Appearance
â:As an organizer of local programming meetups and hackathons, you have noticed that the number of participants attending your events can be modeled by the function ( P(t) = 50e^{0.3t} ), where ( t ) is the number of months since you started organizing these events.1. Determine the time ( t ) in months it will take for the number of participants to reach 300. Express your answer in terms of natural logarithms.2. During a recent hackathon, the participants were divided into teams to solve a complex problem using PHP. The efficiency ( E ) of a team in solving the problem is inversely proportional to the square root of the number of team members ( n ). Given that a team of 4 members has an efficiency of 5, find the efficiency of a team with 9 members.
đ§ :<think>Okay, so I've got these two problems to solve. Let me start with the first one.1. The problem says that the number of participants at the events is modeled by the function ( P(t) = 50e^{0.3t} ), where ( t ) is the number of months since I started organizing. I need to find the time ( t ) when the number of participants reaches 300. They want the answer expressed in terms of natural logarithms, so I don't need to compute a numerical value, just leave it in terms of ln.Alright, so I know that ( P(t) = 50e^{0.3t} ). I need to set this equal to 300 and solve for ( t ). Let me write that equation out:( 50e^{0.3t} = 300 )Hmm, okay. To solve for ( t ), I should first isolate the exponential part. So I'll divide both sides by 50.( e^{0.3t} = frac{300}{50} )Simplifying the right side:( e^{0.3t} = 6 )Now, to solve for ( t ), I need to take the natural logarithm of both sides because the base is ( e ). Remember, ( ln(e^{x}) = x ).So, applying ln to both sides:( ln(e^{0.3t}) = ln(6) )Simplifying the left side:( 0.3t = ln(6) )Now, to solve for ( t ), I'll divide both sides by 0.3.( t = frac{ln(6)}{0.3} )Hmm, 0.3 is the same as ( frac{3}{10} ), so dividing by 0.3 is the same as multiplying by ( frac{10}{3} ). So, I can write this as:( t = frac{10}{3} ln(6) )Let me double-check my steps. Starting from ( P(t) = 50e^{0.3t} ), set equal to 300, divide both sides by 50 to get ( e^{0.3t} = 6 ). Take natural log of both sides, so ( 0.3t = ln(6) ), then ( t = ln(6)/0.3 ), which is the same as ( (10/3)ln(6) ). Yeah, that seems right.So, the time ( t ) is ( frac{10}{3} ln(6) ) months.Moving on to the second problem.2. During a recent hackathon, participants were divided into teams. The efficiency ( E ) of a team is inversely proportional to the square root of the number of team members ( n ). A team of 4 has an efficiency of 5. I need to find the efficiency of a team with 9 members.Alright, so inversely proportional means that ( E ) is proportional to ( 1/sqrt{n} ). So, mathematically, this can be written as:( E = frac{k}{sqrt{n}} )where ( k ) is the constant of proportionality.Given that when ( n = 4 ), ( E = 5 ). So, I can plug these values into the equation to find ( k ).So, substituting:( 5 = frac{k}{sqrt{4}} )Simplify ( sqrt{4} ) to 2:( 5 = frac{k}{2} )Multiply both sides by 2:( k = 10 )Okay, so now we know the constant ( k ) is 10. So the efficiency formula is:( E = frac{10}{sqrt{n}} )Now, we need to find the efficiency when ( n = 9 ). So, plug in 9 for ( n ):( E = frac{10}{sqrt{9}} )Simplify ( sqrt{9} ) to 3:( E = frac{10}{3} )Which is approximately 3.333..., but since the question doesn't specify, I can leave it as a fraction.Let me verify my steps. The efficiency is inversely proportional to the square root of n, so ( E = k/sqrt{n} ). Given E=5 when n=4, so 5 = k/2, so k=10. Then for n=9, E=10/3. Yep, that seems correct.So, the efficiency of a team with 9 members is ( frac{10}{3} ).Final Answer1. The time ( t ) is boxed{dfrac{10}{3} ln 6} months.2. The efficiency of a team with 9 members is boxed{dfrac{10}{3}}.</think>
â:An agricultural scientist is conducting a study on the growth rate of a particular crop variety. The scientist is skeptical about using advanced technology and instead relies on traditional methods of observation and manual data collection. The growth of the crop is modeled by a logistic growth function given by:[ P(t) = frac{K}{1 + frac{K - P_0}{P_0} e^{-rt}}, ]where ( P(t) ) is the population size of the crop at time ( t ), ( K ) is the carrying capacity of the environment, ( P_0 ) is the initial population size at ( t = 0 ), and ( r ) is the intrinsic growth rate.1. The scientist notices that without using technology, there is a delay in recognizing when the population reaches half of the carrying capacity. Calculate the time ( t_h ) when the population size ( P(t) ) reaches (frac{K}{2}). Express ( t_h ) in terms of ( K ), ( P_0 ), and ( r ).2. Due to the scientist's skepticism of technology, they decide to approximate the intrinsic growth rate ( r ) using manual measurements over a period of time. If the scientist records that the population size has doubled from ( P_0 ) to ( 2P_0 ) in ( T ) days, find an expression for the intrinsic growth rate ( r ) in terms of ( T ), ( K ), and ( P_0 ).
đ§ :<think>Okay, so I have this problem about a logistic growth model for a crop. The scientist is using traditional methods, which means he's not using any fancy technology, so he's manually collecting data. The model is given by:[ P(t) = frac{K}{1 + frac{K - P_0}{P_0} e^{-rt}} ]Alright, so part 1 is asking for the time ( t_h ) when the population reaches half the carrying capacity, which is ( frac{K}{2} ). Hmm, okay, so I need to set ( P(t) = frac{K}{2} ) and solve for ( t ).Let me write that down:[ frac{K}{2} = frac{K}{1 + frac{K - P_0}{P_0} e^{-rt}} ]Hmm, okay, so I can start by simplifying this equation. Maybe I can divide both sides by ( K ) to make it simpler.[ frac{1}{2} = frac{1}{1 + frac{K - P_0}{P_0} e^{-rt}} ]Then, taking reciprocals on both sides:[ 2 = 1 + frac{K - P_0}{P_0} e^{-rt} ]Subtract 1 from both sides:[ 1 = frac{K - P_0}{P_0} e^{-rt} ]Hmm, okay, so now I can solve for ( e^{-rt} ). Let me write that:[ e^{-rt} = frac{P_0}{K - P_0} ]Wait, that seems a bit off. Let me double-check my steps.Starting again:[ frac{K}{2} = frac{K}{1 + frac{K - P_0}{P_0} e^{-rt}} ]Divide both sides by ( K ):[ frac{1}{2} = frac{1}{1 + frac{K - P_0}{P_0} e^{-rt}} ]Take reciprocals:[ 2 = 1 + frac{K - P_0}{P_0} e^{-rt} ]Subtract 1:[ 1 = frac{K - P_0}{P_0} e^{-rt} ]So, yes, that seems correct. Then, solving for ( e^{-rt} ):[ e^{-rt} = frac{P_0}{K - P_0} ]Wait, but ( e^{-rt} ) is equal to that fraction. So, to solve for ( t ), I can take the natural logarithm of both sides.Taking ln:[ -rt = lnleft( frac{P_0}{K - P_0} right) ]Therefore, solving for ( t ):[ t_h = -frac{1}{r} lnleft( frac{P_0}{K - P_0} right) ]Hmm, that seems okay. But let me think if there's another way to write this. Maybe factor out the negative sign:[ t_h = frac{1}{r} lnleft( frac{K - P_0}{P_0} right) ]Yes, that looks better because the argument of the logarithm is positive, which makes sense since ( K > P_0 ) in a logistic growth model.So, that's part 1. I think that makes sense. Let me just recap:Set ( P(t) = K/2 ), solved for ( t ), ended up with ( t_h = frac{1}{r} lnleft( frac{K - P_0}{P_0} right) ). That seems correct.Moving on to part 2. The scientist notices that the population doubles from ( P_0 ) to ( 2P_0 ) in ( T ) days. So, we need to find an expression for ( r ) in terms of ( T ), ( K ), and ( P_0 ).Alright, so we have ( P(T) = 2P_0 ). Let's plug that into the logistic growth equation.[ 2P_0 = frac{K}{1 + frac{K - P_0}{P_0} e^{-rT}} ]Hmm, okay, let's solve this for ( r ).First, multiply both sides by the denominator:[ 2P_0 left(1 + frac{K - P_0}{P_0} e^{-rT}right) = K ]Let me distribute the ( 2P_0 ):[ 2P_0 + 2(K - P_0) e^{-rT} = K ]Subtract ( 2P_0 ) from both sides:[ 2(K - P_0) e^{-rT} = K - 2P_0 ]Divide both sides by ( 2(K - P_0) ):[ e^{-rT} = frac{K - 2P_0}{2(K - P_0)} ]Hmm, okay, so now take the natural logarithm of both sides:[ -rT = lnleft( frac{K - 2P_0}{2(K - P_0)} right) ]Therefore, solving for ( r ):[ r = -frac{1}{T} lnleft( frac{K - 2P_0}{2(K - P_0)} right) ]Hmm, let me see if that can be simplified. Maybe factor out the negative sign:[ r = frac{1}{T} lnleft( frac{2(K - P_0)}{K - 2P_0} right) ]Yes, that looks better. Let me check if the argument inside the logarithm is positive. Since ( K > 2P_0 ) (otherwise, the population can't double before reaching carrying capacity), so ( K - 2P_0 > 0 ), and ( K - P_0 > 0 ). Therefore, the fraction is positive, which is good because logarithm of a negative number is undefined.So, that seems to be the expression for ( r ).Wait, let me think again. If ( K ) is the carrying capacity, and the population doubles from ( P_0 ) to ( 2P_0 ), we must have ( 2P_0 < K ), otherwise, the population can't double without exceeding the carrying capacity. So, that condition is necessary for this to hold.Therefore, the expression for ( r ) is:[ r = frac{1}{T} lnleft( frac{2(K - P_0)}{K - 2P_0} right) ]I think that's correct.Let me recap part 2:Set ( P(T) = 2P_0 ), substituted into the logistic equation, solved for ( e^{-rT} ), took natural log, and solved for ( r ). Got to the expression above.So, I think that's the answer.Final Answer1. The time when the population reaches half the carrying capacity is (boxed{t_h = dfrac{1}{r} lnleft( dfrac{K - P_0}{P_0} right)}).2. The intrinsic growth rate is (boxed{r = dfrac{1}{T} lnleft( dfrac{2(K - P_0)}{K - 2P_0} right)}).</think>
â:A farmer, named John, is interested in expanding his agricultural operations by acquiring one of the neighboring parcels of land. John has identified two potential parcels: Parcel A and Parcel B. He wants to assess the value of these parcels based on their productivity potential and market value growth over time.1. Parcel A has a current market value of 500,000 and is expected to appreciate at an annual rate of 4%. However, it requires an initial investment in irrigation infrastructure costing 50,000, which will grow the productivity of the land by 20%. Assuming the productivity gain directly translates to an equivalent increase in market value, calculate the value of Parcel A after 5 years, including the effect of both appreciation and productivity gain.2. Parcel B has a current market value of 400,000 and an annual appreciation rate of 5%. This parcel is already well-irrigated and productive, requiring no additional investment. However, it is adjacent to a river that floods every 10 years on average, causing a 15% reduction in market value each time it floods. Assuming the flooding event occurs exactly every 10 years, calculate the expected market value of Parcel B after 10 years, accounting for both appreciation and the flooding event.Which parcel should John choose based on their respective values after the specified time periods?
đ§ :<think>First, I'll evaluate Parcel A. It currently has a market value of 500,000 and appreciates at 4% annually. Additionally, an initial investment of 50,000 in irrigation infrastructure increases productivity by 20%, which directly translates to a 20% increase in market value. So, the initial value after the productivity gain is 500,000 plus 20% of 500,000, which equals 600,000. Then, I'll calculate the appreciation over 5 years using the formula for compound interest: 600,000 multiplied by (1 + 0.04) raised to the power of 5. This gives the value of Parcel A after 5 years.Next, I'll assess Parcel B. Its current market value is 400,000 with an annual appreciation rate of 5%. Since it's already well-irrigated, there's no additional investment needed. However, it faces a flooding risk every 10 years, which reduces the market value by 15%. I'll calculate the appreciation over 10 years first: 400,000 multiplied by (1 + 0.05) raised to the power of 10. Then, I'll account for the flooding event by reducing the appreciated value by 15%. This will give the expected market value of Parcel B after 10 years.Finally, I'll compare the values of both parcels after their respective time periods to determine which one John should choose.</think>
â:An experienced nurse, working in a busy hospital, is responsible for managing the medication schedules of patients in the Intensive Care Unit (ICU). The nurse has observed that the medication efficacy decays exponentially over time according to the function ( E(t) = E_0 e^{-lambda t} ), where ( E_0 ) is the initial efficacy, ( lambda ) is the decay constant, and ( t ) is the time in hours.Sub-problem 1:One patient requires an antibiotic that initially has an efficacy (( E_0 )) of 100 units. The decay constant (( lambda )) for this medication is 0.3 per hour. The nurse administers the medication at 8:00 AM. Calculate the efficacy of the antibiotic at 2:00 PM on the same day.Sub-problem 2:The nurse needs to ensure that the efficacy of the medication does not drop below 20 units for effective treatment. Determine the latest possible time after 8:00 AM that the nurse can administer the next dose to maintain the required efficacy threshold.
đ§ :<think>Alright, so I've got these two sub-problems to solve about medication efficacy. Let me start with the first one.Sub-problem 1:Okay, the nurse administers an antibiotic at 8:00 AM, and we need to find the efficacy at 2:00 PM. The efficacy decays exponentially according to the function ( E(t) = E_0 e^{-lambda t} ). The initial efficacy ( E_0 ) is 100 units, and the decay constant ( lambda ) is 0.3 per hour.First, I need to figure out how much time has passed between 8:00 AM and 2:00 PM. Let me count the hours: from 8 AM to 12 PM is 4 hours, and then from 12 PM to 2 PM is another 2 hours. So, total time is 6 hours. Therefore, ( t = 6 ) hours.Now, plugging the values into the formula:( E(6) = 100 times e^{-0.3 times 6} )Let me compute the exponent first: ( 0.3 times 6 = 1.8 ). So, it becomes ( e^{-1.8} ).I remember that ( e^{-1} ) is approximately 0.3679, so ( e^{-1.8} ) would be less than that. Maybe I should use a calculator for a more precise value. But since I don't have one handy, perhaps I can approximate it.Alternatively, I can recall that ( e^{-1.8} ) is approximately 0.1653. Let me verify that. Hmm, actually, ( e^{-1} approx 0.3679 ), ( e^{-2} approx 0.1353 ). Since 1.8 is closer to 2 than to 1, maybe 0.1653 is a reasonable approximation.So, ( E(6) approx 100 times 0.1653 = 16.53 ) units.Wait, let me double-check that exponent calculation. 0.3 per hour times 6 hours is indeed 1.8. So, that's correct.Alternatively, if I use a calculator, I can compute ( e^{-1.8} ) more accurately. Let me think. The natural logarithm of 1.8 is approximately 0.5878, but that's not helpful here. Alternatively, using the Taylor series expansion for ( e^{-x} ) around x=0:( e^{-x} = 1 - x + frac{x^2}{2!} - frac{x^3}{3!} + cdots )But for x=1.8, this might not converge quickly. Maybe it's better to use known values.Wait, I think ( e^{-1.8} ) is approximately 0.1653. So, 100 times that is approximately 16.53. So, about 16.5 units.But let me see if I can compute it more precisely. Maybe using the fact that ( e^{-1.8} = e^{-1} times e^{-0.8} ). I know ( e^{-1} approx 0.3679 ) and ( e^{-0.8} approx 0.4493 ). Multiplying these together: 0.3679 * 0.4493 â 0.1653. Yep, that's consistent.So, the efficacy at 2:00 PM is approximately 16.53 units. Since the problem doesn't specify rounding, I can present it as approximately 16.5 units or keep it at 16.53.Wait, but maybe I should use a calculator for more precision. Let me try to compute ( e^{-1.8} ) step by step.Alternatively, I can use the fact that ( e^{-1.8} = 1 / e^{1.8} ). Let me compute ( e^{1.8} ).I know that ( e^1 = 2.71828 ), ( e^{0.8} approx 2.2255 ). So, ( e^{1.8} = e^{1 + 0.8} = e^1 times e^{0.8} â 2.71828 * 2.2255 â 6.05 ). Therefore, ( e^{-1.8} â 1 / 6.05 â 0.165289 ). So, 0.165289 * 100 â 16.5289, which is approximately 16.53 units.So, the efficacy at 2:00 PM is approximately 16.53 units.Sub-problem 2:Now, the second part is to determine the latest possible time after 8:00 AM that the nurse can administer the next dose so that the efficacy doesn't drop below 20 units.So, we need to find the time ( t ) when ( E(t) = 20 ).Given ( E(t) = E_0 e^{-lambda t} ), we have:( 20 = 100 e^{-0.3 t} )Let me solve for ( t ).First, divide both sides by 100:( 20 / 100 = e^{-0.3 t} )Simplify:( 0.2 = e^{-0.3 t} )Take the natural logarithm of both sides:( ln(0.2) = -0.3 t )Solve for ( t ):( t = - ln(0.2) / 0.3 )Compute ( ln(0.2) ). I know that ( ln(1) = 0 ), ( ln(0.5) â -0.6931 ), ( ln(0.2) ) is more negative. Let me recall that ( ln(0.2) â -1.6094 ).So, ( t = - (-1.6094) / 0.3 = 1.6094 / 0.3 â 5.3647 ) hours.So, approximately 5.3647 hours after 8:00 AM.Now, converting 0.3647 hours to minutes: 0.3647 * 60 â 21.88 minutes, which is roughly 21 minutes and 53 seconds.So, 5 hours and approximately 22 minutes after 8:00 AM.Adding 5 hours to 8:00 AM brings us to 1:00 PM. Adding another 22 minutes brings us to approximately 1:22 PM.Therefore, the latest possible time to administer the next dose is around 1:22 PM.Wait, let me verify the calculation for ( t ):( t = ln(20/100) / (-0.3) = ln(0.2) / (-0.3) )Which is ( ln(0.2) â -1.6094 ), so ( t â (-1.6094)/(-0.3) â 5.3647 ) hours.Yes, that's correct.Alternatively, using a calculator for more precision:( ln(0.2) â -1.60943791 )So, ( t â 1.60943791 / 0.3 â 5.364793 ) hours.Convert 0.364793 hours to minutes: 0.364793 * 60 â 21.8876 minutes, which is about 21 minutes and 53 seconds.So, 5 hours and 21.8876 minutes after 8:00 AM.Adding 5 hours to 8:00 AM is 1:00 PM. Adding 21.8876 minutes brings us to 1:21:53 PM.So, approximately 1:22 PM.Therefore, the nurse should administer the next dose by approximately 1:22 PM to ensure the efficacy doesn't drop below 20 units.Wait, but let me think again. The question says "the latest possible time after 8:00 AM". So, it's the time when the efficacy reaches 20 units, which is the threshold. So, the nurse needs to administer the next dose before the efficacy drops below 20, so the latest time is when it's exactly 20, which is at approximately 1:22 PM.So, summarizing:Sub-problem 1: Efficacy at 2:00 PM is approximately 16.53 units.Sub-problem 2: The latest time to administer the next dose is approximately 1:22 PM.But let me check if I did everything correctly.For Sub-problem 1:Time from 8 AM to 2 PM is 6 hours. So, t=6.E(t) = 100 * e^{-0.3*6} = 100 * e^{-1.8} â 100 * 0.1653 â 16.53.Yes, that seems correct.For Sub-problem 2:We set E(t) = 20.20 = 100 * e^{-0.3 t}Divide both sides by 100: 0.2 = e^{-0.3 t}Take natural log: ln(0.2) = -0.3 tSo, t = -ln(0.2)/0.3 â 5.3647 hours â 5 hours 22 minutes.Yes, that's correct.So, the answers are:1. Approximately 16.53 units at 2:00 PM.2. The latest time is approximately 1:22 PM.Wait, but the problem says "the latest possible time after 8:00 AM". So, 5.3647 hours after 8:00 AM is 1:22 PM.Yes, that's correct.I think that's all.</think>