Solve the following: 1) 3(1)/(2) + (3(1)/(2) of 3(1)/(2)) of (1)/(8) 2) (0.325 x 2.76 x 100)/(13 x 0.25 x 92)

Here are the solutions to the problems: 1) Using your knowledge of BODMAS, solve the following: 3(1)/(2) + (3(1)/(2) of 3(1)/(2)) of (1)/(8) Step 1: Convert mixed numbers to improper fractions. 3(1)/(2) = (3 × 2 + 1)/(2) = (7)/(2) The expression becomes: (7)/(2) + ((7)/(2) × (7)/(2)) × (1)/(8) Step 2: Solve the operation inside the parentheses (multiplication). ((7)/(2) × (7)/(2)) = (49)/(4) The expression becomes: (7)/(2) + (49)/(4) × (1)/(8) Step 3: Perform the multiplication ("of") operation. (49)/(4) × (1)/(8) = (49 × 1)/(4 × 8) = (49)/(32) The expression becomes: (7)/(2) + (49)/(32) Step 4: Perform the addition. Find a common denominator, which is 32. (7)/(2) = (7 × 16)/(2 × 16) = (112)/(32) (112)/(32) + (49)/(32) = (112 + 49)/(32) = (161)/(32) Step 5: Convert the improper fraction to a mixed number. (161)/(32) = 5 (1)/(32) The solution is 5(1)/(32). 2) Solve: (0.325 × 2.76 × 100)/(13 × 0.25 × 92) Step 1: Simplify the numerator by multiplying by 100. 0.325 × 2.76 × 100 = 0.325 × 276 Step 2: Rewrite the expression and simplify by grouping terms. (0.325 × 276)/(13 × 0.25 × 92) = (0.325)/(0.25) × (276)/(92) × (1)/(13) Step 3: Calculate each fraction. (0.325)/(0.25) = (325)/(250) = (13 × 25)/(10 × 25) = (13)/(10) = 1.3 (276)/(92) = 3 Step 4: Substitute the simplified values back into the expression and multiply. 1.3 × 3 × (1)/(13) = (13)/(10) × 3 × (1)/(13) = (3)/(10) Step 5: Convert the fraction to a decimal. (3)/(10) = 0.3 The solution is 0.3. 3) Given the universal set U = \1, 2, 3, 4, , 20\, If Set A = Multiples of 3 and B = Factors of 18 Find: a) Elements of Set A Set A consists of multiples of 3 within the universal set U. A = \3, 6, 9, 12, 15, 18\ b) Elements of Set B Set B consists of factors of 18. B = \1, 2, 3, 6, 9, 18\ c) A B The union of A and B includes all unique elements from both sets. A B = \1, 2, 3, 6, 9, 12, 15, 18\ d) (A B)' Step 1: Find the intersection of A and B (A B), which are the elements common to both sets. A B = \3, 6, 9, 18\ Step 2: Find the complement of (A B), denoted as (A B)'. This includes all elements in the universal set U that are not in A B. U = \1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20\ (A B)' = \1, 2, 4, 5, 7, 8, 10, 11, 12, 13, 14, 15, 16, 17, 19, 20\ e) What is the sum of the elements of B? The elements of Set B are \1, 2, 3, 6, 9, 18\. Sum = 1 + 2 + 3 + 6 + 9 + 18 Sum = 39 The sum of the elements of B is 39. Last free one today — make it count tomorrow, or type /upgrade for unlimited.