answer:Let's start by simplifying the given expression for ( x ): [ x = frac{1}{log_{frac{1}{2}} frac{1}{3}} + frac{1}{log_{frac{1}{5}} frac{1}{3}} ] 1. Convert the logarithms using the change of base formula: [ log_{frac{1}{a}} b = frac{log b}{log frac{1}{a}} = frac{log b}{-log a} = -frac{log b}{log a} ] Therefore: [ log_{frac{1}{2}} frac{1}{3} = -frac{log frac{1}{3}}{log 2} = -frac{log 3^{-1}}{log 2} = -frac{-log 3}{log 2} = frac{log 3}{log 2} ] And: [ log_{frac{1}{5}} frac{1}{3} = -frac{log frac{1}{3}}{log 5} = -frac{log 3^{-1}}{log 5} = -frac{-log 3}{log 5} = frac{log 3}{log 5} ] 2. Substitute these results into the original expression for ( x ): [ x = frac{1}{frac{log 3}{log 2}} + frac{1}{frac{log 3}{log 5}} = frac{log 2}{log 3} + frac{log 5}{log 3} ] 3. Combine the fractions over a common denominator: [ x = frac{log 2 + log 5}{log 3} = frac{log (2 times 5)}{log 3} = frac{log 10}{log 3} = log_{3} 10 ] 4. Evaluate the range of ( log_{3} 10 ): begin{itemize} item We know ( 9 < 10 < 27 ). item In logarithmic form: ( 3^2 < 10 < 3^3 ). item Therefore, ( log_3 9 < log_3 10 < log_3 27 ). item Simplifying, ( 2 < log_3 10 < 3 ). end{itemize} So, ( x = log_3 10 ) lies between 2 and 3. # Conclusion Therefore, the value of ( x ) belongs to the interval ((2, 3)). [ boxed{(D) (2, 3)} ]

answer:Let's denote the total number of female officers on the police force as F. According to the information given, half of the 300 officers on duty that night were female. So, the number of female officers on duty that night was: 300 officers / 2 = 150 female officers We are also told that these 150 female officers represent 15% of the total number of female officers on the police force. To find the total number of female officers (F), we can set up the following equation: 15% of F = 150 0.15 * F = 150 Now, we can solve for F: F = 150 / 0.15 F = 1000 So, there are boxed{1000} female officers on the police force.

answer:Let's denote the event "A solves the cube on the i^{th} attempt" as A_i and "B solves the cube on the i^{th} attempt" as B_i. Given, P(A_i) = 0.8 and P(B_i) = 0.6. (1) The event "person A succeeds on their third attempt" can be written as bar{A_1} bar{A_2} A_3. Since each attempt is independent, we have P(bar{A_1} bar{A_2} A_3) = P(bar{A_1})P(bar{A_2})P(A_3) = 0.2 times 0.2 times 0.8 = boxed{0.032}. (2) The event "at least one of them succeeds on their first attempt" is the complement of the event "neither of them succeeds on their first attempt", which can be written as bar{A_1} bar{B_1}. Therefore, P(text{at least one succeeds}) = 1 - P(bar{A_1} bar{B_1}) = 1 - P(bar{A_1})P(bar{B_1}) = 1 - 0.2 times 0.4 = boxed{0.92}.

answer:**Answer**: The proposition P: "All domestic mobile phones have trap consumption" is a universal proposition containing the quantifier "all". The negation of a universal proposition is an existential proposition. To negate it, we need to change the form of the quantifier, which leads us to the answer. Therefore, the correct answer is boxed{text{C: There is a domestic mobile phone that does not have trap consumption}}.