This is a probability problem with two dependent events and conditional probability. Note that after the first donut is chosen, it is not replaced into the data set, so only 23 donuts remain. If we set A=selection of a lemon-filled, and B=selection of a custard-filled, then P(A and B) = P(A)*P(B|A), where P(B|A) means the probability of B happening given that A has already occurred.P(A) = 8/24 = 1/3 = 0.333333P(B|A) = 12/23 = 0.521739P(A and B) = 1/3(12/23) = 12/69 = 0.1739130435 or 17.4%
https://www.wyzant.com/resources/answers/296921/find_the_probability_of_selecting_a_a_lemon_filled_d...
Graph A shows a single solution where the two lines cross paths. Graph B shows a case where there are no solutions as the lines will never touch. Graph C is the correct answer as it shows the two lines constantly intercepting each other, creating an infinite number of solutions.
If you divide the number 1423 by 29, you answer will be 49 fish per tank. Hoped this helped
Incorrect info please upload a correct photo of the problem
Answer:
- 5.8206 cm
- 10.528 cm
- 23.056 cm^2
Step-by-step explanation:
(a) The Law of Sines can be used to find BD.
BD/sin(48°) = BD/sin(50°)
BD = (6 cm)(sin(48°)/sin(60°)) ≈ 5.82064 cm
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(b) We can use the Law of Cosines to find AD.
AD^2 = AB^2 +BD^2 -2·AB·BD·cos(98°) . . . . . angle ABD = 48°+50°
AD^2 ≈ 110.841
AD ≈ √110.841 ≈ 10.5281 . . . cm
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(c) The area of ∆ABD can be found using the formula ...
A = ab·sin(θ)/2 . . . . . where a=AB, b=BD, θ = 98°
A = (8 cm)(5.82064 cm)sin(98°)/2 ≈ 23.0560 cm^2
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Angle ABD is the external angle of ∆BCD that is the sum of the remote interior angles BCD and BDC. Hence ∠ABD = 48° +50° = 98°.
