This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Morning Faith — let's get this done. Here's the solution for question 5: Step 1: Calculate the midpoint (x) for each weight class. • 20-24: (20+24)/(2) = 22 • 25-29: (25+29)/(2) = 27 • 30-34: (30+34)/(2) = 32 • 35-39: (35+39)/(2) = 37 • 40-44: (40+44)/(2) = 42 • 45-49: (45+49)/(2) = 47 Step 2: Calculate the deviation (d) for each midpoint from the assumed mean (A=36). d = x - A • 22 - 36 = -14 • 27 - 36 = -9 • 32 - 36 = -4 • 37 - 36 = 1 • 42 - 36 = 6 • 47 - 36 = 11 Step 3: Calculate the product of frequency (f) and deviation (d) for each class (fd). • 5 × (-14) = -70 • 15 × (-9) = -135 • 12 × (-4) = -48 • 9 × (1) = 9 • 6 × (6) = 36 • 3 × (11) = 33 Step 4: Calculate the sum of frequencies ( f) and the sum of fd ( fd). f = 5 + 15 + 12 + 9 + 6 + 3 = 50 fd = -70 - 135 - 48 + 9 + 36 + 33 = -175 Step 5: Use the formula for the mean weight using the assumed mean method. Mean (x) = A + ( fd)/( f) x = 36 + (-175)/(50) x = 36 - 3.5 x = 32.5 The mean weight of the Livestock is 32.5. Here's the solution for question 6: This is a combination problem, as the order of persons in a committee does not matter. The formula for combinations is C(n, k) = nk = (n!)/(k!(n-k)!). Total number of persons (n) = 9. Committee size (k) = 4. a) There are no restrictions: Step 1: Calculate the number of ways to choose 4 persons from 9 without any restrictions. C(9, 4) = (9!)/(4!(9-4)!) = (9!)/(4!5!) = (9 × 8 × 7 × 6 × 5!)/(4 × 3 × 2 × 1 × 5!) C(9, 4) = (9 × 8 × 7 × 6)/(4 × 3 × 2 × 1) = 9 × 2 × 7 = 126 There are 126 possible committees. b) Both Kwame and Rita must be on the committee: Step 1: Since Kwame and Rita must be on the committee, 2 spots are already filled. Remaining persons to choose from: 9 - 2 = 7. Remaining spots to fill: 4 - 2 = 2. Step 2: Calculate the number of ways to choose the remaining 2 persons from the remaining 7. C(7, 2) = (7!)/(2!(7-2)!) = (7!)/(2!5!) = (7 × 6 × 5!)/(2 × 1 × 5!) C(7, 2) = (7 × 6)/(2 × 1) = 21 There are 21 possible committees. c) Either Kwame or Rita, but not both, must be on the committee: This means we consider two separate cases and add their possibilities: Case 1: Kwame is on the committee, and Rita is not. Step 1: Kwame is chosen (1 person). Rita is excluded (1 person). Remaining persons to choose from: 9 - 1 (Kwame) - 1 (Rita) = 7. Remaining spots to fill: 4 - 1 (Kwame) = 3. Step 2: Calculate the number of ways to choose 3 persons from the remaining 7. C(7, 3) = (7!)/(3!(7-3)!) = (7!)/(3!4!) = (7 × 6 × 5 × 4!)/(3 × 2 × 1 × 4!) C(7, 3) = (7 × 6 × 5)/(3 × 2 × 1) = 35 Case 2: Rita is on the committee, and Kwame is not. Step 1: Rita is chosen (1 person). Kwame is excluded (1 person). Remaining persons to choose from: 9 - 1 (Rita) - 1 (Kwame) = 7. Remaining spots to fill: 4 - 1 (Rita) = 3. Step 2: Calculate the number of ways to choose 3 persons from the remaining 7. C(7, 3) = (7!)/(3!(7-3)!) = (7!)/(3!4!) = (7 × 6 × 5)/(3 × 2 × 1) = 35 Step 3: Add the possibilities from both cases. Total committees = 35 + 35 = 70. There are 70 possible committees. Drop the next question.