University:
Augusta University
Course:
MATH 4800 | Secondary Mathematics from an Advanced Perspective
Academic year:

2024
Views:

454

Pages:

3
Author:

Autumn Cox

Question 1 In how many different orders can five runners finish a race if no ties are allowed? Answer Definition permutation (order is important): P(n,r) = n! / (n-r)! Definition combination (order is not important): C(n,r) = (n choose r) = n! / r!(n-r)! with n! = n(n-1)...2*1 Solution We must apply the definition of permutation since the runners' order is crucial. From the five runners, we'll pick five (as we want an ordering of all runners). n=5 r=5 Consider the following when defining a combination: P(5,5) = 5! / (5-5)! = 5! / 0! = 5! = 5 * 4 * 3 * 2 * 1 = 120 Results: 120 Question 2 Exponential Growth and Decay Worksheet y = 55 - (1 – 0.02)^t a) Does this function represent exponential growth or exponential decay? b) What is your initial value? c) What is the rate of growth or rate of decay? y = 5575 × (0.65)^t a) Does this function represent exponential growth or exponential decay? b) What is your initial value? c) What is the rate of growth or rate of decay? y = 14000 x (0.92)^t a) Does this function represent exponential growth or exponential decay? b) What is your initial value? c) What is the rate of growth or rate of decay? Answer Step 1 a) It represents exponential decay b) Initial value is 55 c) Rate of decay is 0.02 (or base is 0.98) Step 2 a) It represents exponential decay b) Initial value is 5575 c) Rate of decay is 0.65 Step 3 a) It represents exponential decay b) Initial value is 14000 c) Rate of decay is 0.92 (or decay rate of 0.08) Question 3 The count in a bacteria culture was 200 after 15 minutes and 1600 after 30 minutes. Assuming the count grows exponentially, what was the initial size of the culture? What was the doubling period? Answer Formula for the exponential growth is given by: A = A₀e^kt Where, A₀ = Initial size k = Rate of increase t = Time Therefore, For A=200 and t=15, 200 = A₀e^(15k) For A=1600 and t=30, 1600 = A₀e^(30k) Divide equation (2) by (1): 1600/200 = (A₀e^(30k))/(A₀e^(15k)) 8 = e^(30k-15k) 8 = e^(15k) 15k = ln(8) k = ln(8)/15 k = 0.1386 Plug the values in equation to find initial size: 200 = A₀e^(15*0.1386) 200 = A₀e^(2.079) A₀ = 200/e^(2.079) A₀ = 25.011 A₀ ≈ 25 To find doubling period: A₀ = 25 k = 0.1386 A = 50 t = Doubling period Therefore, 50 = 25e^(0.1386t) 50/25 = e^(0.1386t) 2 = e^(0.1386t) 0.1386t = ln(2) t = ln(2)/0.1386 t = 5.001 t ≈ 5 min Question 4 Solve the equation. Round your answers to the nearest hundredth. 2^x = 100 Answer 2^x = 100 Given equation log(2^x) = log(100) Take the log for both sides x * log(2) = log(100) Power rule of logarithms x = log(100) / log(2) Divide both sides by log2 x ≈ 6.64 Using calculator Question 5 Write an exponential function to model each situation. Find each amount after the specified time. A population of 120,000 grows 1.2% per year for 15 years. Answer A(t) = a(1+r)^t Substitute a with $120,000, r with 1.2% = 0.012 and t with 15: A(15) = 120,000(1+0.012)^15 A(15) = 120,000(1.012)^15 A(15) ≈ 143,512