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question:A personal trainer, Jordan, recommends a specific blend of supplements to enhance athletic performance. Jordan notices that the effectiveness of the supplements depends on two key factors: the timing of intake and the dosage relative to the athlete's body mass.1. Jordan conducts an experiment with 10 athletes and finds that the effectiveness ( E ) of the supplements can be modeled by the function ( E(t, d) = 5sin(frac{pi t}{6}) cdot ln(d) ), where ( t ) is the time of the day in hours when the supplements are taken (0 ≤ t ≤ 24), and ( d ) is the dosage in milligrams per kilogram of body mass. Determine the optimal time ( t ) and dosage ( d ) that maximizes the effectiveness ( E ). Use calculus to find the critical points and determine their nature (maximum, minimum, or saddle point).2. Additionally, Jordan observes that the athletes’ performance improvement ( P ) over a month can be modeled as a function of the average daily effectiveness ( bar{E} ), given by ( P(bar{E}) = 3bar{E}^2 + 2bar{E} + 1 ). If the athletes follow Jordan's optimal recommendations from the first sub-problem, calculate the performance improvement ( P ) over a month. Assume that the average daily effectiveness ( bar{E} ) is the same each day and use the results from the first sub-problem.
answer:Let's solve the given problems step-by-step. Problem 1: Determine the optimal time ( t ) and dosage ( d ) that maximizes the effectiveness ( E ).The effectiveness function is given by:[ E(t, d) = 5sinleft(frac{pi t}{6}right) ln(d) ]To find the critical points, we need to take the partial derivatives of ( E ) with respect to ( t ) and ( d ) and set them to zero.1. Partial derivative with respect to ( t ):[ E_t = frac{partial E}{partial t} = 5cosleft(frac{pi t}{6}right) cdot frac{pi}{6} ln(d) ]Set ( E_t = 0 ):[ 5cosleft(frac{pi t}{6}right) cdot frac{pi}{6} ln(d) = 0 ]Since ( ln(d) neq 0 ) (because ( d > 0 )), we have:[ cosleft(frac{pi t}{6}right) = 0 ]The solutions to ( cos(x) = 0 ) are:[ frac{pi t}{6} = frac{pi}{2} + kpi ][ t = 3 + 6k ]where ( k ) is an integer. Since ( 0 leq t leq 24 ), valid values of ( t ) are:[ t = 3, 9, 15, 21 ]2. Partial derivative with respect to ( d ):[ E_d = frac{partial E}{partial d} = 5sinleft(frac{pi t}{6}right) cdot frac{1}{d} ]Set ( E_d = 0 ):[ 5sinleft(frac{pi t}{6}right) cdot frac{1}{d} = 0 ]Since ( frac{1}{d} neq 0 ) (because ( d > 0 )), we have:[ sinleft(frac{pi t}{6}right) = 0 ]The solutions to ( sin(x) = 0 ) are:[ frac{pi t}{6} = kpi ][ t = 6k ]where ( k ) is an integer. Since ( 0 leq t leq 24 ), valid values of ( t ) are:[ t = 0, 6, 12, 18, 24 ]To find the optimal ( t ), we need to evaluate the effectiveness ( E(t, d) ) at the critical points. Evaluate ( E(t, d) ) at the critical points:- For ( t = 3, 9, 15, 21 ):[ sinleft(frac{pi t}{6}right) = pm 1 ]- For ( t = 0, 6, 12, 18, 24 ):[ sinleft(frac{pi t}{6}right) = 0 ]Clearly, ( E(t, d) = 0 ) at ( t = 0, 6, 12, 18, 24 ). Thus, we discard these values.At ( t = 3, 9, 15, 21 ), we have:[ E(t, d) = pm 5 ln(d) ]Since we are maximizing ( E ), consider:[ E(t, d) = 5 ln(d) ]The effectiveness ( E ) is maximized when ( ln(d) ) is maximized. Since ( ln(d) ) is increasing, the dosage ( d ) should be as large as possible. Therefore, the optimal values are:[ t = 3, 9, 15, 21 ][ d to infty ] Problem 2: Calculate the performance improvement ( P ) over a month.Given ( P(bar{E}) = 3bar{E}^2 + 2bar{E} + 1 ).The average daily effectiveness ( bar{E} = E(t, d) ) with optimal ( t ) and ( d ):[ bar{E} = 5 ln(d) ]As ( d to infty ), ( ln(d) to infty ). Therefore, ( bar{E} to infty ).Thus:[ P(bar{E}) = 3(5 ln(d))^2 + 2(5 ln(d)) + 1 ]Let ( x = 5 ln(d) ):[ P(x) = 3x^2 + 2x + 1 ]As ( x to infty ):[ P(x) to infty ] Final Answer:The final answer is ( infty ). I hope it is correct.
question:Mrs. Evelyn is an elderly woman who values the trust and unity of her local community. She has always been fond of seeing the youth repent and make amends for their mistakes, believing it strengthens the community bonds. Recently, she decided to organize a community project to plant trees in the neighborhood park, where each participant (youth) must plant a certain number of trees as a form of repentance.1. Mrs. Evelyn observes that if each of the 𝑛 participants plants 𝑘 trees, the total number of trees planted would be ( T = n cdot k ). However, she wants the total number of trees planted to be a perfect square. Additionally, Mrs. Evelyn believes that the number of participants should be a prime number. Given that the total number of trees planted is 144, find all possible pairs (𝑛, 𝑘) that satisfy these conditions.2. After successfully planting the trees, Mrs. Evelyn wants to create a circular flower bed around each tree. She decides that the radius of each flower bed should be directly proportional to the number of trees planted by each participant, with the proportionality constant being ( c = 0.5 ). If each flower bed must have an integer radius, determine the maximum and minimum possible radii of the flower beds, given the constraints from the first sub-problem.
answer:Let's solve the given problem step-by-step. Part 1: Finding pairs ((n, k))1. The total number of trees planted (T) is 144.2. We want (T = n cdot k) to be a perfect square.3. The number of participants (n) must be a prime number.4. Given (T = 144), we need to express 144 as a product of (n) and (k).First, note that (144 = 12^2), so it is indeed a perfect square. Now, we need to find pairs ((n, k)) such that (n) is a prime number and (k) is an integer.Since (n) is prime and (n cdot k = 144), (n) must be a prime factor of 144. Let's list the prime factors of 144:[ 144 = 2^4 times 3^2 ]The prime factors of 144 are 2 and 3. We need to find pairs ((n, k)) where (n) is one of these prime factors and (k) is an integer such that the product (n cdot k = 144).1. (n = 2) [ k = frac{144}{2} = 72 ] So, one pair is ((2, 72)).2. (n = 3) [ k = frac{144}{3} = 48 ] So, another pair is ((3, 48)).Next, we check larger primes:3. (n = 5), but ( frac{144}{5} ) is not an integer.4. (n = 7), but ( frac{144}{7} ) is not an integer.5. Continue this process, and the next valid prime factor is: - (n = 11), but ( frac{144}{11} ) is not an integer. - (n = 13), but ( frac{144}{13} ) is not an integer. - (n = 17, 19, ldots) (none yield integers).Thus, we only have the pairs ((2, 72)) and ((3, 48)). Part 2: Finding maximum and minimum radii1. The radius (r) of each flower bed is directly proportional to the number of trees planted by each participant, with proportionality constant (c = 0.5).[ r = 0.5 cdot k ]We want the radius (r) to be an integer, so we need (0.5 cdot k) to be an integer. This implies (k) must be an even number.Let's check the pairs:- For ((2, 72)): [ r = 0.5 cdot 72 = 36 ] Radius (r = 36) is an integer.- For ((3, 48)): [ r = 0.5 cdot 48 = 24 ] Radius (r = 24) is also an integer.Thus, the possible radii are 24 and 36.The maximum possible radius is 36, and the minimum possible radius is 24. Final Answer:The final answer is ((2, 72)) and ((3, 48)) for the pairs ((n, k)), with the maximum and minimum possible radii being 36 and 24, respectively. I hope it is correct.
question:A computer science student specializing in software development collaborates with a telecommunication engineering student on optimizing data transmission protocols. They are tasked with devising an efficient algorithm to manage data packets in a network, where the probability of data packet loss follows a Poisson distribution due to random noise.1. Suppose the average rate of packet loss is 3 packets per minute. The probability of a packet being successfully transmitted is given by ( P_t = e^{-lambda} left( sum_{k=0}^{n} frac{lambda^k}{k!} right) ), where (lambda) is the average rate of packet loss, and (n) is the maximum number of packets that can be lost without requiring retransmission. Given that the system can tolerate up to 2 lost packets without requiring retransmission, calculate the probability (P_t) that a packet is successfully transmitted.2. To further optimize the network, the students propose using a redundant packet transmission strategy, where each packet is sent twice to increase reliability. However, redundant transmissions increase network load. The telecommunication student models this load using a queue with Poisson arrivals and exponential service times (M/M/1 queue). If the arrival rate of original packets is 5 packets per minute and the service rate is 8 packets per minute, calculate the expected time a packet (including its redundant copy) spends in the system.
answer:To solve the given problem, we will address each part step-by-step.Part 1: Calculation of the Probability (P_t)We are given that the average rate of packet loss (lambda = 3) packets per minute and the system can tolerate up to (n = 2) lost packets without requiring retransmission. The probability (P_t) of a packet being successfully transmitted is given by:[P_t = e^{-lambda} left( sum_{k=0}^{n} frac{lambda^k}{k!} right)]Substituting (lambda = 3) and (n = 2), we get:[P_t = e^{-3} left( frac{3^0}{0!} + frac{3^1}{1!} + frac{3^2}{2!} right)]Calculating each term individually:- (frac{3^0}{0!} = 1)- (frac{3^1}{1!} = 3)- (frac{3^2}{2!} = frac{9}{2} = 4.5)Now, sum these values:[sum_{k=0}^{2} frac{3^k}{k!} = 1 + 3 + 4.5 = 8.5]Thus, the probability (P_t) is:[P_t = e^{-3} times 8.5]Calculating (e^{-3}):[e^{-3} approx 0.0498]Therefore:[P_t approx 0.0498 times 8.5 approx 0.4233]Part 2: Expected Time in the System for Redundant TransmissionsFor the M/M/1 queue, the arrival rate (lambda = 5) packets per minute and the service rate (mu = 8) packets per minute. The utilization factor (rho) is given by:[rho = frac{lambda}{mu} = frac{5}{8} = 0.625]The expected time (W) a packet spends in the system (waiting time plus service time) for an M/M/1 queue is given by:[W = frac{1}{mu - lambda} = frac{1}{8 - 5} = frac{1}{3} text{ minutes}]Since each packet is sent twice, we need to consider the processing time for both the original packet and its redundant copy. However, the M/M/1 model does not directly affect the time for individual packets due to redundancy, as the model considers the combined arrival rate already. Therefore, the expected time a packet (including its redundant copy) spends in the system remains:Final Answer: (P_t approx 0.4233) and (W = frac{1}{3} text{ minutes}). I hope it is correct.
question:Dr. Murray's old friend and colleague, an accomplished mathematician, often enjoyed exploring complex mathematical theories and concepts with him. They both shared a particular interest in topology and number theory. Inspired by their past collaborations, consider the following problem:1. Let ( X ) be a compact Hausdorff space, and let ( f: X rightarrow X ) be a continuous function. Prove that if ( f ) is homotopic to the identity map ( text{id}_X ), then there exists a point ( x_0 in X ) such that ( f(x_0) = x_0 ).2. Dr. Murray's colleague had a fascination with prime numbers and often challenged Dr. Murray with problems involving them. If ( p ) is an odd prime, show that the sum of the Legendre symbols ( left( frac{a}{p} right) ) for ( a = 1, 2, ldots, p-1 ) is zero. That is, prove that:[sum_{a=1}^{p-1} left( frac{a}{p} right) = 0]
answer:Let's approach each part of the problem step-by-step: Part 1: Existence of a Fixed PointTo prove that if ( f: X rightarrow X ) is homotopic to the identity map ( text{id}_X ), then there exists a point ( x_0 in X ) such that ( f(x_0) = x_0 ):# Step 1: Understand the Given Conditions- ( X ) is a compact Hausdorff space.- ( f: X rightarrow X ) is continuous.- ( f ) is homotopic to the identity map ( text{id}_X ), i.e., there exists a continuous map ( H: X times [0, 1] rightarrow X ) such that: - ( H(x, 0) = f(x) ) for all ( x in X ), - ( H(x, 1) = x ) for all ( x in X ).# Step 2: Apply the Lefschetz Fixed Point TheoremThe Lefschetz Fixed Point Theorem states that if ( L(f) neq 0 ), then ( f ) has a fixed point, where ( L(f) ) is the Lefschetz number of ( f ).- Since ( f ) is homotopic to ( text{id}_X ), the maps ( f ) and ( text{id}_X ) induce the same maps in homology, hence ( L(f) = L(text{id}_X) ).- For any space, the Lefschetz number of the identity map ( text{id}_X ) is equal to the Euler characteristic of ( X ), which is non-zero for compact spaces unless ( X ) is empty.- Therefore, ( L(f) = L(text{id}_X) neq 0 ).Thus, ( f ) must have a fixed point. Part 2: Sum of Legendre SymbolsTo prove that the sum of the Legendre symbols ( left( frac{a}{p} right) ) for ( a = 1, 2, ldots, p-1 ) is zero:# Step 1: Understand Legendre SymbolsThe Legendre symbol ( left( frac{a}{p} right) ) is defined as:[left( frac{a}{p} right) = begin{cases} 1 & text{if } a text{ is a quadratic residue modulo } p, -1 & text{if } a text{ is not a quadratic residue modulo } p, 0 & text{if } p mid a.end{cases}]Since ( a ) ranges from 1 to ( p-1 ), ( p nmid a ).# Step 2: Properties of Quadratic Residues- The number of quadratic residues modulo ( p ) among ( 1, 2, ldots, p-1 ) is ( frac{p-1}{2} ).- The number of non-quadratic residues is also ( frac{p-1}{2} ).# Step 3: Calculate the Sum- For each quadratic residue ( a ), ( left( frac{a}{p} right) = 1 ).- For each non-quadratic residue ( a ), ( left( frac{a}{p} right) = -1 ).Therefore, the sum is:[sum_{a=1}^{p-1} left( frac{a}{p} right) = left(frac{p-1}{2}right) cdot 1 + left(frac{p-1}{2}right) cdot (-1) = 0.]Final Answer: The final answer is that both parts are proved as requested. I hope it is correct.
question:A business magazine columnist is analyzing the growth patterns of two competing industries, Industry A and Industry B, over a decade. The columnist uses a critical yet optimistic perspective to model this growth with the following assumptions:1. Industry A's revenue ( R_A(t) ) in billions of dollars at year ( t ) can be modeled by the differential equation:[ frac{dR_A}{dt} = k_1 R_A left(1 - frac{R_A}{K}right) - m_1 R_A^2 ]where ( k_1 ), ( K ), and ( m_1 ) are positive constants.2. Industry B's revenue ( R_B(t) ) in billions of dollars at year ( t ) can be modeled by the differential equation:[ frac{dR_B}{dt} = k_2 R_B left(1 - frac{R_B}{L}right) + m_2 sin(omega t) ]where ( k_2 ), ( L ), ( m_2 ), and ( omega ) are positive constants.Sub-problems:1. Determine the equilibrium points for both industries' revenue models and analyze their stability. What do these equilibrium points represent in the context of industry growth?2. Assuming initial revenues ( R_A(0) = R_{A0} ) and ( R_B(0) = R_{B0} ), solve the differential equations for ( R_A(t) ) and ( R_B(t) ) given the specific constants ( k_1 = 0.1 ), ( K = 100 ), ( m_1 = 0.01 ), ( k_2 = 0.08 ), ( L = 80 ), ( m_2 = 5 ), and ( omega = frac{pi}{4} ). Plot the revenue growth over the decade for both industries and compare their long-term behavior.
answer:Let's solve each sub-problem step by step. Sub-problem 1: Determine the equilibrium points and their stabilityIndustry A:The differential equation for Industry A is:[ frac{dR_A}{dt} = k_1 R_A left(1 - frac{R_A}{K}right) - m_1 R_A^2 ]To find the equilibrium points, set (frac{dR_A}{dt} = 0):[ k_1 R_A left(1 - frac{R_A}{K}right) - m_1 R_A^2 = 0 ]Factor out (R_A):[ R_A left( k_1 left(1 - frac{R_A}{K}right) - m_1 R_A right) = 0 ]This gives us two equilibrium points:1. ( R_A = 0 )2. ( k_1 left(1 - frac{R_A}{K}right) = m_1 R_A )Solving for (R_A):[ k_1 - frac{k_1 R_A}{K} = m_1 R_A ][ k_1 = R_A left( m_1 + frac{k_1}{K} right) ][ R_A = frac{k_1 K}{k_1 + m_1 K} ]Thus, the equilibrium points for Industry A are:[ R_A = 0 quad text{and} quad R_A = frac{k_1 K}{k_1 + m_1 K} ]Stability Analysis for Industry A:1. At ( R_A = 0 ): - Linearize the equation around ( R_A = 0 ): [ frac{dR_A}{dt} approx k_1 R_A ] - The coefficient ( k_1 > 0 ), so ( R_A = 0 ) is unstable.2. At ( R_A = frac{k_1 K}{k_1 + m_1 K} ): - Linearize the equation around ( R_A = frac{k_1 K}{k_1 + m_1 K} ): [ R_A = frac{k_1 K}{k_1 + m_1 K} ] Let ( R_A = frac{k_1 K}{k_1 + m_1 K} + epsilon ) Substituting and simplifying, the term ( epsilon ) will decay if ( k_1 - left(m_1 + frac{k_1}{K}right) left(frac{k_1 K}{k_1 + m_1 K}right) < 0 ). - This condition holds, so ( R_A = frac{k_1 K}{k_1 + m_1 K} ) is stable.Industry B:The differential equation for Industry B is:[ frac{dR_B}{dt} = k_2 R_B left(1 - frac{R_B}{L}right) + m_2 sin(omega t) ]To find the equilibrium points, set (frac{dR_B}{dt} = 0):Since ( m_2 sin(omega t) ) is time-dependent and periodic, there are no fixed equilibrium points in the traditional sense. The term ( m_2 sin(omega t) ) induces oscillatory behavior around the equilibrium of the logistic part without the oscillatory term.Stability Analysis for Industry B:Without the ( m_2 sin(omega t) ) term, the equilibrium points are:[ k_2 R_B left(1 - frac{R_B}{L}right) = 0 ][ R_B = 0 quad text{or} quad R_B = L ]Including the oscillatory term, ( R_B ) will oscillate around ( L ). Sub-problem 2: Solve the differential equations and plot the revenue growthFor Industry A:Given ( k_1 = 0.1 ), ( K = 100 ), and ( m_1 = 0.01 ):[ frac{dR_A}{dt} = 0.1 R_A left(1 - frac{R_A}{100}right) - 0.01 R_A^2 ]This simplifies to:[ frac{dR_A}{dt} = 0.1 R_A - 0.001 R_A^2 - 0.01 R_A^2 ][ frac{dR_A}{dt} = 0.1 R_A - 0.011 R_A^2 ]This is a logistic equation with an added term. Using standard logistic growth solution techniques:[ R_A(t) = frac{R_{A0} K}{R_{A0} + (K - R_{A0}) e^{-k_1 t}} ]where ( K = frac{k_1 K}{k_1 + m_1 K} = frac{0.1 times 100}{0.1 + 0.01 times 100} = frac{10}{2} = 5 ).But since this is more complex, numerical methods or software like MATLAB/Mathematica can be used for exact solutions.For Industry B:Given ( k_2 = 0.08 ), ( L = 80 ), ( m_2 = 5 ), and ( omega = frac{pi}{4} ):[ frac{dR_B}{dt} = 0.08 R_B left(1 - frac{R_B}{80}right) + 5 sinleft(frac{pi}{4} tright) ]This differential equation can be solved numerically using software.Final Answer:The final equilibrium points are:- Industry A: ( R_A = 0 ) (unstable) and ( R_A = frac{k_1 K}{k_1 + m_1 K} = 5 ) (stable)- Industry B: ( R_B = 0 ) (unstable) and ( R_B = L = 80 ) (stable, oscillatory behavior around ( L ) due to the sinusoidal term)The detailed numerical solution for ( R_A(t) ) and ( R_B(t) ) requires numerical computation.Final Answer: The final answer is ( R_A = 0 ) (unstable), ( R_A = 5 ) (stable), and ( R_B ) oscillating around ( 80 ). I hope it is correct.
question:An active member of the local community, Jamie, is organizing a fundraising event for the local animal shelter. Jamie has collaborated with a veterinarian to spread awareness through various campaigns targeted at different demographics. They have decided to use a combination of social media advertisements and physical flyers to maximize outreach and donations.1. Optimization Problem: Jamie has a budget of 10,000 for the campaign. The cost of a social media advertisement is 50 per ad, and each ad is expected to reach an average of 1,000 people. The cost of printing and distributing a physical flyer is 1 per flyer, and each flyer is expected to reach an average of 50 people. Jamie wants to reach at least 50,000 people. Formulate and solve a linear programming problem to determine the optimal number of social media advertisements and physical flyers Jamie should use to maximize the number of people reached within the budget constraints.2. Probability and Expected Value: During the event, Jamie plans to hold a raffle to encourage donations. Every participant who donates 20 gets a raffle ticket. Jamie expects to sell 500 raffle tickets. The prizes are as follows: one first prize of 1,000, two second prizes of 500 each, and five third prizes of 100 each. Calculate the expected value of a single raffle ticket and determine if the expected value is more or less than the cost of the ticket. Use this information in your calculations:- Define the decision variables for the linear programming problem.- Set up the objective function and constraints.- Use appropriate probability formulas to find the expected value.
answer: Part 1: Optimization ProblemFirst, let's define the decision variables:- Let ( x ) be the number of social media advertisements.- Let ( y ) be the number of physical flyers.# Objective Function:We want to maximize the number of people reached. Each social media advertisement reaches 1,000 people, and each physical flyer reaches 50 people. Thus, our objective function is:[ text{Maximize } Z = 1000x + 50y ]# Constraints:1. Budget Constraint: The total cost of social media advertisements and physical flyers should not exceed 10,000. [ 50x + y leq 10,000 ]2. Reach Constraint: Jamie wants to reach at least 50,000 people. [ 1000x + 50y geq 50,000 ]3. Non-negativity Constraints: The number of advertisements and flyers cannot be negative. [ x geq 0 ] [ y geq 0 ]Now we solve the linear programming problem.# Step 1: Convert inequalities to equalities using slack variables.[ 50x + y + s_1 = 10,000 ][ 1000x + 50y - s_2 = 50,000 ][ x, y, s_1, s_2 geq 0 ]# Step 2: Graph the inequalities.We will graph the constraints and find the feasible region.1. Budget Constraint: [ 50x + y = 10,000 ] [ y = 10,000 - 50x ]2. Reach Constraint: [ 1000x + 50y = 50,000 ] [ y = 1,000 - 20x ]# Step 3: Identify corner points.To find the intersection of the lines ( y = 10,000 - 50x ) and ( y = 1,000 - 20x ):Set ( 10,000 - 50x = 1,000 - 20x ):[ 10,000 - 50x = 1,000 - 20x ][ 9,000 = 30x ][ x = 300 ]Substitute ( x = 300 ) into ( y = 10,000 - 50(300) ):[ y = 10,000 - 15,000 = -5,000 ]Since ( y ) cannot be negative, this point is not feasible.Check the boundary points:1. If ( x = 0 ): [ y = 10,000 ] [ y = 1,000 ] (doesn’t satisfy the reach constraint)2. If ( y = 0 ): [ 50x = 10,000 ] [ x = 200 ]Check if ( x = 200 ) and ( y = 0 ) satisfies the reach constraint:[ 1000(200) + 50(0) = 200,000 ] (satisfies the reach constraint)Since the feasible region is unbounded, we check the feasible points and constraints to find the optimal solution:- ( (x, y) = (200, 0) )Calculate the objective function at the feasible points:[ Z = 1000(200) + 50(0) = 200,000 ]Final Answer for Part 1: The optimal number of social media advertisements is 200, and the number of physical flyers is 0. Part 2: Probability and Expected ValueThe expected value ( EV ) of a single raffle ticket is calculated as follows:- Probability of winning the first prize (1,000) = ( frac{1}{500} )- Probability of winning a second prize (500) = ( frac{2}{500} )- Probability of winning a third prize (100) = ( frac{5}{500} )- Probability of winning nothing = ( 1 - left( frac{1}{500} + frac{2}{500} + frac{5}{500} right) )Calculate the expected value:[ EV = left( frac{1}{500} times 1000 right) + left( frac{2}{500} times 500 right) + left( frac{5}{500} times 100 right) + left( frac{492}{500} times 0 right) ][ EV = left( frac{1}{500} times 1000 right) + left( frac{2}{500} times 500 right) + left( frac{5}{500} times 100 right) ][ EV = 2 + 2 + 1 = 5 ]Final Answer for Part 2: The expected value of a single raffle ticket is 5. The expected value is less than the cost of the ticket (20).