Answer:
The correct statements are;
1) ΔBCD is similar to ΔBSR
2) BR/RD = BS/SC
3) (BR)(SC) = (RD)(BS)
Step-by-step explanation:
1) Since RS is parallel to DC, we conclude that;
∠BDC = ∠BRS (Angles formed on the same side of the transversal)
Furthermore;
∠BCD = ∠BSR (Angles formed on the same side of the transversal)
∠CBD = ∠CBD (Reflexive property)
Thus;
ΔBCD ~ ΔBSR by the Angle-Angle-Angle (AAA) similarity criterion.
2) Given that ΔBCD ~ ΔBSR, we obtain;
BC/BS = BD/BR → (BS + SC)/BS = (BR + RD)/BR = 1 + SC/BS = RD/BR + 1
1 + SC/BS = 1 + RD/BR thus, SC/BS = 1 + BR/RD - 1
SC/BS = RD/BR
By inverting both sides we find;
BR/RD = BS/SC
3) From BR/RD = BS/SC, we apply cross multiplication;
BR/RD = BS/SC leads to;
BR × SC = RD × BS → (BR)(SC) = (RD)(BS).
There are several possible outcomes. The initial composition of the urns is as follows: Urn 1 contains 2 red chips and 4 white chips, totaling 6 chips, whereas Urn 2 has 3 red and 1 white, amounting to 4 chips. When a chip is drawn from the first urn, the probabilities are as follows: for a red chip, it is probability is (2 red from 6 chips = 2/6 = 1/2); for a white chip, it is (4 white from 6 chips = 4/6 = 2/3). After the chip is transferred to the second urn, two scenarios can arise: if the chip drawn from the first urn is white, then Urn 2 will contain 3 red and 2 white chips, making a total of 5 chips, creating a 40% chance for drawing a white chip. Conversely, if a red chip is drawn first, Urn 2 will contain 4 red and 1 white chip, which results in a 20% chance of drawing a white chip. This scenario exemplifies a dependent event, as the outcome hinges on the type of chip drawn first from Urn 1. For the first scenario, the combined probability is (the probability of drawing a white chip from Urn 1) multiplied by (the probability of drawing a white chip from Urn 2), equaling 26.66%. For the second scenario, the probabilities yield a value of 6%.
