Courtney Brewer Finite Mathematics Task 2 (1)

Courtney Brewer Finite Mathematics Task 2 Part 1: A: − [ 4 2 − | − 3 | 3 ] ( 6 − 8 ) 2 + 15 ÷ 5 × [ ( 4 + 17 ) −− 2 3 ] − [ 4 2 − 3 3 ] ( 6 − 8 ) 2 + 15 ÷ 5 × [ ( 4 + 17 ) −(− 2 ) 3 ] Step 1: Start in the numerator. The first step of order of operations is to calculate the parenthesis. Inside the parentheses is an absolute value that you must calculate first before moving onto exponents. Find the absolute value of -3. The absolute value of -3 is 3 because the distance -3 is away from zero is 3. − [ 16 − 27 ] ( 6 − 8 ) 2 + 15 ÷ 5 × [ ( 4 + 17 ) −(− 2 ) 3 ] Step 2: The next step is to solve for any exponents. Calculate the powers given. 4 2 = 16 3 3 = 27 − [ − 11 ] ( 6 − 8 ) 2 + 15 ÷ 5 × [ ( 4 + 17 ) −(− 2 ) 3 ] Step 3: The last step for solving within the parentheses of the numerator is to subtract 16 and 27. We do this because there are no other operations inside the parentheses. 16 - 27 = -11 − [ − 11 ] ( 6 − 8 ) 2 + 15 ÷ 5 × [ ( 4 + 17 ) −(− 2 ) 3 ] Step 4: All that is left in the numerator is to simplify –(-11). This will become 11 because -1 x -11 is 11. 11 ( 6 − 8 ) 2 + 15 ÷ 5 × ¿¿ Step 5: We now move down into our denominator. We must first solve what is inside of the brackets: ( 4 + 17 ) − ¿ . Within this expression, we must calculate 4 + 17 first because it is in parentheses. 4 + 17 = 21 11 ( 6 − 8 ) 2 + 15 ÷ 5 × [ 21 −(− 8 ) ] Step 6: Continuing inside the brackets, solve for the exponent next. (− 2 ) 3 = -8 11 ( 6 − 8 ) 2 + 15 ÷ 5 × [ 21 + 8 ] Step 7: In the brackets –(-8) becomes an 8 because it means the same as -1 x -8 or the opposite of a -8. 11 ( 6 − 8 ) 2 + 15 ÷ 5 × 29 Step 7: Lastly in the brackets, solve 21-8. 21 + 8 = 29 11 (− 2 ) 2 + 15 ÷ 5 × 29 Step 8: Now that we have solved everything in the brackets, we can solve what is in the parentheses. 6 – 8 = -2 11 4 + 15 ÷ 5 × 29 Step 9: Now that all grouping brackets and parentheses are calculated, we must now solve for any exponents.

(− 2 ) 2 = 4 11 4 + 3 × 29 Step 10: Continuing in the denominator, our next step is to solve any multiplication or division. Since there is both multiplication and division, we must calculate whichever comes first. In this problem, division comes before multiplication. 15 ÷ 5 = 3 11 4 + 87 Step 11: After completing division because it came before multiplication from left to right, now you must solve the multiplication 3 x 13. 3 x 29 = 87 11 91 Step 12: The last calculation left in the denominator is 4 + 39 which is 91. B:

43÷21= 2 r1 This correlates to a mixed number of 2 1 21 Part 2:

A local business has hats printed with its logo to give away at a booth at the town’s summer festival. For last year’s festival, the business ordered 512 hats and paid $1,341.67 for them. This year, the business plans to order 725 hats. Assuming the prices have not changed, how much will these 725 hats cost in dollars and cents? C: A local business has hats printed with its logo to give away at a booth at the town’s summer festival. For last year’s festival, the business ordered 512 hats and paid $1,341.67 for them. This year, the business plans to order 725 hats. Assuming the prices have not changed, how much will these 725 hats cost in dollars and cents? Identify the ratio used. The ratio is between the number of hats and their cost. Define the unknown variable. The unknown variable, x, will represent the cost for this year’s hats. Provide a proportion to solve the problem that uses the unknown variable. 512 hats $ 1,341.67 = 725 hats x D: A local business has hats printed with its logo to give away at a booth at the town’s summer festival. For last year’s festival, the business ordered 512 hats and paid $1,341.67 for them. This year, the business plans to order 725 hats. Assuming the prices have not changed, how much will these 725 hats cost in dollars and cents? Solve the problem set up in part C, including the following information: •   an explanation of  each  step •   work for  all  steps •   a summary sentence that answers the problem within the context of the scenario The proportion: 512 hats $ 1,341.67 = 725 hats x Step 1: Cross-multiply 512 ×x = 725 × 1341.67 512 x = 972710.75 Step 2: Divide each side by 512 in order to get the variable by itself. 512 x 512 = 972710.75 512 X = 1899.825684 or $1,899.83 If a local business paid $1,341.67 for 512 hats last year and the prices have not changed since then they would’ve paid $1,899.83 for 725 hats this year for the town’s summer festival. Part 3:

divides two quantities evenly. This is the only way that you would be able to get both the largest number that you can divide both numbers by without having any remainders. We find the GCF by using the smallest of the exponents in each prime number appearing in the prime factorization. B: There is a factory that makes widgets in different colors. The production time for each color is different. At the end of

each day, all the widgets produced that day are placed in boxes. On a certain day, the factory made 2,400 silver widgets and 4,950 green widgets. It takes 32 minutes to produce a batch of silver widgets and 22 minutes to produce a batch of green widgets. Determine the answer to the following question: “In a factory, if both colors of widgets start production at the same time and run continuously, after how many minutes will their starting time align again?” Explain the steps and problem-solving strategy used to determine the answer. If it takes 32 minutes to produce a batch of silver widgets and 22 minutes to produce a batch of green widgets, then we can list out the following times for each widget: Silver: 32, 64, 96, 128, 160, 192, 224, 256, 288, 320, 352 … Green: 22, 44, 66, 88, 110, 132, 154, 176, 198, 220, 242, 264, 286, 308, 330, 352 … This shows that the next time that they will produce a widget at the same time again will be at 352 minutes after the starting time. Explain why the final answer is a greatest common divisor, least common multiple, or neither. The final answer was the least common multiple because it represents the first time (or least/smallest number) that both 32 and 22 share in common between their multiples (which is their times it takes to produce repeated over and over. Sources: Sources: Curtis, M., Morris, E., & Frank Vahid. (2020). 10.1 Prime factorizations. In QTT2: Finite Math . zybook, Zyante Inc.