UNIT 5 QZ, Q-20 UNIT 5 / UNIT 8 QUESTIONS KEEP ANSWERS SEPARATE BY UNITS UNIT II STUDY GUIDE Consumer Mathematics Reading Assignment See information below. Key Terms 1. Add-on interest method 2. Annual percentage rate 3. Average daily balance method 4. Common logarithmic function 5. Compound interest 6. Finance charge 7. Future value 8. Inflation 9. Interest rate 10. Percent 11. Present value 12. Principle 13. Simple interest 14. Unpaid balance method Course Learning Outcomes for Unit II Upon completion of this unit, students should be able to: 1. Apply mathematical principles used in real-world situations. 1.1 Apply percent concepts and equations to solve problems. 1.2 Utilize the average daily balance method to compute credit card charges. 1.3 Utilize the simple and compound interest formula to solve applied problems. 1.4 Calculate the annual percentage rate. 1.5 Compute finance charges on a credit card using the unpaid balance method. Unit Lesson Chapter 8: Consumer Mathematics Assume that you are buying a car. Before paying for it, you research different loans and calculate the amount the interest you will owe if you chose a particular loan. Once a loan is chosen, you write a check for the down payment and balance your checkbook accordingly. Although this scenario may not be a common occurrence, the use of mathematics it portrays is common. The basic math skills required to manage money is called consumer mathematics. Consumer mathematics is frequently used every day. The information shared in this lesson will help you understand the basis behind taxes, raises, interest payments, loans, and other financial situations that arise. 8.1 Percents, Taxes, and Inflation: In this section, we will learn how to use percentages to calculate taxes and percent of inflation. Percent Many monetary calculations use the notion of percentages. Percentages are an easy way to describe the amount of something. For example, you may be required to pay off 10% of your $1,000 loan each month. This means that you will be paying $100 each month. We say the percentage instead of the actual amount for simplicity and because the amount of the loan may vary. Writing Percent Using Decimals The word \"percent\" means \"per hundred\". Therefore, one percent is one part to a hundred parts. We write a percent as a fraction where one hundred is the denominator. MAT 1301, Liberal Arts Math 1 Any fraction with a denominator of 100 can easily be changed to a decimal by dividing by hundred or moving the decimal two places to the left. This is shown in the example below: Example: Convert each percent to a decimal: a) 78% b) 3% c) 27.35% d) 0.08% Solution: a) Convert 78% to a decimal. 78 We can think of 78% as seventy-eight hundredths or 100. Therefore, we write 78 and move the decimal to the left two places. The decimal form of seventy-eight hundredths is 0.78. b) Convert 3% to a decimal. MAT 1301, Liberal Arts Math 2 We can think of 3% as three-hundredths. Therefore, we write 3.and move the decimal to the left two places. 3% = 0.03 The decimal form of three-hundredths is 0.03. c) Convert 27.35% to a decimal. We can write 27.35% as Since we are dividing by 100, we will move the decimal two places to the left. The decimal form of 27.35% is 0.2735. d) Convert 0.08% to a decimal. We can write 0.08% as Since we are dividing by 100, we will move the decimal two places to the left. 0.08% = 0.0008 The decimal form of 0.08% is 0.0008. Converting a Decimal to a Percent We have learned to move the decimal of a number two places to the left when converting a percent to a decimal. We can do the opposite when converting a decimal to a percent. Example: Write 0.0035 as a percent. Solution: We can write 0.0035 as a percent by multiplying the decimal by 100 and adding a % symbol at the end of the number. When we multiply by 100, we are really moving the decimal two places to the right. MAT 1301, Liberal Arts Math 3 Converting a Fraction to a Percent Fractions can be converted to percentages by dividing the denominator into the numerator and then multiplying the decimal by 100. For example: Example: Convert each fraction to a percent: a) b) c) 3 4 9 25 11 8 Solution: 3 a) Convert 4 to a percent. We must first divide the denominator into the numerator and then convert the decimal form into a percent. 3 4 = 0.75 We can now multiply 0.75 by 100 and add the % symbol. 9 b) Convert 25 to a percent. We must first divide the denominator into the numerator and then convert the decimal form into a percent. 9 25 = 0.36 We can now multiply 0.36 by 100 and add the % symbol. c) Convert MAT 1301, Liberal Arts Math 11 8 to a percent. 4 We must first divide the denominator into the numerator and then convert the decimal form into a percent. 11 8 = 1.375 We can now multiply 1.375 by 100 and add the % symbol. Percent of Change The percent of change refers to an increase or decrease in an amount. For example, assume that there are 50 students in a course. It is projected that the percent of change will increase by 50% next year. This means that an additional 25 students will take the course next year making the total number of students to be 75. We can use the following formula when calculating a percent of change: The base amount is the amount that we started with. It is sometimes referred to as our original or initial amount. Example: Dealer markup on a car: If a dealer buys a car from the manufacturer for $19,875 and then sells it for $21,065, what is his markup? Solution: For this problem, we will use the following formula: We substitute our given information and solve, converting the final decimal answer to a percent: This tells us that the markup (percent of change) is 6%. The Percent Equation The percent equation is used by taking some percent of a base quantity and setting it equal to an amount. The percent equation is: We will learn how to use this equation by doing the following examples. MAT 1301, Liberal Arts Math 5 Example: a) 12 is what percent of 80? b) 77 is 22% of what number? c) What is 23% of 140? Solution: For each problem, we will use the percent equation. The first step in solving these problems is to identify what information we are given and what information we are asked to find. a) 12 is what percent of 80? We will use the formula: b) 77 is 22% of what number? We will use the formula: MAT 1301, Liberal Arts Math 6 First, convert 22% into a decimal by dividing by 100 or by moving the decimal two places to the left. a) What is 23% of 140? We will use the formula; First, convert 23% into a decimal by dividing by 100 or by moving the decimal two places to the left. Example: In a certain fiscal quarter, 40% of McDonald's profits came from Happy Meals. If Happy Meals' profits were $508 million, how much profit came from selling other items? Solution: We will use the formula: MAT 1301, Liberal Arts Math 7 In this problem, $508 million accounted for 40% of McDonald's profits. Therefore, the base is not known. First, convert 40% to a decimal, then substitute the given information into the formula and solve for the base. Therefore, $1,270 million were total profits realized by McDonald's that quarter. To find the profits that came from items other than Happy Meals we will need to subtract the Happy Meal profits from the total base profits. $1,270 - $508 = $762, so $762 million in profits came from other items. Taxes Taxes are usually given in the form of a percent. We will use the same concepts learned above to solve tax-related problems. Example: Use the following table to calculate the federal income tax due for a taxable income of $148,000: Line 1 Line 2 Line 3 Line 4 Line 5 Line 6 If your taxable income is over $0 7,550 30,650 74,200 154,800 336,550 But not over The tax is Of the amount over 7,550 30,650 74,200 154,800 336,550 ................ ..................10% $755.00 + 15% $4,220.00 + 25% $15,107.50 + 28% $37,675.50 + 33% $97,653.00 + 38% 0$ 7,550 30,650 74,200 154,800 336,550 Federal income taxes due for a single person. Solution: We will use Line 4 because the taxable income is greater than $74,200 but less than $154,800. Line 4 states that the tax is $15,107.50 + 28% for any amount over 74,200. Therefore, we will subtract 74,200 from our amount (148,000) to find the amount that will be taxed. MAT 1301, Liberal Arts Math 8 Therefore, $35,771.50 is owed. 8.2 Interest: Interest is the amount of money that is accumulated over time from a base amount. In today's world, people can earn interest from a savings account or can pay interest for a new car, house, or loan. In this section, we will learn how to use simple interest and compound interest formulas. We will need to know some key terms before calculating interest. The chart below contains some terms that will be used in both simple interest and compound interest formulas. Letter Used in Equation Term P Principal r Interest rate t Time (in years) The length of time that the deposit remains in the bank. The time is usually stated in years. I Interest earned The amount of money earned on your deposit. A Future value Definition The amount of money that is deposited. This is sometimes called the present value. The percentage earned on the deposit. The rate is given as a percent. The amount that will be in your account at some point in the future. Simple Interest The most basic way to compute interest is by using the simple interest formula. This formula calculates the amount of interest earned from depositing a principal (P) amount over a certain length of time (t) for a defined interest rate (r). Formula for Computing Simple Interest Rate - We calculate simple interest using the following formula: Where I is the interest earned, P is the principal, r is the annual interest rate, and t is the time in years. As shown by the formula, the interest earned (I) is found by multiplying the principal (P), the annual interest rate (r), and the time (t) together. MAT 1301, Liberal Arts Math 9 Note: The annual interest rate (r) must be converted to a decimal before it is substituted into the formula. The time (t) must be converted to years before it is substituted into the formula. For example, 6 months equals 0.5 years. Example: Use the simple interest formula, = and elementary algebra to find the missing quantity: I = ?, P = $1,000, r = 8%, t = 3 years Solution: First, convert the interest rate (r) into decimal form. To do this, divide 8 by 100 or move the decimal two places to the left. r = 8% = .08 Next, substitute the given values into the simple interest formula and solve. Example: Use the simple interest formula = and elementary algebra to find the missing quantity: I = $700, P = $3,500, r = ?, t = 4 years Solution: We are asked to find the interest rate (r). We will substitute the values given for I, P, and t and solve for r. = 700 = 3,500 4 Multiply: 3500 4 = 14,000 700 = 14,000 700 14,000 = Divide: 700 14,000 = 0.05 0.05 = Next, convert r = 0.05 to a percent by multiplying by 100 or by moving the decimal two places to the right. Therefore, r = 5%. The simple interest formula allowed us to calculate the interest earned from our deposit. The next formula will allow us to calculate the future amount of our deposit over a certain number of years. MAT 1301, Liberal Arts Math 10 Computing the Future Value using Simple Interest -To find the future value of an account that pays simple interest, use the formula: Where A is the future value, P is the principal, r is the annual interest, and t is the time in years. Example: Use the future value formula missing quantity. and elementary algebra to find the A = $1,770, P = ?, r = 6%, t = 3 years Solution: First, we need to convert the rate into decimal form. To do this, divide 6 by 100 or move the decimal two places to the left. r = 6% = .06 Now, substitute and solve. Therefore, P = $1,500. Example: Use the future value formula quantity. and elementary algebra to find the missing A = $966, P = $840, r = 5%, t =? Years Solution: First, we need to convert the rate into decimal form. To do this, divide 6 by 100 or move the decimal two places to the left. r = 5% = .05 Now, substitute and solve. MAT 1301, Liberal Arts Math 11 Therefore, t = 3 years Compound Interest Compounding interest means the bank is paying interest on the previously earned interest. For example, assume that we deposited $100 at 10% interest. Simple interest only applies the 10% to the $100. Compound interest will apply the 10% to the 100 plus any additional money that was earned through the interest. Compound Interest Formula - Assume that an account with principal P is paying an annual interest r and compounding is being done m times per year. If the money remains in the account for n time periods, then the future value, A, of the account is given by the formula: This formula introduces two new variables: m = number of times compounding is done per year (compounding period) n = number of compounding periods Example: Use the formula for computing future value using compound interest to determine the value of the account at the end of the specified time period. MAT 1301, Liberal Arts Math 12 Solution: First, convert 7% to a decimal. To do this, divide by 100 or move the decimal two places to the left. 7% = .07 Example: Use the formula for computing future value using compound interest to determine the value of the account at the end of the specified time period. MAT 1301, Liberal Arts Math 13 Solution: First, convert 4% to a decimal. To do this, divide by 100 or move the decimal two places to the left. 4% = .04 Next, substitute and solve: Example: For this exercise, you are given an annual interest rate and the compounding period for two investments. Decide which the better investment is. 4.75% compounded monthly; 4.70% compounded daily Solution: You must compare (1 + ) for each case where m is the number of times the money compounds in a single year. Solving for both we get, MAT 1301, Liberal Arts Math 14 Therefore, we can see that 4.75% compounded monthly is the better choice. Solving for Unknowns in the Compound Interest Formula We will be asked to solve for unknown variables using the compound interest formula. This formula contains exponential expressions. Therefore, we need to learn the properties of exponents and how to use these properties to solve for an unknown variable in the exponential position The diagram below lists the basic components of an exponential equation. As shown, the base is the number that is being multiplied, the exponent is the small number attached to the base which tells us how many times the base is being multiplied to itself, and the answer is the solution after the exponent is applied to the base. Note: The exponent is sometimes referred to as a power. The common logarithmic function is used to solve for an exponent. The logarithmic function has a base of 10. This function is located on your calculator and is usually represented by the word \"log.\" Try practicing with your calculator by pressing the \"log\" key and then entering 100,000. You should receive an answer of 5. This means that 10 raised to the 5th power is 100,000. The exponential problem, 105 = 100,000 is related to the following logarithmic problem: Now that we know some key terms, we will introduce a property that will enable us to solve for exponents: MAT 1301, Liberal Arts Math 15 Exponent Property of the Log Function - This property enables us to solve for the exponent by taking it out of its original position. Example: Solve: Solution: X is in the exponent position. Therefore, we will use the exponent property of the log function to solve for x. Steps: Example: Solve: (1.15) = 3 Solution: X is in the exponent position. Therefore, we will use the exponent property of the log function to solve for x. MAT 1301, Liberal Arts Math 16 Steps: We use the exponent property of log functions to solve for the time variable in compound interest problems. Example: You have $1,500 to invest. Find the time in years that it takes your investment to double with annual compounding if your annual interest rate is 3.5%. Solution: We will use the compound interest formula, = (1 + ) , to solve. The problem states that we want to find the number of years it will take to double our investment. Therefore, our investment ($1,500) is our principle (P) and our future value is double our investment (2 1500 = $3,000). The problem also states that compounding is done once a year, so m=1. Our annual interest rate is 3.5%. The variables are as follows: A = 2 x $1,500 = $3,000 P = $1,500 r = 3.5% = .035 m=1 MAT 1301, Liberal Arts Math 17 Substituting we have: Example: Use the compound interest formula = (1 + ) and the given information to solve for t. A = $1,500 P = $1,000 r = 4% = .04 MAT 1301, Liberal Arts Math 18 Solution: Substituting we have: 8.3 Consumer Loans: A loan is sum of money that is borrowed and then paid back within a specified period. Loans that have a fixed number of payments are called installment loans. Each payment is called an installment. For example, suppose you borrowed $1,000 and were required to pay $100 a month for 10 months. This is an installment loan with an installment of $100. Loans are usually borrowed with interest. This means the amount of the loan will be charged with an interest rate determined by the person you are borrowing money from. The interest rate on a loan is often called a finance charge. This charge will determine the payments of each installment. The Add-On Interest Method The add-on interest payment method is used to determine the monthly payment of an installment loan. This method uses the simple interest formula to calculate the interest of an installment loan. Once the interest is calculated, we will add the interest to the purchase or base price and divide by the number of monthly payments. The formula for calculating the monthly payment using the add-on method is: Where is the amount of the loan, is the amount of interest due on the loan, and is the number of monthly payments. Example: MAT 1301, Liberal Arts Math 19 Compute the monthly payments for the add-on interest loan. The amount of the loan, the annual interest rate, and the term of the loan are given. Solution: We will use the add-on method to solve. First, use the simple interest formula to solve for : Next, find the number of monthly payments (n): The problem states that we will pay off the loan in t=3 years. Convert years to months by multiplying by 12 because there are 12 months in a year. Lastly, determine the monthly payments by substituting into the formula. Therefore, the monthly payments will be $30.34. Example: Angela's bank gave her a 4-year add-on interest loan for $6,480 to pay for new equipment for her antiques restoration business. The annual interest rate is 11.65%. How much interest will she pay? What are her monthly payments? Solution: From the problem, we know the following: P = $6,480 r = 11.65% = 0.1165 t = 4 years First, use the simple interest formula, = to solve for : MAT 1301, Liberal Arts Math 20 Therefore, Angela will pay $3,019.68 in interest. Next, find the number of monthly payments (n): The problem states that we will pay off the loan in 4 years. Convert years to months by multiplying by 12 because there are 12 months in a year. Lastly, determine the monthly payments by substituting into the + formula = . Therefore, Angela will pay $197.91 a month. The Unpaid Balance Method The unpaid balance method is used for loans where the initial borrowed amount and the monthly payment varies from month to month. This is called an open credit loan. Credit cards provide an open credit loan because the interest charged depends on the amount of purchases that were charged to the card. Unpaid Balance Method for Computing the Finance Charge on a Credit Card - Where P = previous month's balance + finance charge + purchases made - returns - payments r = annual interest rate 1 t = 12 Example: Use the unpaid balance method to find the finance charge on the credit card account. Last month's balance, the payment, the annual interest rate, and any other transactions are given. Last month's balance = $475 Payment = $225 Interest rate = 18% = 0.18 Bought ski jacket = $180 Returned camera = $145 Solution: First, calculate the interest rate (finance charge) for last month's balance using the simple interest formula: MAT 1301, Liberal Arts Math 21 Next, calculate the current principle: P = previous month's balance + finance charge + purchases made - returns -payments Lastly, use the simple interest formula to find your new finance charge: The Average Daily Balance Method The average daily balance method is another method that calculates the monthly payment for open-ended credit or credit card loan. Use the following steps to compute the finance charge on a credit card loan (open ended credit) - 1. Add the outstanding balance for your account for each day of the month. 2. Divide by the total found in step 1 by the number of days in the month to find the average daily balance. 3. Find the finance charge by using the simple interest formula , = , where P is the average daily balance found in step 2, r is the annual interest rate, and t is the number of days in the month divided by 365. MAT 1301, Liberal Arts Math 22 Example: Use the average daily balance method to compute the finance charge on the credit card account for the previous month. The starting balance and transactions on the account for the month are given. Assume an annual interest rate of 21%. Month: October (31 days); previous month's balance: $190 Date Transaction October 9 Charged $35 for a book October 11 Charged $20 for gasoline October 20 Made payment of $110 October 26 Charged $13 for lunch Solution: 1. Add the outstanding balance for your account for each day of the month. We will construct a table: Transaction Balance on October 1st Charged $35 for a book on Oct. 9th Charged $20 for gasoline Oct. 11th Made payment of $110 on Oct. 20th Charged $13 for lunch on Oct. 26th Number of days that passed until next activity Oct. 1st, 2nd, 3rd, 4th, 5th, 6th, 7th, 8th = 8 days Oct. 9th, 10th = 2 days Oct. 11th , 12th, 13th, 14th, 15th, 16th, 17th, 18th, 19th = 9 days Oct. 20th, 21st, 22nd, 23rd, 24th, 25th = 6 days Oct. 26th, 27th, 28th, 29th, 30th, 31st = 6 days Balance Number of days Balance $190 8190 = $1520 190+35 = $225 225+20 = $245 2225 = $450 245-110= $135 135+13 = $148 6135 = $810 9245 = $2205 6148 = $888 Total = $5,873 2. Divide by the total found in step 1 by the number of days in the month to find the average daily balance. 5873 = = 189.451613 $. 31 3. Find the finance charge by using the simple interest formula, = , where P is the average daily balance found in step 2, r is the annual interest rate, and t is the number of days in the month divided by 365. 31 Recall that P = $189.45, r = 21% = 0.21, t = 365 = = (189.45)(0.21) ( 31 ) 365 31 ) 365 1233.33 = 365 = 3.38 = 39.7848 ( Therefore, your finance charge will be $3.38. MAT 1301, Liberal Arts Math 23 8.6 Annual Percentage Rate: The interest rate methods introduced in section 8.3 discussed monthly interest rates. In this section, we will discuss a method for finding an annual interest rates. This is called the annual percentage rate or APR. Previously, we are given monthly interest rates. However, as payments are made and our principle becomes smaller, our initial interest rate will not truly reflect the monthly payments. For example, assume that you had a loan of $1000 with a fixed payment of $100 over 10 months. This would be a 10% interest rate for the first month. Our remaining balance is $900. Our monthly payment is $100 for the second month; however, our interest rate is no longer 10%. Our interest rate for the second month is 100 = 0.11 or 11%. If we continue, we will find that our interest rate increases each 900 month. Therefore, stating that our interest rate 10% each month is not a true statement. The \"true\" interest rate is represented by the annual interest rate. We will first calculate the APR using the simple interest formula and elementary algebra. Example: Find the APR of the loan by using the given information. Loan amount = $8,000 Yearly payments = 4 Rate = 12% = 0.12 Solution: First, solve for the total interest paid: Next, we can divide $8,000 by 4 yearly payments to give us a re-payment of $2,000 per year. Note t = 1 because the payments are made once a year. Therefore, we can use the interest formula = re-written as: Combine like terms and solve for r: MAT 1301, Liberal Arts Math 24 The APR is 19.2%. Finding the APR using Tables Standard financial tables have been constructed to find the APR. The table that will be used throughout this section is located on page 426 of your textbook and is shown below. It provides the annual interest for per $100 on a loan. Annual Interest Table, (Pirnot, 2014, p. 426) Steps for finding the APR using the table on page 426 - 1. Find the finance charge on the loan if it is not already given. 2. Determine the finance charge per $100 on the loan. 3. Use the line of the table that corresponds to the number of payments to find the number closest to the amount found in step 2. 4. The column heading for the column containing the number found in step 3 is the APR. Example: Use the following table to find the APR to the nearest whole percent for the given situation. Assume that all interest rates are annual rates. MAT 1301, Liberal Arts Math 25 Annual Interest Table, (Pirnot, 2014, p. 426) Jonathan has agreed to pay off a $4,500 loan by making 24 monthly payments. The total finance charge on his loan is $600. Solution: 1. Find the finance charge on the loan if it is not already given. The finance charge is $600. 2. Determine the finance charge per $100 on the loan. 3. Use the line of the table that corresponds to the number of payments to find the number closest to the amount found in step 2. We are making 24 monthly payments. Therefore, we will use line that corresponds with 24 in the table. 4. The column heading for the column containing the number found in step 3 is the APR. From the table, we can see that for a 24-payment loan, this corresponds to an APR of roughly 12%. Example: Use the following table to find the APR to the nearest whole percent for the given situation. Assume that all interest rates are annual rates. Annual Interest Table, (Pirnot, 2014, p. 426) MAT 1301, Liberal Arts Math 26 Megan pays a finance charge of $310 on a 12-month, $5,000 loan. Solution: 1. Find the finance charge on the loan if it is not already given. The finance charge is $310. 2. Determine the finance charge per $100 on the loan. 3. Use the line of the table that corresponds to the number of payments to find the number closest to the amount found in step 2. We are making 12 monthly payments. Therefore, we will use line that corresponds with 12 in the table. 4. The column heading for the column containing the number found in step 3 is the APR. From the table, we can see that for a 12-payment loan, this corresponds to an APR of roughly 11%. Estimating the APR The APR can be calculated by using a formula for add-on interest loans. This estimate does not require the use of a table as the previous examples did. Formula to Approximate the APR for the Add-On Interest Loan- Where is the annual interest rate and is the number of payments. Example: Estimate the annual percentage rate for the add-on loan using the given number of payments and annual interest rate. n = 36 r = 6.4% = 0.064 Solution: Use the following formula: Reference MAT 1301, Liberal Arts Math 27 Pirnot, T. L. (2014). Mathematics all around (5th ed.). Boston, MA: Pearson Education. Reading Assignment Chapter 8: Consumer Mathematics: The Mathematics of Everyday Life Section 8.1 Percents, Taxes, and Inflation, pp. 379-387 Section 8.2 Interest, pp. 388-398 Section 8.3 Consumer Loans, pp. 399-407 Section 8.6 Looking Deeper: Annual Percentage Rate, pp. 424-430 MAT 1301, Liberal Arts Math 28 UNIT VIII STUDY GUIDE Review for the Final: Chapters 6, 8, 9, 12, 13, and 14 Reading Assignment Course Learning Outcomes for Unit VIII Upon completion of this unit, students should be able to: See information below. Learning Activities (Non-Graded) See information below. Key Terms Review all Key Terms within each unit. 1. Apply mathematical principles used to real-world situations. 2. Relate number theory, integer computation, and rational number concepts to problem-solving applications. 3. Perform computations involving exponents, scientific notations, and sequences within the real number system. 4. Apply geometric principles and formulas to solve problems. 5. Demonstrate counting techniques. 6. Calculate basic statistical measures and analyze distribution graphs. Unit Lesson Review for Final Exam Unit VIII focuses on the information that was presented in Chapters 6, 8, 9, 12, 13, and 14 of the textbook. The purpose of this unit is to allow you time to review and prepare for the Final Exam. The Final Exam will be comprised of 50 multiple-choice questions. You will have four hours in which to complete this exam. Preparation You must complete the Unit VIII Homework assignment. This assignment is comprised of questions from all chapters. In addition to the required homework assignment, the following actions will help you prepare for the Final Exam: Review the previous unit lessons. Review your previous assessments. Determine the areas that were your weakest. Review those weak areas in the textbook and create a study plan to master those concepts. Review your homework. Go into your final well rested. Additional Study Materials As you review and determine weak areas that you might need to work on, it would be a good idea to go back and review the examples in each unit lesson and review the summaries for each of the chapters that were covered in the course. Summaries may be found on the following pages: MAT 1301, Liberal Arts Math 1 Chapter 6.1, 6.2, 6.3 o Summary and Review, pages 295 and 297 Chapter 8.1, 8.2, 8.3, 8.6 o Summary and Review, pages 431-434 Chapter 6.4, 6.5, 6.6 o Summary and Review, pages 296-298 Chapter 9.1, 9.2, 9.3, 9.5 o Summary and Review, pages 509 - 512 Chapter 12.1, 12.2, 12.3 o Summary and Review, pages 640 - 641 Chapter 13.1, 13.2, 13.3 o Summary and Review, pages 697 - 699 Chapter 14.1, 14.2, 14.3 o Summary and Review, pages 759 - 761 After each summary and review section, there is a chapter test with which you can test yourself and make sure you have mastered the chapter content. The answers for all test questions are available in the back of the textbook or your Student's Solutions Manual. If you have difficulty with any of the test questions, go back and re-read the chapter content and review the examples provided. If you are having difficulty with the order of operations or solving equations, you may go to Appendix A on page 764 of your textbookfor a quick review. Taking the Final Exam You will have four hours during which you are to complete your Final Exam. It is a proctored exam. It is suggested that you work each problem 2-3 times on paper before submitting your final exam in MyMathLab. Work through the problems on the final separately, allowing time in between each one. Reconcile any differences between the answer you get the first and second time you work each problem. If you need to work the test a third time, do so, and compare your answers one more time. Doing this will help you to avoid careless mistakes, skipped problems, and calculator keystroke mistakes. Double check your work on the final exam to be sure that each question answer is the one that you want checked before you click the 'Submit' button. Re-read the directions and the problem after you work it to be sure that you have done everything and to make sure that you followed the directions correctly. Note: When it is time to take your final, disregard the instructions about printing the assessment. This course is in a lab, and none of the assessments will print. The final exam will not print. Do not waste any of your four hours trying to print it. The final exam will not print! Reading Assignment Review: Required Reading Units I-VII to prepare for the comprehensive Final Exam. MAT 1301, Liberal Arts Math 2 Learning Activities (Non-Graded) Practice What You Have Learned Chapter tests and cumulative tests will provide you with good practice for the final exam. You can work any problems that you feel you have not quite mastered, and then check your answers in the back of the book. These activities are non-graded learning activities, which mean that you do not have to submit them. If you experience difficulty in mastering any of the concepts, contact your instructor for additional information and guidance. Review What You Have Learned The homework assignment for Unit VIII is cumulative, allowing you to review everything you have learned in the course. As the final exam is also cumulative, the homework will be a good practice run for you. You will need to obtain a minimum 70% grade on the homework in order to be able to take the final exam. Before attempting the homework and the final exam, study the chapter summaries, review the concepts taught in the chapters, and work the odd-numbered problems in the review exercises. Study Plan for Unit VIII Access the MA 1130 course in MyMathLab, click on \"Unit VIII,\" scroll down to the bottom of the page to find and select \"MYMATHLAB COURSE MATERIAL (REQUIRED),\" and open your chapter study plan to see your progress and do practice exercises. Your study plan is updated each time you take an assessment. The study plan is optional and is generated specifically for you based on your results to provide practice where it is needed to obtain mastery of the unit concepts. The study plan will be very beneficial in studying for your final exam. Other Resources and Activities You may click on \"Tools for Success\" on the toolbar on the left. By doing that, you will bring up the Multimedia Library which will have video lectures given by the authors of the book, PowerPoints, animation, and interactive figures on a variety of topics. You will also be able to view the Answers, Glossary, and Index from your textbook. Non-graded Learning Activities are provided to aid students in their course of study. You do not have to submit them. If you have questions, contact your instructor for further guidance and information. MAT 1301, Liberal Arts Math 3 UNIT V STUDY GUIDE Counting Reading Assignment See information below. Key Terms 1. Combination 2. Fundamental counting principle 3. Listing 4. Permutation 5. Tree diagrams Course Learning Outcomes for Unit V Upon completion of this unit, students should be able to: 1. Apply mathematical principles used in real-world situations. 1.1 Draw tree diagrams to represent counting situations graphically. 1.2 Apply counting techniques to solve applied problems. 1.3 Apply the theory of permutations and combinations to solve counting problems. 5. Demonstrate counting techniques. 5.1 Count elements in a set systematically. 5.2 Apply the fundamental counting principle. 5.3 Demonstrate how to solve counting problems with special conditions. 5.4 Calculate the number of permutations and combinations of n objects taken r at a time. 5.5 Use factorial notation to represent the number of permutations of a set of objects. 5.6 Create slot diagrams to organize information in counting problems. Unit Lesson Numbers and counting are some of the first mathematical principles that were introduced to you at a young age. People first learn how to count to ten, then to a hundred, and eventually by even and odd numbers. This unit introduces more principles about counting. In particular, some mathematical properties and formulas that will help in counting larger sets will be learned. 12.1 Systematic Counting: Systematic counting is the act of counting objects in an organized way. For example, if you wanted to list the all the ways that a die could be rolled, you would write \"1, 2, 3, 4, 5, 6\" because a die has six sides each consisting of a number 1 through 6. Listing these methods in numeric order is a systematic way of finding the answer. The systematic approach will be used to solve the following example. Example: You are selecting from the set of W = {Carrie (U)nderwood, Kelly (C)larkson, Chris (D)aughtry, Fantasia (B)arrino, and Clay (A)iken} MAT 1301, Liberal Arts Math 1 This set consists of American Idol contestants who have had successful careers. List all the ways that you can select two singers without repetition; order is not important. For example, BB is not allowed and UC is the same as CU. Solution: A set is a list of items in a list. Set W in the example consists of 5 singers and each singer is denoted by a letter: U, C, D, B, and A. Sets are usually named with a capital letter such as a W. Sets are also usually written with brackets, { }, around the list of items. W = {U, C, D, B, and A} A systematic list will be written to show all the ways in which two singers can be selected. First, pair the first singer (U) with the other singers. This results in the following: UC, UD, UB, UA Next, pair the second singer (C) with the other singers. The problem states that repetition is not allowed in our list. Therefore, CU will not be included in the list below because UC was previously identified. CD, CB, CA Next, pair the third singer (D) with the other singers. This will give result in the following: DB, DA Now we will pair the fourth singer (B) with the other singers. This will result in the following: BA We cannot pair the last singer (A) with other singers without repeating what was already listed. Thus, the list is complete! There are 10 ways to select two singers from set W. Dice are often used when learning about different counting strategies. The next example uses a systematic counting approach to list the different ways that an event can occur when rolling two dice. Example: Assume that you are rolling two dice: the first one is red and the second is green. Use a systematic listing to determine the number of ways a total of 5 can be rolled. Hint: A total can be rolled two ways: (1, 2) and (2, 1). Solution: A dice has six sides and each side is numbered with either a 1, 2, 3, 4, 5, or 6. To differentiate the two dice, we will assume that one is red and the other green. MAT 1301, Liberal Arts Math 2 Dice One: R(red) = {1,2,3,4,5,6} Dice Two: G(green) = {1,2,3,4,5,6} We will use ordered pairs in the form (red, green) to list the number of ways that we can roll a total of 5. First, choose the first number in set R. This number is 1. 1 can only be paired with 4 to make a total of 5. (1,4) Next, choose the second number in set R. This number is 2. The number 2 can only be paired with a 3 to make a total of 5. (2,3) Next, choose the third number in set R. This number is 3. The numbers 3 can only be paired with 2 to make a total of 5. (3,2) Next, choose the third number in set R. This number is 4. The number 4 can only be paired with 1 to make a total of 5. (4,1) Next, choose the fourth number in set R. This number is 5. The number 5 cannot be paired with any number in set G to make a total of 5. The systematic approach ensured that all ordered pairs were identified. Thus, the list is complete. There are 4 ways to list a total of 5 when two dice are rolled. Tree Diagrams A tree diagram illustrates the different ways in which an event can occur and is a graphical approach to counting. Example: Draw a tree diagram that illustrates the different ways to flip a penny, nickel, dime, and quarter. Use this diagram to solve Exercises 5-8 on page 612. In how many ways can exactly two tails be obtained? Solution: To draw the tree, start by writing out the potential outcomes from flipping a penny: heads or tails. Next, branch from those options and list the possible outcomes for flipping a nickel. This process for flipping a dime is continued following by a quarter. Then list the combined sequence of heads and tails for each path of the tree and highlight those that satisfy the condition of getting exactly two tails. MAT 1301, Liberal Arts Math 3 The final answer is 6 ways. Example: If a tree diagram was drawn showing how many ways five coins could be flipped, how many branches would it have? Solution: The tree begins at a single point, and then two branches will lead to the results of the first flip. Each of those branches will split into two branches on the second flip for a total of four branches. The third flip will branch from each of those four branches for a total of eight branches. The next flip, flip four, will double the number to sixteen branches. The final flip will double this number one last time for a total of thirty-two ways for five coins being flipped. Example: Assume that a triple-deck ice cream cone with vanilla, strawberry, and chocolate as possible flavors is purchased. The flavors can be repeated or not. Two cones will be considered different if the flavors are the same but occur in different order. How many different flavors are possible? Solution: Since repetition is allowed, each time a choice of flavor is made, there will be three options. Since the order of the flavors do not matter, the multiplication principle is used to solve. = 3 3 3 = 27 There are 27 different flavor possibilities. MAT 1301, Liberal Arts Math 4 12.2 The Fundamental Counting Principle: A new car is being purchased, and the individual purchasing the car needs to decide on the features he/she wants to include in the car. The car cannot be built without first deciding on a model, a color, and an audio system. Therefore, the overall task cannot be completed without finishing a series of tasks. The number of ways to build the car could be identified by drawing a tree diagram; however, the fundamental counting principle allows one to identify the number of ways a series of tasks can be completed by a using a simple formula. The Fundamental Counting Principle (FCP) - If you wants to perform a series of tasks, and the first task can be done in a ways, the second can be done in b ways, the third can be done in c ways, and so on, then all tasks can be done in a b c ... ways. Assume that you can select among 3 models, 5 colors, and 2 audio systems when building a car. By using the FCP formula, your know that there are 3 5 2 = 30 ways to build your new car. Example: The Equestrian Club has eight members. If the club wants to select a president, vice president, and treasurer (all of whom must be different), in how many ways can this be done? Solution: By applying the fundamental counting principle, we can see that there are eight ways that the first position can be filled (one of the eight members). Once that position is filled, there are seven ways that the second position can be filled (one of the seven remaining members). Moving on to the third position, there are only six ways that it can be filled (one of the six remaining members). Therefore, those possibilities can be multiplied to obtain the result: 8 7 6 = different ways Example: The early bird special at TGI Friday's features an appetizer, soup or salad, entre, and dessert. If there are 5 appetizers, 6 choices for soup or salad, 13 entres, and 4 desserts, how many different meals are possible? (It is assumed that one makes a selection from each category.) Solution: To solve this problem, the number of ways that each food category may be selected needs to be determined. There are 5 appetizers, 6 choices for soup or salad, 13 entres, and 4 desserts. Therefore, the fundamental counting principle can be used, and all choices can be multiplied together to find the result. 5 6 13 4 = 1,560 There are 1,560 different ways to choose a meal. 12.3 Permutations and Combinations: Permutations and combinations are mathematical terms that are used when solving detailed counting problems. MAT 1301, Liberal Arts Math 5 Factorial Notation A factorial is a mathematical operation used when computing permutations and combinations. It is denoted by an exclamation mark (!). For example, n! means that the factorial operation will be applied to n and is read \"n factorial\". Definition - If n is a counting number, the symbol n! stands for the product n(n-1)(n2)(n-3)21. One defines 0! = 1. Example: Solve: 9! Solution: The definition of a factorial will be used to solve. n! = n(n-1)(n-2)(n-3)21 = 9(9-1)(9-2)(9-3)21 = 987654321 Note: Keep subtracting until 1 is obtained as the last digit in the sequence. = 563,760 Example: Solve: 10! 7! Solution: 10! 7! = 10987654321 = 10987654321 7654321 7654321 Write the definition of each factorial. Cancel like factors on the top and bottom of the fraction. = 10 9 8 Multiply. = Example: Solve: 10! 7! 3! MAT 1301, Liberal Arts Math 6 Solution: 10! 7! 3! 10987654321 = (7654321)(321) Write the definition of each factorial. 10987654321 = (7654321)(321) Cancel like factors on the top and bottom of the fraction. = 1098 = 720 Multiply. 321 Divide. 6 = Permutations A permutation is the act of rearranging a set of numbers or items in an order or sequence without repetition. For example, suppose you have the following set of letters: S = {a, b, c} This set has 6 permutations because the letters can be arranged 6 different ways without repetition. These ways are illustrated below: abc, acb, bac, bca, cba, cab The mathematical notation used to express permutations is denoted by (, ). This is read as \"the number of permutations of n objects taken r at a time.\" The figure below illustrates this and may be found on page 623 of the textbook. There were 3 letters selected from set S in the example above. All 3 letters were chosen to make our permutations. Therefore, it was found that (, ) = (3, 3) = 6. The knowledge of factorials can be used to compute permutations using a formula. Formula for computing (, ): (, ) = MAT 1301, Liberal Arts Math ! ( )! 7 Example: Find the permutation. Eight objects taken three at a time. Solution: First, identify n and r. There are eight objects, so n = 8. Three objects are taken at a time, so r = 3. Next, solve (, ) = (8,3). (8,3) 8! = (83)! Plug in the values for n and r in the permutation formula. 8! = 5! Subtract the values in the parentheses. = 87654321 = 87654321 Write the definition of each factorial. 54321 Cancel like factors on the top and bottom of the 54321 fraction. = 876 Multiply. = Example: Solve. (10,3) Solution: (10,3) 10! = (103)! Plug in the values for n and r in the permutation formula. = 10! = 10987654321 = 10987654321 Subtract the values in the parentheses. 7! 7654321 7654321 Write the definition of each factorial. Cancel like factors on the top and bottom of the fraction = 10 9 8 MAT 1301, Liberal Arts Math Multiply. = 8 Identifying Permutations in Word Problems The following permutation problems will help to identify when permutations should be calculated. Remember that a permutation will be calculated when order matters, and items in a list cannot be repeated. Example: On a biology quiz, a student must match eight terms with their definitions. Assume that the same term cannot be used twice. Write the formula that will be used. Solving the problem is not required. Solution: The key to this problem is to know that the order of matching the terms with their definitions matters, thus this is a permutation. Therefore, it is denoted as P(8,8). Example: A password for a computer consists of three different letters of the alphabet followed by four different digits from 0 to 9. How many different passwords are possible? Solution: In this situation the order matters, so this is a permutation. Three letters will be chosen from a possible 26, and 4 digits from a possible 10. The two results will then be multiplied by each other for the final result. (26,3) = (10,4) = 26! 26! 26 25 24 23! = = (26 3)! 23! 23! = 26 25 24 = 15,600 10! 10! 10 9 8 7 6! = = (10 4)! 6! 6! = 10 9 8 7 = 5040 Therefore, multiplying both results together results in the total number of passwords possible: 15,600 5040 = , , possible passwords. Combinations Combinations can be described as selecting items from a group when the order of the selection does not matter. In other words, group (abc) would by the same as (bca), because the order does not matter. The mathematical notation used to express combinations is denoted by (, ). This is expressed as follows: \"We are forming a combination of n objects taken r at a time.\" Formula for computing (, ): (, ) = MAT 1301, Liberal Arts Math (, ) ! = ! ! ( )! 9 Example: Find the number of combinations. Eight objects taken three at a time. Solution: C(8,3) 8! = 3!(83)! Plug in the values for n and r in the combination formula. 8! = 3!5! Subtract the values in the parentheses. 87654321 = (321)(54321) 87654321 = (321)(54321) Write the definition of each factorial. Cancel like factors on the top and bottom of the fraction. 876 = 321 = Multiply. 336 Divide. 6 = combinations Example: Solve. (6,2) Solution: (6,2) 6! = 2!(62)! Plug in the values for n and r in the combination formula. 6! = 2!4! 654321 = (21)(4321) 654321 = (21)(4321) Subtract the values in the parentheses. Write the definition of each factorial. Cancel like factors on the top and bottom of the fraction. 65 = 21 MAT 1301, Liberal Arts Math Multiply. 10 = 30 Divide. 2 = Identifying Combinations in Word Problems The following combination problems will help you identify when combinations should be calculated. Remember that a combination will be calculated when the order of the items does not matter. Example: A committee is to be formed consisting of 5 men and 4 women. If the membership is to be chosen from 12 men and 10 women, how many different committees are possible? Solution: This is a combination problem so the formula below will be used: (, ) = ! ! ( )! Where \"n\" is the number of elements taken \"r\" at a time. Step 1: Find the number of ways to choose the men. Out of the 12 possible men, 5 of them should be chosen. 12! (12 5)! 5! 12 11 10 9 8 7 6 5 4 3 2 1 = (7 6 5 4 3 2 1)(5 4 3 2 1) 12 11 10 9 8 95040 = = = 792 54321 120 (12,5) = Step 2: Find the number of ways to choose the women. Out of the 10 possible women, 4 of them should be chosen. (10,4) = 10! 10 9 8 7 6 5 4 3 2 1 = (10 4)! 4! (6 5 4 3 2 1)(4 3 2 1) 10 9 8 7 5040 = = = 210 4321 24 Step 3: Multiply the number of ways to choose 5 men and 4 women. (12,5) (10,4) = 792 210 = 166,320 Thus, there are 166,320 ways to choose 5 men from 12 men and to choose 4 women from 10 women. Example: A pet store has 6 fluffy bunnies and 9 hamsters. In how many ways can 3 animals be selected if there can be at most 1 fluffy bunny? MAT 1301, Liberal Arts Math 11 Solution: To answer this, all the ways that one or less fluffy bunnies can be chosen need to be looked at. 1. If one chooses 0 fluffy bunnies, then 3 hamsters are chosen. 2. If one chooses 1 fluffy bunny, then 2 hamsters are chosen. So for each one, the formula for combinations should be used: (, ) = ! ( )! ! Where \"n\" is the number of elements taken \"r\" at a time. 1. 0 fluffy bunnies are chosen out of 6 fluffy bunnies, and 3 hamsters are chosen out of 9 hamsters. 6! 9! (6 0)! 0! (9 3)! 3! 654321 = (6 5 4 3 2 1)(1) 987654321 (6 5 4 3 2 1)(3 2 1) = 1 84 = 84 (6,0) (9,3) = 2. 1 fluffy bunny is chosen out of 6 fluffy bunnies, and 2 hamsters are chosen out of 9 hamsters. 6! 9! (6 1)! 1! (9 2)! 2! 654321 = (5 4 3 2 1)(1) 987654321 (7 6 5 4 3 2 1)(2 1) = 6 36 = 216 (6,1) (9,2) = Now, the possible combinations need to be added together. 84 + 216 = Reading Assignment Chapter 12: Counting: Just How Many Are There? Section 12.1 Introduction to Counting Methods, pp. 606-613 Section 12.2 The Fundamental Counting Principle, pp. 614-621 Section 12.3 Permutations and Combinations, pp. 622-634 MAT 1301, Liberal Arts Math 12