A ball selected at random from a bag containing 3 red balls and 4 black balls. The first ball is not replaced and a second ball is selected. Find the probability of selecting two different balls.

Here are the solutions for the questions. b) A ball selected at random from a bag containing 3 red balls and 4 black balls. The first ball is not replaced and a second ball is selected. Find the probability of selecting two different balls. Step 1: Determine the total number of balls. Total balls = 3 red + 4 black = 7 balls. Step 2: Identify the two possible scenarios for selecting two different balls. Scenario 1: First ball is Red (R), second ball is Black (B). Scenario 2: First ball is Black (B), second ball is Red (R). Step 3: Calculate the probability for Scenario 1 (R then B). • Probability of drawing a Red ball first: P(R_1) = Number of red ballsTotal number of balls = (3)/(7). • After drawing one red ball, there are 2 red balls and 4 black balls left, making a total of 6 balls. • Probability of drawing a Black ball second (given the first was red): P(B_2 | R_1) = Number of black ballsRemaining total balls = (4)/(6). • Probability of Scenario 1: P(R_1 and B_2) = P(R_1) × P(B_2 | R_1) = (3)/(7) × (4)/(6) = (12)/(42). Step 4: Calculate the probability for Scenario 2 (B then R). • Probability of drawing a Black ball first: P(B_1) = Number of black ballsTotal number of balls = (4)/(7). • After drawing one black ball, there are 3 red balls and 3 black balls left, making a total of 6 balls. • Probability of drawing a Red ball second (given the first was black): P(R_2 | B_1) = Number of red ballsRemaining total balls = (3)/(6). • Probability of Scenario 2: P(B_1 and R_2) = P(B_1) × P(R_2 | B_1) = (4)/(7) × (3)/(6) = (12)/(42). Step 5: Add the probabilities of the two scenarios to find the total probability of selecting two different balls. P(two different balls) = P(R_1 and B_2) + P(B_1 and R_2) = (12)/(42) + (12)/(42) = (24)/(42) Step 6: Simplify the fraction. (24)/(42) = (4 × 6)/(7 × 6) = (4)/(7) The probability of selecting two different balls is (4)/(7). 3. a) Given that _x 2(1)/(4) = 2, calculate the value of x. Step 1: Convert the mixed fraction to an improper fraction. 2(1)/(4) = (2 × 4 + 1)/(4) = (9)/(4) Step 2: Rewrite the logarithmic equation in exponential form. The definition of a logarithm states that if _b a = c, then b^c = a. Applying this to the given equation: _x (9)/(4) = 2 x^2 = (9)/(4) Step 3: Solve for x. Take the square root of both sides: x = ±sqrt((9)/(4)) x = ±sqrt(9)sqrt(4) x = ±(3)/(2) Step 4: Consider the domain restrictions for the base of a logarithm. The base of a logarithm (x) must be positive and not equal to 1. Therefore, x = (3)/(2) is the valid solution. The value of x is (3)/(2). 3. b) Prove that the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle of the cyclic quadrilateral. Step 1: Consider a cyclic quadrilateral ABCD. Let's extend side BC to a point E, forming an exterior angle DCE. We need to prove that DCE = BAD. Step 2: Use the property of angles on a straight line. Angles BCD and DCE form a linear pair on the straight line BCE. Therefore, their sum is 180^. BCD + DCE = 180^ (1) Step 3: Use the property of a cyclic quadrilateral. In a cyclic quadrilateral, the sum of opposite interior angles is 180^. Angles BCD and BAD are opposite interior angles. Therefore, their sum is 180^. BCD + BAD = 180^ (2) Step 4: Compare the two equations. From equation (1), we can express DCE as: DCE = 180^ - BCD From equation (2), we can express BAD as: BAD = 180^ - BCD Step 5: Conclude the proof. Since both DCE and BAD are equal to the same expression (180^ - BCD), they must be equal to each other. DCE = BAD Thus, the exterior angle of a cyclic quadrilateral is equal to its interior opposite angle. Last free one today — make it count tomorrow, or type /upgrade for unlimited.