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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You're on a roll — Here are the solutions to questions 11, 12, 13, and 36. Question 11: A bag contains 8 black, 5 white, and 7 pink identical bottles. Total number of bottles = 8 + 5 + 7 = 20. Two bottles are picked at random one after the other with replacement. a) They are all pink? Step 1: Calculate the probability of picking a pink bottle. P(pink) = Number of pink bottlesTotal bottles = (7)/(20) Step 2: Since the picking is with replacement, the probability of picking two pink bottles is the product of their individual probabilities. P(pink and pink) = P(pink) × P(pink) = (7)/(20) × (7)/(20) = (49)/(400) The probability that they are all pink is (49)/(400). b) The first is white and the second is white? Step 1: Calculate the probability of picking a white bottle. P(white) = Number of white bottlesTotal bottles = (5)/(20) = (1)/(4) Step 2: Since the picking is with replacement, the probability of picking two white bottles is the product of their individual probabilities. P(white and white) = P(white) × P(white) = (1)/(4) × (1)/(4) = (1)/(16) The probability that the first is white and the second is white is (1)/(16). c) The first is pink and the second is white? Step 1: Calculate the probability of picking a pink bottle first. P(first pink) = (7)/(20) Step 2: Calculate the probability of picking a white bottle second (with replacement, so the total number of bottles and the number of white bottles remain the same). P(second white) = (5)/(20) = (1)/(4) Step 3: Multiply the probabilities. P(first pink and second white) = P(first pink) × P(second white) = (7)/(20) × (1)/(4) = (7)/(80) The probability that the first is pink and the second is white is (7)/(80). Question 12: A motorist travels from town R to town S on a bearing of 065^ for 54 km. He then travels from town S to town Q on a bearing of 155^ for 80 km. i) The distance between town R and town Q, correct to one decimal place. Step 1: Draw a diagram representing the journey. Let N_R be the North line at R and N_S be the North line at S. The bearing of S from R is 065^. So, the angle between N_R and RS is 65^. The line SR makes an angle of 65^ with the South line at S (alternate interior angles). The angle from the North line at S, clockwise to SR, is 180^ + 65^ = 245^. The bearing of Q from S is 155^. This is the angle from the North line at S, clockwise to SQ. Step 2: Calculate the interior angle RSQ of the triangle RSQ. RSQ = 245^ - 155^ = 90^ So, triangle RSQ is a right-angled triangle with the right angle at S. Step 3: Use the Pythagorean theorem to find the distance RQ. RQ^2 = RS^2 + SQ^2 RQ^2 = (54 km)^2 + (80 km)^2 RQ^2 = 2916 + 6400 RQ^2 = 9316 RQ = sqrt(9316) ≈ 96.519 km Correct to one decimal place, the distance between town R and town Q is 96.5 km. ii) The bearing of town Q from town R. Step 1: Find the angle SRQ in the right-angled triangle RSQ. ( SRQ) = OppositeAdjacent = (SQ)/(RS) = (80)/(54) SRQ = ((80)/(54)) ≈ (1.4815) ≈ 55.99^ Step 2: The bearing of Q from R is the angle from the North line at R, clockwise to RQ. This is the initial bearing of S from R plus SRQ. Bearing of Q from R = 065^ + SRQ Bearing of Q from R = 65^ + 55.99^ = 120.99^ Correct to one decimal place, the bearing of town Q from town R is 121.0^. Question 13: Points A, B, and C have position vectors: OA = (2, 3), OB = (6, -1), and OC = (1, 5). a) Find the vectors AB and AC. Step 1: Calculate AB. AB = OB - OA = (6, -1) - (2, 3) = (6-2, -1-3) = (4, -4) Step 2: Calculate AC. AC = OC - OA = (1, 5) - (2, 3) = (1-2, 5-3) = (-1, 2) The vectors are AB = (4, -4) and AC = (-1, 2). b) Determine the position vector of point D such that AD = AB + AC. Step 1: Calculate AD. AD = AB + AC = (4, -4) + (-1, 2) = (4-1, -4+2) = (3, -2) Step 2: Find the position vector of D, OD. We know AD = OD - OA, so OD = OA + AD. OD = (2, 3) + (3, -2) = (2+3, 3-2) = (5, 1) The position vector of point D is (5, 1). c) Show whether points A, B, C are collinear. Step 1: For points A, B, C to be collinear, vector AB must be a scalar multiple of vector AC (i.e., AB = k AC for some scalar k). We have AB = (4, -4) and AC = (-1, 2). Step 2: Check if there is a scalar k such that (4, -4) = k(-1, 2). From the x-components: 4 = k(-1) k = -4. From the y-components: -4 = k(2) k = -2. Since the value of k is not consistent (-4 ≠ -2), AB is not a scalar multiple of AC. Therefore, points A, B, C are not collinear. Question 36: Solve the equation: (7x)/(2) + (2)/(3x) = 34. Step 1: Find a common denominator for the fractions, which is 6x. Multiply the entire equation by 6x to eliminate the denominators. 6x ( (7x)/(2) ) + 6x ( (2)/(3x) ) = 6x (34) Step 2: Simplify the terms. 3(7x^2) + 2(2) = 204x 21x^2 + 4 = 204x Step 3: Rearrange the equation into the standard quadratic form ax^2 + bx + c = 0. 21x^2 - 204x + 4 = 0 Step 4: Use the quadratic formula x = -b ± sqrt(b^2 - 4ac)2a to solve for x. Here, a=21, b=-204, c=4. x = -(-204) ± sqrt((-204)^2 - 4(21)(4))2(21) x = 204 ± sqrt(41616 - 336)42 x = 204 ± sqrt(41280)42 Step 5: Calculate the square root and simplify. sqrt(41280) ≈ 203.1743 x_1 = (204 - 203.1743)/(42) = (0.8257)/(42) ≈ 0.01966 x_2 = (204 + 203.1743)/(42) = (407.1743)/(42) ≈ 9.6946