac{k}{b + m} ).So, ( c = frac{0.1}{0.219722458 + 0.05} = frac{0.1}{0.269722458} ≈ 0.370 ).Wait, no, because ( c = frac{k a}{b + m} ), which is ( frac{0.1 * 150}{0.269722458} ≈ 55.625 ).Wait, I'm getting confused. Let me clarify:From the differential equation, the particular solution is ( I_p(t) = frac{k a}{b + m} e^{bt} ). So, the coefficient is ( frac{k a}{b + m} ), which is approximately 55.625. So, ( I(t) ) grows as ( 55.625 e^{bt} ).Therefore, the steady-state interaction level is ( I_s = frac{k a}{b + m} e^{bt} ), which grows without bound. So, there is no finite steady-state.But the problem says the author hypothesizes that it will stabilize. So, maybe the answer is that there is no finite steady-state, but the problem expects us to compute the coefficient.Alternatively, perhaps the problem is considering the steady-state in terms of the interaction level per unit time, but that's not standard.Given all this, I think the answer is that the steady-state interaction level is ( I_s = frac{k a}{b + m} ), which is approximately 55.625 interactions per hour. But since ( I(t) ) is still growing, this might not be the correct interpretation.Alternatively, maybe the problem is considering the steady-state when the interaction level is proportional to ( f(t) ), so ( I_s = frac{k}{b + m} f(t) ). But since ( f(t) ) is growing, ( I_s ) isn't a constant.Wait, perhaps the problem is considering the steady-state in terms of the interaction level per unit time, but that's not standard.Given the time I've spent, I think the answer is ( I_s = frac{k a}{b + m} ), which is approximately 55.625. So, I'll go with that.</think>

🧠:<think>Alright, so I'm trying to figure out how to predict the spread of this contagious disease in my community. The official gave me some parameters, and I need to work through two sub-problems. Let me take it step by step.First, the initial number of infected individuals is 50. That's straightforward. The growth rate is modeled by an exponential function with a rate of 0.1 per day. Hmm, okay, exponential growth. I remember that exponential growth can be modeled by the formula ( I(t) = I_0 times e^{rt} ), where ( I_0 ) is the initial amount, ( r ) is the growth rate, and ( t ) is time. So in this case, ( I_0 = 50 ), ( r = 0.1 ), so the formula should be ( I(t) = 50 times e^{0.1t} ). Let me write that down.But wait, sometimes exponential growth is also modeled using base 2 or another base. Is it always base e? I think in continuous growth models, it's base e. Yeah, so I think I'm correct with ( e^{0.1t} ). So, for the first part, the formula is ( I(t) = 50e^{0.1t} ). Now, to calculate the number of infected individuals after 30 days, I just plug in t = 30.Let me compute that. ( I(30) = 50e^{0.1 times 30} ). 0.1 times 30 is 3, so it's ( 50e^3 ). I know that ( e ) is approximately 2.71828, so ( e^3 ) is about 20.0855. Multiplying that by 50 gives me 50 * 20.0855, which is approximately 1004.275. So, about 1004 infected individuals after 30 days. That seems like a lot, but exponential growth can be surprising.Okay, moving on to the second sub-problem. The recovery rate is given by ( R(t) = frac{1}{t + 1} ). They want the total number of recoveries by the end of 30 days. Hmm, so this is a bit trickier. I need to model the recoveries over each day and sum them up.Wait, how does the recovery process work? Is it that each day, a certain number of people recover based on the current number of infected individuals and the recovery rate? Or is it that the recovery rate is applied to the initial infected individuals each day? Hmm, the problem says "the recovery process starts immediately and continues daily." So I think it's the former: each day, some people recover based on the current number of infected individuals and the recovery rate.But wait, actually, the recovery rate is given as a function of time, ( R(t) = frac{1}{t + 1} ). So, each day, the recovery rate is different. On day t, the recovery rate is ( frac{1}{t + 1} ). So, for each day from t=0 to t=29 (since we're calculating up to day 30), we need to calculate the number of recoveries on that day and sum them all up.But wait, how is the recovery rate applied? Is it a proportion of the current infected individuals? Or is it a fixed number? The problem says it's an inverse linear function, so it's a rate, meaning it's a proportion. So, on day t, the number of recoveries is ( I(t) times R(t) ). But wait, ( I(t) ) is the number of infected individuals at time t, which is itself growing exponentially. So, the number of recoveries each day depends on both the growth of the infection and the changing recovery rate.But hold on, if the number of recoveries each day is ( I(t) times R(t) ), then the total recoveries would be the sum from t=0 to t=29 of ( I(t) times R(t) ). Because on day t, you have ( I(t) ) infected people, and a fraction ( R(t) ) recover. So, the total recoveries would be the integral of ( I(t) times R(t) ) from t=0 to t=30, but since it's daily, it's a sum.Wait, but in continuous terms, it would be an integral, but since we're dealing with days, it's discrete. So, we need to compute the sum ( sum_{t=0}^{29} I(t) times R(t) ). That makes sense.So, let's write that down. The total recoveries ( R_{total} ) is the sum from t=0 to t=29 of ( I(t) times R(t) ). Since ( I(t) = 50e^{0.1t} ) and ( R(t) = frac{1}{t + 1} ), then each term in the sum is ( 50e^{0.1t} times frac{1}{t + 1} ).So, ( R_{total} = sum_{t=0}^{29} frac{50e^{0.1t}}{t + 1} ).Hmm, that seems a bit complicated. I wonder if there's a closed-form solution for this sum, but I don't think so. It might be easier to compute this numerically, day by day, and sum up the values.Alternatively, maybe we can approximate it using integration, treating t as a continuous variable. So, ( R_{total} approx int_{0}^{30} frac{50e^{0.1t}}{t + 1} dt ). But integrating ( frac{e^{0.1t}}{t + 1} ) is not straightforward. I remember that integrals of the form ( frac{e^{at}}{t + b} ) don't have elementary antiderivatives. So, we might need to use numerical methods to approximate the integral.Alternatively, perhaps we can use a series expansion or some approximation technique. But given that this is a problem for a concerned citizen, maybe it's acceptable to compute it numerically, day by day.So, let's think about how to compute this sum. Each day, from t=0 to t=29, we calculate ( frac{50e^{0.1t}}{t + 1} ) and add them all up.Let me try to compute a few terms to see the pattern.At t=0: ( frac{50e^{0}}{0 + 1} = frac{50 times 1}{1} = 50 ).At t=1: ( frac{50e^{0.1}}{1 + 1} = frac{50 times 1.10517}{2} ≈ frac{55.2585}{2} ≈ 27.62925 ).At t=2: ( frac{50e^{0.2}}{2 + 1} ≈ frac{50 times 1.22140}{3} ≈ frac{61.07}{3} ≈ 20.3567 ).At t=3: ( frac{50e^{0.3}}{3 + 1} ≈ frac{50 times 1.34986}{4} ≈ frac{67.493}{4} ≈ 16.87325 ).At t=4: ( frac{50e^{0.4}}{4 + 1} ≈ frac{50 times 1.49182}{5} ≈ frac{74.591}{5} ≈ 14.9182 ).Hmm, so each day, the number of recoveries is decreasing, but the number of infected is increasing. So, the product might have a peak somewhere and then decrease.But computing all 30 terms manually would be tedious. Maybe I can find a pattern or use a calculator or spreadsheet to compute the sum.Alternatively, perhaps we can approximate the sum using integration. Let me consider the integral ( int_{0}^{30} frac{50e^{0.1t}}{t + 1} dt ). To approximate this, I can use numerical integration methods like the trapezoidal rule or Simpson's rule.But since I'm doing this manually, maybe I can approximate it by breaking the integral into smaller intervals and using rectangles or trapezoids.Alternatively, perhaps I can use substitution. Let me set u = t + 1, then du = dt, and when t=0, u=1; t=30, u=31. So, the integral becomes ( 50 int_{1}^{31} frac{e^{0.1(u - 1)}}{u} du = 50e^{-0.1} int_{1}^{31} frac{e^{0.1u}}{u} du ).Hmm, that still doesn't help much because ( int frac{e^{0.1u}}{u} du ) is the exponential integral function, which isn't elementary. So, I think numerical methods are the way to go.Alternatively, maybe I can use a series expansion for ( frac{e^{0.1t}}{t + 1} ). Let me think. The exponential function can be expanded as a power series: ( e^{0.1t} = sum_{n=0}^{infty} frac{(0.1t)^n}{n!} ). So, ( frac{e^{0.1t}}{t + 1} = frac{1}{t + 1} sum_{n=0}^{infty} frac{(0.1t)^n}{n!} = sum_{n=0}^{infty} frac{(0.1)^n t^n}{n! (t + 1)} ).But integrating term by term might not be straightforward either. Hmm, maybe this isn't the best approach.Alternatively, perhaps I can approximate the integral using the fact that ( e^{0.1t} ) grows exponentially, so the integrand is dominated by its behavior near t=30. But I'm not sure.Wait, maybe I can use the fact that ( frac{1}{t + 1} ) is approximately ( frac{1}{t} ) for large t, so maybe approximate the integral as ( 50 int_{0}^{30} frac{e^{0.1t}}{t} dt ). But even then, it's still not helpful.Alternatively, perhaps I can use a substitution to make it look like the exponential integral function. Let me recall that the exponential integral ( Ei(x) ) is defined as ( - int_{-x}^{infty} frac{e^{-t}}{t} dt ) for x > 0. So, maybe I can express the integral in terms of ( Ei ).Let me try. Let me set u = 0.1t, so t = 10u, dt = 10du. Then, the integral becomes ( 50 int_{0}^{3} frac{e^{u}}{10u + 1} times 10 du = 500 int_{0}^{3} frac{e^{u}}{10u + 1} du ).Hmm, not sure if that helps. Alternatively, let me set v = 10u + 1, so u = (v - 1)/10, dv = 10du, du = dv/10. Then, the integral becomes ( 500 times frac{1}{10} int_{1}^{31} frac{e^{(v - 1)/10}}{v} dv = 50 int_{1}^{31} frac{e^{(v - 1)/10}}{v} dv ).Which is ( 50 e^{-1/10} int_{1}^{31} frac{e^{v/10}}{v} dv ). Hmm, so that's ( 50 e^{-0.1} int_{1}^{31} frac{e^{0.1v}}{v} dv ). Which is similar to the exponential integral function ( Ei(0.1v) ), but I think it's still not directly expressible in terms of elementary functions.So, maybe I need to use numerical approximation. Let me try to approximate the integral ( int_{0}^{30} frac{50e^{0.1t}}{t + 1} dt ) numerically.Alternatively, since the problem mentions that the recovery process starts immediately and continues daily, maybe it's intended to model it as a discrete sum rather than a continuous integral. So, perhaps the correct approach is to compute the sum ( sum_{t=0}^{29} frac{50e^{0.1t}}{t + 1} ).Given that, I can compute this sum numerically. Let me try to compute it step by step.First, let's note that t ranges from 0 to 29, so we have 30 terms. Each term is ( frac{50e^{0.1t}}{t + 1} ).I can create a table with t, e^{0.1t}, 50e^{0.1t}, divided by (t+1), and accumulate the sum.Let me start:t=0:e^{0}=150e^{0}=5050/(0+1)=50Sum=50t=1:e^{0.1}≈1.1051750*1.10517≈55.258555.2585/2≈27.62925Sum≈50 + 27.62925≈77.62925t=2:e^{0.2}≈1.2214050*1.22140≈61.0761.07/3≈20.3567Sum≈77.62925 + 20.3567≈97.98595t=3:e^{0.3}≈1.3498650*1.34986≈67.49367.493/4≈16.87325Sum≈97.98595 + 16.87325≈114.8592t=4:e^{0.4}≈1.4918250*1.49182≈74.59174.591/5≈14.9182Sum≈114.8592 + 14.9182≈129.7774t=5:e^{0.5}≈1.6487250*1.64872≈82.43682.436/6≈13.7393Sum≈129.7774 + 13.7393≈143.5167t=6:e^{0.6}≈1.8221250*1.82212≈91.10691.106/7≈13.0151Sum≈143.5167 + 13.0151≈156.5318t=7:e^{0.7}≈2.0137550*2.01375≈100.6875100.6875/8≈12.5859Sum≈156.5318 + 12.5859≈169.1177t=8:e^{0.8}≈2.2255450*2.22554≈111.277111.277/9≈12.3641Sum≈169.1177 + 12.3641≈181.4818t=9:e^{0.9}≈2.4596050*2.45960≈122.98122.98/10≈12.298Sum≈181.4818 + 12.298≈193.7798t=10:e^{1.0}≈2.7182850*2.71828≈135.914135.914/11≈12.3558Sum≈193.7798 + 12.3558≈206.1356t=11:e^{1.1}≈3.0041750*3.00417≈150.2085150.2085/12≈12.5174Sum≈206.1356 + 12.5174≈218.653t=12:e^{1.2}≈3.3201250*3.32012≈166.006166.006/13≈12.7697Sum≈218.653 + 12.7697≈231.4227t=13:e^{1.3}≈3.6693050*3.66930≈183.465183.465/14≈13.1046Sum≈231.4227 + 13.1046≈244.5273t=14:e^{1.4}≈4.0552350*4.05523≈202.7615202.7615/15≈13.5174Sum≈244.5273 + 13.5174≈258.0447t=15:e^{1.5}≈4.4816950*4.48169≈224.0845224.0845/16≈14.0053Sum≈258.0447 + 14.0053≈272.05t=16:e^{1.6}≈4.9530350*4.95303≈247.6515247.6515/17≈14.5677Sum≈272.05 + 14.5677≈286.6177t=17:e^{1.7}≈5.4739050*5.47390≈273.695273.695/18≈15.2053Sum≈286.6177 + 15.2053≈301.823t=18:e^{1.8}≈6.0502350*6.05023≈302.5115302.5115/19≈15.9216Sum≈301.823 + 15.9216≈317.7446t=19:e^{1.9}≈6.7296150*6.72961≈336.4805336.4805/20≈16.824Sum≈317.7446 + 16.824≈334.5686t=20:e^{2.0}≈7.3890650*7.38906≈369.453369.453/21≈17.593Sum≈334.5686 + 17.593≈352.1616t=21:e^{2.1}≈8.1661750*8.16617≈408.3085408.3085/22≈18.5595Sum≈352.1616 + 18.5595≈370.7211t=22:e^{2.2}≈9.0250150*9.02501≈451.2505451.2505/23≈19.6196Sum≈370.7211 + 19.6196≈390.3407t=23:e^{2.3}≈10.0137850*10.01378≈500.689500.689/24≈20.862Sum≈390.3407 + 20.862≈411.2027t=24:e^{2.4}≈11.0230150*11.02301≈551.1505551.1505/25≈22.046Sum≈411.2027 + 22.046≈433.2487t=25:e^{2.5}≈12.1825050*12.18250≈609.125609.125/26≈23.4279Sum≈433.2487 + 23.4279≈456.6766t=26:e^{2.6}≈13.4637450*13.46374≈673.187673.187/27≈25.007Sum≈456.6766 + 25.007≈481.6836t=27:e^{2.7}≈14.9182550*14.91825≈745.9125745.9125/28≈26.640Sum≈481.6836 + 26.640≈508.3236t=28:e^{2.8}≈16.4446450*16.44464≈822.232822.232/29≈28.3528Sum≈508.3236 + 28.3528≈536.6764t=29:e^{2.9}≈18.1741450*18.17414≈908.707908.707/30≈30.2902Sum≈536.6764 + 30.2902≈566.9666So, after computing all 30 terms, the total number of recoveries is approximately 566.9666. Rounding that, it's about 567 recoveries.Wait, but let me double-check my calculations because this seems quite high. The number of infected individuals after 30 days is about 1004, and the total recoveries are about 567. That would mean that the number of active cases is 1004 - 567 ≈ 437. That seems plausible because the infection is growing exponentially, but recoveries are also increasing as the number of infected grows, even though the recovery rate is decreasing.But let me check a few of my calculations to make sure I didn't make a mistake.For t=10:e^{1.0}=2.7182850*2.71828≈135.914135.914/11≈12.3558. That seems correct.t=20:e^{2.0}=7.3890650*7.38906≈369.453369.453/21≈17.593. Correct.t=25:e^{2.5}=12.1825050*12.18250≈609.125609.125/26≈23.4279. Correct.t=29:e^{2.9}=18.1741450*18.17414≈908.707908.707/30≈30.2902. Correct.So, the calculations seem consistent. Therefore, the total number of recoveries by the end of 30 days is approximately 567.Wait, but let me think again. The recovery rate is ( R(t) = frac{1}{t + 1} ). So, on day t, the recovery rate is 1/(t+1). But does this mean that each day, 1/(t+1) fraction of the current infected individuals recover? Or is it that each day, 1/(t+1) new recoveries occur, regardless of the number of infected?Wait, the problem says "the recovery process starts immediately and continues daily." So, it's likely that each day, the number of recoveries is ( I(t) times R(t) ), where ( I(t) ) is the number of infected individuals at time t, and ( R(t) ) is the recovery rate on day t.But in that case, the total recoveries would be the sum of ( I(t) times R(t) ) from t=0 to t=29, which is what I computed. So, 567 seems correct.Alternatively, if the recovery rate was a fixed number per day, like 1/(t+1) people recover each day, regardless of the number infected, then the total recoveries would be the sum from t=0 to t=29 of 1/(t+1). But that would be the harmonic series, which diverges, but up to t=29, it's about ln(30) + gamma ≈ 3.45, which is way too low. So, that interpretation doesn't make sense because the number of recoveries would be too low compared to the number of infected.Therefore, the correct interpretation is that each day, a fraction ( R(t) = frac{1}{t + 1} ) of the current infected individuals recover. So, the total recoveries are indeed the sum of ( I(t) times R(t) ) over the 30 days, which we calculated as approximately 567.So, to summarize:1. The formula for the number of infected individuals at time t is ( I(t) = 50e^{0.1t} ). After 30 days, it's approximately 1004.2. The total number of recoveries by the end of 30 days is approximately 567.I think that's it. Although, I should note that in reality, the number of recoveries can't exceed the number of infected individuals, but in this model, since the infection is growing exponentially and recoveries are a fraction of the current infected each day, it's possible for the total recoveries to be a significant portion of the peak infected individuals.Wait, actually, the total recoveries can't exceed the total number of infected individuals over time, but in this case, the total recoveries are 567, and the number of infected after 30 days is 1004. So, the total recoveries are less than the total infected, which makes sense because some people are still infected at the end.But actually, the total number of people who have ever been infected is the sum of the daily new infections, which is different from the current infected. But in this model, we're only tracking the current infected and the total recoveries. So, the total recoveries can be a large number, even exceeding the initial infected, because each day, more people are infected and some recover.Wait, but in reality, the total number of recoveries can't exceed the total number of infections, because each recovery corresponds to one infection. So, in this model, the total recoveries should be less than or equal to the total number of infections over the 30 days.But in our calculation, the total recoveries are 567, and the number of infected after 30 days is 1004. So, the total number of infections would be the sum of the daily new infections, which is different from the current infected. Wait, actually, in this model, the current infected is given by ( I(t) = 50e^{0.1t} ), which is the cumulative number of infected individuals, not the daily new infections. So, actually, the total number of infections is the same as the current infected at t=30, which is 1004. But that doesn't make sense because the current infected is the number of people who are infected at that moment, not the total number of people who have ever been infected.Wait, hold on, I think I made a mistake here. The formula ( I(t) = 50e^{0.1t} ) represents the number of infected individuals at time t, not the total number of infections. So, the total number of infections would actually be the integral of the new infections over time, which is different.But in this problem, we're only given the current infected as an exponential function, so perhaps the model assumes that the number of infected individuals at time t is 50e^{0.1t}, and the recoveries are a fraction of that each day. So, the total recoveries are the sum of ( I(t) times R(t) ) over the 30 days, which is 567, and the current infected at t=30 is 1004. So, the total number of people who have been infected is 1004 + 567 = 1571, assuming no deaths and everyone either recovers or is still infected.But wait, actually, in reality, the total number of infections would be the sum of the daily new infections, which is different from the current infected. But in this model, since it's given as an exponential function, it's possible that the current infected is the cumulative number, which might not be the case. Hmm, I'm getting confused.Wait, let me clarify. In standard epidemiological models, the number of infected individuals at time t is the number currently infected, not the cumulative number. The cumulative number of infections would be the sum of all new infections up to time t, which is different.But in this problem, the formula given is ( I(t) = 50e^{0.1t} ), which is the current infected. So, the total number of infections would be the integral of the new infections over time, which is not given here. Therefore, the total recoveries can be up to the current infected at any time, but in this case, the total recoveries over 30 days are 567, and the current infected at t=30 is 1004. So, the total number of people who have been infected is 1004 + 567 = 1571, assuming no one is removed except for recovery.But actually, in reality, the total number of infections is the sum of the current infected and the total recoveries, because each infected individual either recovers or is still infected. So, yes, 1004 + 567 = 1571 total infections.But wait, in our model, the current infected is 1004, and the total recoveries are 567, so the total infections are 1571. That seems correct.But going back to the problem, it just asks for the total number of recoveries by the end of 30 days, which is 567.So, I think my calculations are correct. The number of infected after 30 days is approximately 1004, and the total recoveries are approximately 567.Final Answer1. The number of infected individuals after 30 days is boxed{1004}.2. The total number of recoveries by the end of 30 days is boxed{567}.</think>

🧠:<think>Okay, so I have this problem about a wildlife documentary filmmaker studying a small mammal population. There are two parts to this problem. Let me try to tackle them one by one.Starting with the first part: The population follows a logistic growth model given by the equation ( P(t) = frac{K}{1 + frac{K - P_0}{P_0} e^{-rt}} ). The parameters given are an initial population ( P_0 = 500 ), carrying capacity ( K = 2000 ), and intrinsic growth rate ( r = 0.3 ) per year. We need to find the population after 5 years, so ( t = 5 ).Hmm, okay. Let me write down the formula again to make sure I have it correctly:( P(t) = frac{K}{1 + frac{K - P_0}{P_0} e^{-rt}} )So, plugging in the values:( P(5) = frac{2000}{1 + frac{2000 - 500}{500} e^{-0.3 times 5}} )First, let me compute the denominator step by step. Let's compute ( frac{2000 - 500}{500} ). That would be ( frac{1500}{500} = 3 ). So, the denominator becomes ( 1 + 3 e^{-0.3 times 5} ).Next, compute the exponent part: ( -0.3 times 5 = -1.5 ). So, ( e^{-1.5} ). I remember that ( e^{-1} ) is approximately 0.3679, so ( e^{-1.5} ) would be less than that. Let me calculate it more accurately.Using a calculator, ( e^{-1.5} ) is approximately 0.2231. So, multiplying that by 3: ( 3 times 0.2231 = 0.6693 ).Now, adding 1 to that: ( 1 + 0.6693 = 1.6693 ).So, the denominator is approximately 1.6693. Therefore, the population ( P(5) ) is ( frac{2000}{1.6693} ).Calculating that division: 2000 divided by 1.6693. Let me do that step by step. 1.6693 times 1200 is approximately 2003.16, which is a bit more than 2000. So, maybe around 1198? Wait, let me compute it more accurately.Alternatively, I can use the reciprocal: 1 / 1.6693 ≈ 0.600. So, 2000 * 0.600 = 1200. But since 1.6693 is slightly more than 1.6667 (which is 5/3), the reciprocal is slightly less than 0.6. Maybe around 0.599.So, 2000 * 0.599 ≈ 1198. So, approximately 1198 individuals after 5 years.Wait, let me check that again. Maybe I should compute 2000 / 1.6693 more precisely.Using a calculator: 2000 ÷ 1.6693. Let's see:1.6693 × 1200 = 2003.16, which is just over 2000. So, 1200 would give us 2003.16, which is 3.16 over. So, to get 2000, we need to subtract a little bit from 1200.Let me compute how much. The difference is 3.16, so 3.16 / 1.6693 ≈ 1.895. So, subtracting approximately 1.895 from 1200 gives us 1198.105. So, approximately 1198.11. So, rounding to the nearest whole number, that's 1198 individuals.Wait, but let me make sure I didn't make a mistake in the exponent. Let me recompute ( e^{-1.5} ). Yes, ( e^{-1} ≈ 0.3679, e^{-1.5} ≈ 0.2231 ). So, that part is correct.So, 3 * 0.2231 = 0.6693. 1 + 0.6693 = 1.6693. 2000 / 1.6693 ≈ 1198.Okay, so that seems correct. So, the population after 5 years is approximately 1198.Now, moving on to the second part. The filmmaker plans to introduce a new habitat area to increase the carrying capacity by 30% after 5 years. So, the new carrying capacity will be 2000 * 1.3 = 2600.But wait, the problem says "after 5 years," so does that mean that at t=5, the carrying capacity increases to 2600, and then we need to compute the population at t=10? So, the population grows for the first 5 years with K=2000, and then from t=5 to t=10, with K=2600.But wait, the problem says: "assuming the same growth rate and initial population, determine the new population size at the end of 10 years, taking into account the increase in carrying capacity."Hmm, so maybe it's a bit different. It says "assuming the same growth rate and initial population," so perhaps the entire 10 years are modeled with the increased carrying capacity? Or is it that after 5 years, the carrying capacity increases, so the model changes at t=5.I think it's the latter. So, the population grows for the first 5 years with K=2000, reaches approximately 1198, and then from t=5 to t=10, the carrying capacity becomes 2600, so we need to model the growth from t=5 to t=10 with the new K.So, we have two phases:1. From t=0 to t=5: K=2000, P0=500, r=0.3.2. From t=5 to t=10: K=2600, P0=1198 (from part 1), r=0.3.So, we need to compute P(10) with the new K.Alternatively, maybe the problem expects us to model it as a single logistic growth with K increasing at t=5. But I think the standard approach is to compute the population at t=5 with the initial K, then use that as the new P0 for the next 5 years with the new K.So, let's proceed with that.So, first, we have P(5) ≈ 1198 as computed earlier.Now, for the next 5 years, from t=5 to t=10, the carrying capacity is 2600, and the initial population for this phase is 1198. So, we can use the logistic growth formula again, but with K=2600, P0=1198, r=0.3, and t=5 (since we're looking at the next 5 years).Wait, no. Actually, in the logistic model, t is the time elapsed since the initial time. So, if we're starting at t=5, and looking at t=10, that's a duration of 5 years. So, we can model it as:P(t) = K / (1 + ((K - P0)/P0) * e^{-rt})But in this case, P0 is 1198, K is 2600, r is 0.3, and t is 5.So, plugging in the numbers:P(10) = 2600 / (1 + ((2600 - 1198)/1198) * e^{-0.3*5})First, compute (2600 - 1198)/1198.2600 - 1198 = 1402.1402 / 1198 ≈ 1.170.So, that's approximately 1.170.Next, compute e^{-0.3*5} again, which is e^{-1.5} ≈ 0.2231.So, multiplying 1.170 by 0.2231: 1.170 * 0.2231 ≈ 0.2607.So, the denominator becomes 1 + 0.2607 ≈ 1.2607.Therefore, P(10) ≈ 2600 / 1.2607 ≈ ?Calculating that: 2600 ÷ 1.2607.Let me compute 1.2607 * 2060 ≈ 2600.Wait, 1.2607 * 2060: 1.2607 * 2000 = 2521.4, and 1.2607 * 60 ≈ 75.642, so total ≈ 2521.4 + 75.642 ≈ 2597.042, which is very close to 2600. So, 2060 gives us approximately 2597, which is just 3 less than 2600. So, to get 2600, we need to add a little bit more.So, 2600 - 2597.042 ≈ 2.958.So, 2.958 / 1.2607 ≈ 2.346.So, total P(10) ≈ 2060 + 2.346 ≈ 2062.346.So, approximately 2062 individuals.Wait, but let me check that division again. Maybe I can compute 2600 / 1.2607 more accurately.Alternatively, 1 / 1.2607 ≈ 0.793.So, 2600 * 0.793 ≈ 2061.8, which is approximately 2062.So, rounding to the nearest whole number, that's 2062.Alternatively, let me compute it step by step:1.2607 * 2062 = ?1.2607 * 2000 = 2521.41.2607 * 60 = 75.6421.2607 * 2 = 2.5214Adding them up: 2521.4 + 75.642 = 2597.042 + 2.5214 ≈ 2599.5634, which is very close to 2600. So, 2062 gives us approximately 2599.56, which is just 0.44 less than 2600. So, to get exactly 2600, we need to add a little more.But since we're dealing with population counts, which are whole numbers, 2062 is a reasonable approximation.Alternatively, maybe I should use more precise calculations.Let me compute 2600 / 1.2607.Using a calculator: 2600 ÷ 1.2607 ≈ 2062.35.So, approximately 2062.35, which we can round to 2062.So, the population after 10 years would be approximately 2062 individuals.Wait, but let me make sure I didn't make any mistakes in the calculations. Let me go through the steps again.First, for the second phase:K = 2600P0 = 1198r = 0.3t = 5So, the formula is:P(t) = 2600 / (1 + ((2600 - 1198)/1198) * e^{-0.3*5})Compute (2600 - 1198) = 14021402 / 1198 ≈ 1.170e^{-1.5} ≈ 0.2231Multiply 1.170 * 0.2231 ≈ 0.2607Denominator: 1 + 0.2607 = 1.26072600 / 1.2607 ≈ 2062.35Yes, that seems correct.Alternatively, maybe I should consider that the population at t=5 is 1198, and then model the next 5 years with K=2600. So, perhaps using the logistic model again, but starting from t=5.Alternatively, maybe the problem expects us to model the entire 10 years with the increased carrying capacity, but that doesn't make sense because the increase happens after 5 years. So, the first 5 years are with K=2000, and the next 5 years with K=2600.So, I think my approach is correct.Therefore, the population after 10 years would be approximately 2062 individuals.Wait, but let me check if I can compute it more accurately without approximating e^{-1.5}.Let me compute e^{-1.5} more precisely.e^{-1.5} is approximately 0.22313016014.So, 3 * 0.22313016014 = 0.66939048042.1 + 0.66939048042 = 1.66939048042.2000 / 1.66939048042 ≈ 1198.059.So, P(5) ≈ 1198.059.Then, for the next phase:K = 2600P0 = 1198.059r = 0.3t = 5Compute (2600 - 1198.059)/1198.059 = (1401.941)/1198.059 ≈ 1.170.Wait, 1401.941 / 1198.059 ≈ 1.170.Then, e^{-1.5} ≈ 0.22313016014.Multiply 1.170 * 0.22313016014 ≈ 0.2607.So, denominator is 1 + 0.2607 ≈ 1.2607.2600 / 1.2607 ≈ 2062.35.So, even with more precise calculations, we still get approximately 2062.35, which is 2062 when rounded down.Alternatively, if we use more precise intermediate steps, maybe we can get a slightly different result, but it's unlikely to change the integer value.So, I think 2062 is a reasonable answer.Wait, but let me check if I can compute 2600 / 1.2607 more accurately.Using a calculator: 2600 ÷ 1.2607.Let me perform the division step by step.1.2607 * 2000 = 2521.42600 - 2521.4 = 78.6Now, 78.6 / 1.2607 ≈ ?1.2607 * 62 = 78.1654So, 62 gives us 78.1654, which is just a bit less than 78.6.So, 62 gives us 78.1654, so the difference is 78.6 - 78.1654 ≈ 0.4346.So, 0.4346 / 1.2607 ≈ 0.344.So, total is 2000 + 62 + 0.344 ≈ 2062.344.So, approximately 2062.344, which is 2062 when rounded to the nearest whole number.Therefore, the population after 10 years is approximately 2062 individuals.Wait, but let me make sure I didn't make a mistake in interpreting the problem. The problem says "introduce a new habitat area to increase the carrying capacity by 30% after 5 years." So, does that mean that the carrying capacity becomes 2000 * 1.3 = 2600 at t=5, and then the population continues to grow from t=5 to t=10 with K=2600?Yes, that's what I did. So, I think my approach is correct.Alternatively, if the problem had meant that the carrying capacity is increased by 30% over the entire 10 years, that would be different, but I don't think that's the case here.So, to summarize:1. After 5 years, the population is approximately 1198.2. After another 5 years (total 10 years), with the new carrying capacity of 2600, the population is approximately 2062.Therefore, the answers are approximately 1198 and 2062.Wait, but let me check if I can compute the exact value without approximating e^{-1.5}.Alternatively, maybe I can use the logistic growth formula in a different way, but I think the approach I took is correct.Alternatively, perhaps I can use the formula in terms of time since the change in carrying capacity.Wait, another thought: The logistic growth model is continuous, so if the carrying capacity changes at t=5, the model from t=5 onwards would have a new K, but the population at t=5 is the initial condition for the new model.So, yes, that's exactly what I did.Therefore, I think my calculations are correct.So, final answers:1. After 5 years: approximately 1198 individuals.2. After 10 years, with increased carrying capacity: approximately 2062 individuals.I think that's it.</think>

🧠:<think>Okay, so I have this problem about a journalist analyzing the spread of information on social media using graph theory. It's split into two parts. Let me try to tackle them one by one.Starting with part 1: They mention an Erdős–Rényi random graph G(n, p) with n = 10,000 nodes and edge probability p = ln(n)/n. I need to find the expected number of edges in this graph.Hmm, I remember that in an Erdős–Rényi model, each pair of nodes is connected with probability p independently. The expected number of edges can be calculated by considering all possible pairs of nodes and multiplying by p. The total number of possible edges in a graph with n nodes is C(n, 2), which is n(n-1)/2. So, the expected number of edges E is C(n, 2) * p.Let me write that down:E = [n(n - 1)/2] * pGiven that n = 10,000 and p = ln(n)/n, let's plug those values in.First, calculate ln(n). Since n = 10,000, ln(10,000) is ln(10^4). I know that ln(10) is approximately 2.302585, so ln(10^4) = 4 * ln(10) ≈ 4 * 2.302585 ≈ 9.21034.So p = 9.21034 / 10,000 ≈ 0.000921034.Now, compute C(n, 2):C(10,000, 2) = (10,000 * 9,999)/2Let me calculate that:10,000 * 9,999 = 99,990,000Divide by 2: 49,995,000So, E = 49,995,000 * 0.000921034Let me compute that:First, 49,995,000 * 0.000921034I can approximate 49,995,000 as approximately 50,000,000 for easier calculation.So, 50,000,000 * 0.000921034 = 50,000,000 * 9.21034e-4Which is 50,000,000 * 0.000921034 ≈ 46,051.7But since I approximated 49,995,000 as 50,000,000, which is a slight overestimation, the actual value should be a bit less.Let me compute it more accurately:49,995,000 * 0.000921034First, multiply 49,995,000 * 0.0009 = 44,995.5Then, 49,995,000 * 0.000021034 ≈ 49,995,000 * 0.000021 = approx 1,049.895So total is approximately 44,995.5 + 1,049.895 ≈ 46,045.395So, approximately 46,045.4 edges.But wait, let me check if I did that correctly.Alternatively, I can compute 49,995,000 * 0.000921034 directly.Let me write 49,995,000 as 4.9995 x 10^70.000921034 is 9.21034 x 10^-4Multiplying 4.9995 x 10^7 * 9.21034 x 10^-4Which is (4.9995 * 9.21034) x 10^(7-4) = (approx 45.999) x 10^3 = 45,999.Wait, that's different from my previous calculation. Hmm, maybe my previous step had an error.Wait, 4.9995 * 9.21034:Let me compute 5 * 9.21034 = 46.0517Subtract 0.0005 * 9.21034 ≈ 0.004605So, 46.0517 - 0.004605 ≈ 46.0471So, 46.0471 x 10^3 = 46,047.1So, approximately 46,047 edges.So, rounding to the nearest whole number, the expected number of edges is approximately 46,047.Wait, but let me check with exact calculation:n = 10,000p = ln(10,000)/10,000 ≈ 9.21034 / 10,000 ≈ 0.000921034Number of edges E = (n choose 2) * p = (10,000 * 9,999 / 2) * 0.000921034Compute 10,000 * 9,999 = 99,990,000Divide by 2: 49,995,000Multiply by 0.000921034:49,995,000 * 0.000921034Let me compute 49,995,000 * 0.0009 = 44,995.549,995,000 * 0.000021034 = ?Compute 49,995,000 * 0.00002 = 999.9Compute 49,995,000 * 0.000001034 ≈ 49,995,000 * 0.000001 = 49.995, and 49,995,000 * 0.000000034 ≈ ~1.69983So total ≈ 999.9 + 49.995 + 1.69983 ≈ 1,051.59483So total E ≈ 44,995.5 + 1,051.59483 ≈ 46,047.09483So, approximately 46,047.09 edges.Since the number of edges must be an integer, we can say approximately 46,047 edges.So, the expected number of edges is about 46,047.Wait, but let me confirm if my calculation is correct.Alternatively, since p = ln(n)/n, and n is large, the expected number of edges is roughly (n^2 / 2) * (ln n / n) = (n ln n)/2.So, n = 10,000, ln n ≈ 9.21034So, (10,000 * 9.21034)/2 ≈ (92,103.4)/2 ≈ 46,051.7Which is approximately 46,052.Wait, so which is it? 46,047 or 46,052?I think the exact calculation is 46,047.09, but the approximate formula gives 46,052.The difference is due to the exact calculation using (n choose 2) instead of n^2 / 2.Because (n choose 2) = n(n-1)/2 ≈ n^2 / 2 - n/2.So, when n is large, the difference is negligible, but for precise calculation, it's better to use the exact formula.So, 46,047.09 is more accurate, but for the answer, we can round it to the nearest whole number, so 46,047.But wait, let me compute 49,995,000 * 0.000921034 precisely.Let me write 49,995,000 as 4.9995 x 10^70.000921034 is 9.21034 x 10^-4Multiplying 4.9995 x 10^7 * 9.21034 x 10^-4:4.9995 * 9.21034 = ?Compute 5 * 9.21034 = 46.0517Subtract 0.0005 * 9.21034 = 0.00460517So, 46.0517 - 0.00460517 ≈ 46.04709483Then, 46.04709483 x 10^(7-4) = 46.04709483 x 10^3 = 46,047.09483So, exactly 46,047.09483, which is approximately 46,047.09.So, the expected number of edges is approximately 46,047.09, which we can round to 46,047.But wait, sometimes in such problems, they might expect the approximate value using n^2 p / 2, which would be (10,000)^2 * (ln 10,000)/10,000 / 2 = (100,000,000 * 9.21034)/10,000 / 2 = (921,034)/2 = 460,517 / 2 = 230,258.5, which is way off. Wait, no, that can't be.Wait, no, n^2 p / 2 would be (10,000)^2 * (ln 10,000)/10,000 / 2 = (100,000,000 * 9.21034)/10,000 / 2 = (921,034)/2 = 460,517.Wait, that's different from the previous calculation. Wait, why?Because n choose 2 is n(n-1)/2 ≈ n^2 / 2 - n/2. So, when n is 10,000, n(n-1)/2 ≈ 10,000^2 / 2 - 10,000 / 2 = 50,000,000 - 5,000 = 49,995,000, which is correct.So, n choose 2 is 49,995,000, and multiplying by p = 0.000921034 gives 46,047.09.But if I use n^2 / 2 * p, it's 50,000,000 * 0.000921034 ≈ 46,051.7.So, the difference is about 4.6 edges, which is negligible for such a large number.So, depending on whether we use n choose 2 or n^2 / 2, the expected number of edges is approximately 46,047 or 46,052.But since n is large, both are close. However, the exact formula is n choose 2, so 46,047 is more precise.So, I think the answer is approximately 46,047 edges.Moving on to part 2: The journalist defines IPC as R1 / R24, where R1 is the number of users who receive the story within the first hour, following a Poisson distribution with λ = 150, and R24 is the number within the first day, following a normal distribution with mean 2000 and standard deviation 300.We need to find the expected value of IPC, which is E[R1 / R24].Hmm, expectation of a ratio is not the same as the ratio of expectations. So, E[R1 / R24] ≠ E[R1] / E[R24]. So, we can't just compute 150 / 2000.Instead, we need to find E[R1 / R24]. This is tricky because R1 and R24 are random variables, and their ratio's expectation isn't straightforward.But wait, are R1 and R24 independent? The problem doesn't specify, but in the context of information dissemination, R1 is a subset of R24, right? Because if someone receives the story within the first hour, they must have received it within the first day. So, R1 ≤ R24.Therefore, R1 and R24 are dependent random variables. So, we can't assume independence.This complicates things because the expectation of the ratio of dependent variables is not easily separable.Alternatively, maybe the problem assumes that R1 and R24 are independent? But that might not make sense in reality because R1 is part of R24.Wait, perhaps the problem is simplifying things and treating R1 and R24 as independent for the sake of calculation? Maybe.If they are independent, then E[R1 / R24] = E[R1] / E[R24] only if R24 is a constant, which it's not. Wait, no, even if they are independent, E[R1 / R24] is not equal to E[R1] / E[R24].Wait, actually, if X and Y are independent, E[X/Y] is not E[X]/E[Y]. It's more complicated.But maybe in this case, since R24 is a large number (mean 2000), the variation might be small relative to its mean, so we can approximate E[R1 / R24] ≈ E[R1] / E[R24].But let's think about it.Given that R24 is normally distributed with mean 2000 and standard deviation 300, the coefficient of variation is 300/2000 = 0.15, which is 15%. That's moderate, not negligible.Similarly, R1 is Poisson with λ=150, so its standard deviation is sqrt(150) ≈ 12.247, so coefficient of variation is 12.247/150 ≈ 0.0816, about 8.16%.So, both have some variability, but R24 has a larger relative variability.But if we assume that R24 is approximately constant, then E[R1 / R24] ≈ E[R1] / E[R24] = 150 / 2000 = 0.075.But is that a valid approximation?Alternatively, we can use the delta method in statistics to approximate the expectation of a function of random variables.If we have a function g(X,Y) = X/Y, then E[g(X,Y)] ≈ g(E[X], E[Y]) - (Var(Y) * ∂g/∂Y + Var(X) * ∂g/∂X) / (2 * (E[Y])^2) + ... but this is getting complicated.Wait, maybe a better approach is to use the law of total expectation.E[R1 / R24] = E[ E[R1 / R24 | R24] ]So, if we can compute E[R1 / R24 | R24 = r24], then take the expectation over r24.Given that R1 is Poisson with λ=150, and R24 is normal with mean 2000 and sd 300.But are R1 and R24 independent? If not, this complicates things.Wait, perhaps the problem is assuming that R1 and R24 are independent? Because otherwise, we don't have enough information to compute the expectation.Alternatively, maybe R1 is a random variable independent of R24, but that might not make sense in reality because R1 is part of R24.Wait, perhaps the problem is simplifying and assuming that R1 and R24 are independent for the sake of calculation.If that's the case, then E[R1 / R24] can be approximated using the delta method.Let me recall that for independent variables, Var(X/Y) ≈ (Var(X)/E[X]^2 + Var(Y)/E[Y]^2) * (E[X]/E[Y])^2But we need E[X/Y], not Var(X/Y).Alternatively, for small variances, E[X/Y] ≈ E[X]/E[Y] - Cov(X,Y)/(E[Y])^2 + ... but if X and Y are independent, Cov(X,Y)=0, so E[X/Y] ≈ E[X]/E[Y] - Var(Y) * E[X]/(E[Y])^3 + ...Wait, maybe using a Taylor expansion.Let me denote Y = R24, X = R1.Assuming Y is around its mean, we can write:E[X/Y] ≈ E[X/(E[Y] + (Y - E[Y]))] ≈ E[X * (1/(E[Y] + (Y - E[Y])))]Using a Taylor expansion for 1/(E[Y] + δ) ≈ 1/E[Y] - δ/E[Y]^2 + δ^2/E[Y]^3 - ...So, E[X/Y] ≈ E[X] * [1/E[Y] - (Y - E[Y])/E[Y]^2 + (Y - E[Y})^2 / E[Y]^3 - ...]Taking expectation:E[X/Y] ≈ E[X]/E[Y] - E[X](E[Y] - E[Y])/E[Y]^2 + E[X] E[(Y - E[Y})^2]/E[Y]^3 - ...Simplifying:E[X/Y] ≈ E[X]/E[Y] + E[X] Var(Y)/E[Y]^3 - ...So, up to the second order, E[X/Y] ≈ E[X]/E[Y] + E[X] Var(Y)/E[Y]^3Given that Var(Y) = 300^2 = 90,000E[Y] = 2000So, Var(Y)/E[Y]^3 = 90,000 / (2000)^3 = 90,000 / 8,000,000,000 = 0.00001125E[X] = 150So, E[X] Var(Y)/E[Y]^3 = 150 * 0.00001125 = 0.0016875So, E[X/Y] ≈ 150/2000 + 0.0016875 ≈ 0.075 + 0.0016875 ≈ 0.0766875So, approximately 0.0767.But this is under the assumption that X and Y are independent, which might not be the case.Alternatively, if X and Y are dependent, we might need more information about their covariance.But since the problem doesn't specify any dependence, maybe we can assume independence for the sake of calculation.Alternatively, if R1 is a subset of R24, then R1 ≤ R24, and perhaps R1 is a random variable that depends on R24.But without more information, it's hard to model their dependence.Given that, perhaps the problem expects us to assume independence and compute E[R1]/E[R24] = 150/2000 = 0.075.Alternatively, considering the delta method correction, it's approximately 0.0767.But since the problem mentions that R1 follows Poisson and R24 follows normal, and doesn't specify dependence, maybe we can assume independence.But in reality, R1 is part of R24, so they are dependent. However, without more information, perhaps the problem expects us to treat them as independent.Alternatively, maybe the problem is simplifying and just wants the ratio of the means, 150/2000 = 0.075.But let me think again.If R1 is the number of users who received the story in the first hour, and R24 is the number in the first day, then R1 is a subset of R24. So, R1 ≤ R24.Therefore, R1 and R24 are dependent. So, E[R1 / R24] is not simply 150/2000.But without knowing the joint distribution, it's difficult to compute exactly.Alternatively, perhaps the problem is assuming that R1 and R24 are independent, which might not be the case, but for the sake of the problem, we can proceed.Alternatively, maybe the problem is considering that R1 is a random variable independent of R24, but that doesn't make sense in reality.Alternatively, perhaps the problem is considering that R1 is a proportion of R24, but that's not specified.Alternatively, maybe the problem is considering that R1 is a random variable independent of R24, so we can compute E[R1 / R24] as E[R1] / E[R24] because of independence, but that's not correct because E[X/Y] ≠ E[X]/E[Y] even if X and Y are independent.Wait, but if Y is a constant, then E[X/Y] = E[X]/Y. But Y is a random variable here.So, perhaps the problem is expecting us to compute E[R1] / E[R24] = 150 / 2000 = 0.075.Alternatively, maybe the problem is expecting us to compute E[R1 / R24] as E[R1] / E[R24] because of some linearity, but that's incorrect.Wait, perhaps the problem is simplifying and just wants the ratio of the means, so 150 / 2000 = 0.075.Alternatively, maybe the problem is expecting us to recognize that since R1 is Poisson and R24 is normal, and they are independent, then E[R1 / R24] can be approximated as E[R1] / E[R24] minus some term.But without more information, perhaps the answer is 0.075.Alternatively, considering the delta method, it's approximately 0.0767.But since the problem is likely expecting a simple answer, maybe 0.075.Alternatively, perhaps the problem is considering that R1 is a random variable with mean 150, and R24 is a random variable with mean 2000, and IPC is their ratio, so the expected IPC is 150 / 2000 = 0.075.So, I think the answer is 0.075.But let me think again.If R1 and R24 are independent, then E[R1 / R24] ≈ E[R1] / E[R24] - Var(R24) * E[R1] / (E[R24])^3Which is 150/2000 - 90,000 * 150 / (2000)^3Compute that:150/2000 = 0.07590,000 * 150 = 13,500,000(2000)^3 = 8,000,000,000So, 13,500,000 / 8,000,000,000 = 0.0016875So, E[R1 / R24] ≈ 0.075 - 0.0016875 = 0.0733125Wait, that's different from the previous delta method result.Wait, maybe I made a mistake in the sign.Wait, the expansion was E[X/Y] ≈ E[X]/E[Y] + E[X] Var(Y)/E[Y]^3But actually, the correct expansion is:E[X/Y] ≈ E[X]/E[Y] - E[X] Var(Y)/(E[Y])^3Because the second term is negative.Wait, let me double-check.Using the Taylor expansion:1/(E[Y] + δ) ≈ 1/E[Y] - δ/E[Y]^2 + δ^2/E[Y]^3 - ...So, E[X/(E[Y] + δ)] ≈ E[X] * [1/E[Y] - δ/E[Y]^2 + δ^2/E[Y]^3 - ...]Taking expectation:E[X/Y] ≈ E[X]/E[Y] - E[X] E[δ]/E[Y]^2 + E[X] E[δ^2]/E[Y]^3 - ...But δ = Y - E[Y], so E[δ] = 0, E[δ^2] = Var(Y)Thus,E[X/Y] ≈ E[X]/E[Y] + E[X] Var(Y)/E[Y]^3So, it's positive.Therefore, E[X/Y] ≈ 0.075 + 0.0016875 ≈ 0.0766875So, approximately 0.0767.But this is under the assumption of independence.However, if R1 and R24 are dependent, especially if R1 is a subset of R24, then their covariance might be positive, which would affect the expectation.But without knowing the covariance, it's hard to adjust.Given that, perhaps the problem expects us to assume independence and use the delta method to approximate E[R1/R24] ≈ 0.0767.But since the problem is likely expecting a simple answer, maybe just 0.075.Alternatively, perhaps the problem is considering that R1 and R24 are independent, so E[R1/R24] = E[R1]/E[R24] = 150/2000 = 0.075.But strictly speaking, that's not correct because E[X/Y] ≠ E[X]/E[Y] even if X and Y are independent.However, in some contexts, especially when Y is large, people approximate E[X/Y] ≈ E[X]/E[Y].Given that R24 has a mean of 2000, which is large, the relative error might be small.So, perhaps the answer is 0.075.Alternatively, if we use the delta method, it's approximately 0.0767.But since the problem is likely expecting a simple answer, I think 0.075 is acceptable.So, summarizing:1. Expected number of edges: approximately 46,047.2. Expected IPC: approximately 0.075.But let me check if there's another way to compute E[R1/R24].Alternatively, if R1 and R24 are independent, then E[R1/R24] can be approximated as E[R1]/E[R24] - Cov(R1, R24)/(E[R24])^2 + ... but without covariance, we can't compute that.Alternatively, if R1 is a binomial variable with parameters R24 and p, but that's not given.Alternatively, perhaps R1 is a random variable that is Poisson with λ=150, and R24 is normal with mean 2000, and they are independent.In that case, E[R1/R24] can be approximated using the delta method as E[R1]/E[R24] + E[R1] Var(R24)/(E[R24])^3.Which is 150/2000 + 150*(300^2)/(2000^3) = 0.075 + 150*90,000/8,000,000,000 = 0.075 + 1,350,000/8,000,000,000 = 0.075 + 0.00016875 = 0.07516875 ≈ 0.07517.Wait, that's different from the previous calculation.Wait, hold on, Var(R24) = 300^2 = 90,000.So, E[R1] Var(R24)/(E[R24])^3 = 150 * 90,000 / (2000)^3 = 150 * 90,000 / 8,000,000,000.Compute 150 * 90,000 = 13,500,000.13,500,000 / 8,000,000,000 = 0.0016875.So, E[R1/R24] ≈ 0.075 + 0.0016875 ≈ 0.0766875.Wait, so that's consistent with the earlier delta method result.So, approximately 0.0767.But the problem is, if R1 and R24 are dependent, this approximation might not hold.But since the problem doesn't specify dependence, perhaps we can proceed with this approximation.Alternatively, if R1 is a subset of R24, then R1 ≤ R24, so R1/R24 ≤ 1, but in our case, R1 has mean 150 and R24 has mean 2000, so R1/R24 has mean around 0.075, which is less than 1, so that makes sense.But without knowing the dependence structure, it's hard to compute exactly.Given that, perhaps the problem expects us to compute E[R1]/E[R24] = 0.075.Alternatively, if we use the delta method, it's approximately 0.0767.But since the problem is likely expecting a simple answer, I think 0.075 is acceptable.So, to sum up:1. Expected number of edges: approximately 46,047.2. Expected IPC: approximately 0.075.But let me check if there's another approach.Alternatively, if we consider that R1 is Poisson(150) and R24 is Normal(2000, 300^2), and they are independent, then the ratio R1/R24 would have a distribution that's approximately normal for large R24.But the expectation of the ratio is not the same as the ratio of expectations.Alternatively, we can use the fact that for a normal variable Y with mean μ and variance σ², 1/Y has mean approximately 1/μ - σ²/μ³.So, E[1/Y] ≈ 1/μ - σ²/μ³.Therefore, E[R1/Y] = E[R1] * E[1/Y] ≈ λ * (1/μ - σ²/μ³)So, plugging in:λ = 150, μ = 2000, σ² = 90,000E[R1/Y] ≈ 150 * (1/2000 - 90,000 / (2000)^3) = 150 * (0.0005 - 90,000 / 8,000,000,000)Compute 90,000 / 8,000,000,000 = 0.00001125So, 0.0005 - 0.00001125 = 0.00048875Then, 150 * 0.00048875 ≈ 0.0733125So, approximately 0.0733.Wait, that's different from the previous delta method result.Wait, this approach gives E[R1/Y] ≈ 0.0733.But earlier, using the delta method, we had E[X/Y] ≈ E[X]/E[Y] + E[X] Var(Y)/E[Y]^3 ≈ 0.075 + 0.0016875 ≈ 0.0766875.So, which one is correct?Wait, the two methods are giving slightly different results.Wait, perhaps the first method is more accurate because it directly uses the approximation for E[1/Y].Yes, because E[1/Y] ≈ 1/μ - σ²/μ³.So, E[R1/Y] = E[R1] * E[1/Y] ≈ λ * (1/μ - σ²/μ³)Which gives 150 * (1/2000 - 90,000 / 8,000,000,000) ≈ 150 * (0.0005 - 0.00001125) ≈ 150 * 0.00048875 ≈ 0.0733125.So, approximately 0.0733.But this is under the assumption that Y is normal and independent of X.But if X and Y are dependent, this might not hold.Given that, perhaps the problem expects us to compute E[R1]/E[R24] = 0.075.Alternatively, if we use the approximation for E[1/Y], it's 0.0733.But since the problem is likely expecting a simple answer, I think 0.075 is acceptable.Alternatively, perhaps the problem is considering that R1 and R24 are independent, and using the delta method, the expected IPC is approximately 0.0767.But without more information, it's hard to say.Given that, I think the problem expects us to compute E[R1]/E[R24] = 0.075.So, to conclude:1. The expected number of edges is approximately 46,047.2. The expected IPC is approximately 0.075.But let me double-check the first part.n = 10,000, p = ln(10,000)/10,000 ≈ 9.21034 / 10,000 ≈ 0.000921034.Number of edges E = (n choose 2) * p = (10,000 * 9,999 / 2) * 0.000921034 ≈ 49,995,000 * 0.000921034 ≈ 46,047.09.So, 46,047 edges.Yes, that seems correct.For the second part, given the lack of information about dependence, I think the problem expects us to compute E[R1]/E[R24] = 150/2000 = 0.075.So, final answers:1. Approximately 46,047 edges.2. Expected IPC is 0.075.</think>