Appearance
question:Rock, Paper, Scissors, also known as Guessing Game, is a hand game that has been popular for many years, originating in China and then spreading to Japan, Korea, and other places. With the continuous development of AsiaEurope trade, it spread to Europe and gradually became popular around the world. The rules of the game are: before showing their hands, both parties shout the command together, and then at the moment the command ends, they simultaneously show their hands. A clenched fist represents rock, the index and middle fingers extended represent scissors, and an open hand represents paper. Rock beats scissors, scissors beat paper, and paper beats rock. If the hands shown are the same, then it is a draw. If Xiao Qian and Da Nian, two students, play a best of five game of Rock, Paper, Scissors, the probability of Xiao Qian winning by the fourth round is ( ) A: boxed{frac{1}{27}} B: boxed{frac{2}{27}} C: boxed{frac{2}{81}} D: boxed{frac{8}{81}}
answer:**Analysis** This question examines the method of calculating probability, which is a basic question. When solving the problem, it is important to carefully read the question and correctly apply the formula for the probability of an event A occurring exactly k times in n independent repeated trials. **Solution** According to "rock" beats "scissors," "scissors" beat "paper," and "paper" beats "rock," the probability of Xiao Qian winning against Da Nian, drawing, or losing to Da Nian in each round is frac{1}{3} each. Therefore, in the "best of five" game of "Rock, Paper, Scissors" between Xiao Qian and Da Nian, for Xiao Qian to win by the fourth round, it means that in the first three rounds, Xiao Qian wins 2 rounds and does not win 1 round, and wins the fourth round. Therefore, the probability of Xiao Qian winning by the fourth round is: P=C_{3}^{2}left( frac{1}{3}right)^{2}left( frac{2}{3}right) times frac{1}{3}= boxed{frac{2}{27}}. Hence, the correct answer is B.
question:Let f(x)=frac{4x^2 + 6x + 9}{x^2 2x + 5} and g(x) = 2x 3. Find f(g(x)) + g(f(x)), evaluated when x=2.
answer:To solve f(g(x)) + g(f(x)) when x=2: 1. Evaluate g(2): [ g(2) = 2(2) - 3 = 4 - 3 = 1 ] 2. Evaluate f(g(2)) = f(1): [ f(1) = frac{4(1)^2 + 6(1) + 9}{(1)^2 - 2(1) + 5} = frac{4 + 6 + 9}{1 - 2 + 5} = frac{19}{4} ] 3. Find f(2): [ f(2) = frac{4(2)^2 + 6(2) + 9}{(2)^2 - 2(2) + 5} = frac{16 + 12 + 9}{4 - 4 + 5} = frac{37}{5} ] 4. Evaluate g(f(2)) = gleft(frac{37}{5}right): [ gleft(frac{37}{5}right) = 2left(frac{37}{5}right) - 3 = frac{74}{5} - 3 = frac{74}{5} - frac{15}{5} = frac{59}{5} ] 5. Finally, add f(g(2)) and g(f(2)): [ f(g(2)) + g(f(2)) = frac{19}{4} + frac{59}{5} text{. To add, find common denominator: } frac{95}{20} + frac{236}{20} = frac{331}{20} ] So, boxed{frac{331}{20}} is the result.
question:Which of the following statements is correct? ( ) A: If event (A) and event (B) are mutually exclusive, then (P(A) + P(B) = 1) B: If (P(A) + P(B) = 1), then event (A) and event (B) are mutually exclusive C: If (P(A) = 0), then event (A) is definitely an impossible event D: If (P(B) = 1), then event (B) is definitely a certain eve
answer:**Analysis** This question mainly examines the concepts of mutually exclusive events, impossible events, and certain events, and is considered an easy question. **Answer** By using the fact that the sum of the probabilities of two mutually exclusive events is (1), we can determine that option A is correct. Therefore, the answer is boxed{text{A}}.
question:Below is the graph of (y = a sec bx) for some positive constants (a) and (b.) The graph shows that the maximum value of (y) when positive is 3, and period of the function is (2pi). Find (a).
answer:1. **Function Setup**: The function given is (y = a sec bx). 2. **Maximum Value Analysis**: The maximum value of the function (y = a sec bx) when (y) is positive corresponds to (a). 3. **Period Analysis**: The standard period of (sec x) is (2pi). Given the period of (y = a sec bx) is (2pi), set the standard period (2pi) equal to (frac{2pi}{b}) to find (b). [ frac{2pi}{b} = 2pi implies b = 1 ] 4. **Calculation of (a)**: Since the maximum value observed is 3 and (a) corresponds to this maximum, we have (a = 3). [ boxed{a = 3} ]