answer:To solve this problem, let's first understand what each event represents:- Event A is "getting an even number less than 5". On a standard six-sided die, the even numbers less than 5 are 2 and 4. Therefore, there are 2 outcomes that satisfy event A.- Event B is "getting a number less than 5". The numbers less than 5 on a die are 1, 2, 3, and 4. Therefore, there are 4 outcomes that satisfy event B.- The complementary event overline{B} represents "getting a number 5 or greater". On a six-sided die, the numbers 5 and 6 satisfy this condition. Therefore, there are 2 outcomes that satisfy event overline{B}.The probability of an event is calculated as the number of favorable outcomes divided by the total number of outcomes. Since a standard die has 6 faces, the total number of outcomes is 6.- The probability of event A happening is P(A) = frac{text{Number of outcomes in } A}{text{Total number of outcomes}} = frac{2}{6} = frac{1}{3}.- The probability of the complementary event overline{B} happening is P(overline{B}) = frac{text{Number of outcomes in } overline{B}}{text{Total number of outcomes}} = frac{2}{6} = frac{1}{3}.Since event A and event overline{B} are mutually exclusive (no outcome can satisfy both A and overline{B} at the same time), the probability of either event A or event overline{B} occurring is the sum of their individual probabilities:[P(A cup overline{B}) = P(A) + P(overline{B}) = frac{1}{3} + frac{1}{3} = frac{2}{3}]Therefore, the probability of the occurrence of event A cup overline{B} in one experiment is boxed{frac{2}{3}}.Hence, the correct answer is: boxed{text{A}}.

answer:1. Identify the function and the range: The function given is: [ f(x) = left[x right] + left[2 cdot x right] + left[frac{5 cdot x}{3} right] + left[3 cdot x right] + left[4 cdot x right] ] We need to find the total number of different integers this function takes for (0 leq x leq 100).2. Analyze the intervals: The function (f(x)) changes its value when any of the floor functions (left[xright]), (left[2xright]), (left[frac{5x}{3}right]), (left[3xright]), or (left[4xright]) changes its value. This happens at specific points where (x) is a rational number that causes any of these expressions to be an integer.3. Determine the critical points in the interval ([0, 3)): The critical points in ([0, 3)) are: [ 0, frac{1}{4}, frac{1}{3}, frac{1}{2}, frac{3}{5}, frac{2}{3}, frac{3}{4}, 1, frac{6}{5}, frac{5}{4}, frac{4}{3}, frac{3}{2}, frac{5}{3}, frac{7}{4}, frac{9}{5}, 2, frac{9}{4}, frac{7}{3}, frac{12}{5}, frac{5}{2}, frac{8}{3}, frac{11}{4} ] These points are where the value of (f(x)) changes. There are 22 such points.4. Calculate the number of different integers in ([0, 3)): Since there are 22 critical points, there are 22 different integers generated by (f(x)) in the interval ([0, 3)).5. Extend the interval to ([0, 99)): The interval ([0, 99)) can be divided into 33 subintervals of length 3 each: ([0, 3), [3, 6), [6, 9), ldots, [96, 99)). Each of these subintervals will generate 22 different integers. Therefore, the total number of different integers generated in ([0, 99)) is: [ 33 times 22 = 726 ]6. Consider the interval ([99, 100]): The interval ([99, 100]) is analogous to the interval ([0, 1]) because the floor functions will behave similarly. We need to determine the number of different integers generated in ([0, 1]). The critical points in ([0, 1]) are: [ 0, frac{1}{4}, frac{1}{3}, frac{1}{2}, frac{3}{5}, frac{2}{3}, frac{3}{4}, 1 ] There are 8 such points.7. Calculate the total number of different integers: Adding the number of different integers generated in ([0, 99)) and ([99, 100]): [ 726 + 8 = 734 ]The final answer is (boxed{734})

answer:1. Since the absolute value of the difference between any two of the numbers 1,3,6,8 is a prime number, according to the problem, f(1), f(3), f(6), f(8) are four distinct elements in A. Therefore, |A| geqslant 4. On the other hand, if we let A={0,1,2,3}, and the mapping f: mathbf{N}^{*} rightarrow A is defined as: if x in mathbf{N}^{*}, x=4 k + r, then f(x)=r, where k in mathbf{N}, r=0,1,2,3, then for any x, y in mathbf{N}^{*}, if |x-y| is a prime number, assuming f(x) = f(y), then x equiv y(bmod 4), thus 4 mid |x-y|, which contradicts the fact that |x-y| is a prime number. Hence, the set A contains at least 4 elements.

answer:B26. Connect CM, using the equal areas of triangle AMC and triangle BMC we get the area of triangle BMC is 12. Then, using the equal areas of triangle DME and triangle DMC we get the area of triangle SED is 12, i.e., (B).

answer:1. Let ( m = d m' ) and ( n = d n' ) where ( gcd(m, n) = d ). We know that ( operatorname{lcm}(m, n) = frac{m cdot n}{gcd(m, n)} ). Therefore, [ operatorname{lcm}(m, n) = frac{d m' cdot d n'}{d} = d m' n'. ] Given that ( operatorname{lcm}(m, n) - gcd(m, n) = 103 ), we substitute to get: [ d m' n' - d = 103. ] Factoring out ( d ) from the left-hand side, we have: [ d (m' n' - 1) = 103. ] Since 103 is a prime number, the possible values for ( d ) are 1 and 103.2. Case 1: ( d = 103 ) [ 103 (m' n' - 1) = 103 implies m' n' - 1 = 1 implies m' n' = 2. ] The pairs ((m', n')) that satisfy ( m' n' = 2 ) and are relatively prime are ((1, 2)) and ((2, 1)). Thus, the corresponding values of ( m ) and ( n ) are: [ m = 103 cdot 1, quad n = 103 cdot 2 quad text{or} quad m = 103 cdot 2, quad n = 103 cdot 1. ] In both cases, ( m + n = 103 + 206 = 309 ).3. Case 2: ( d = 1 ) [ 1 (m' n' - 1) = 103 implies m' n' - 1 = 103 implies m' n' = 104. ] The pairs ((m', n')) that satisfy ( m' n' = 104 ) and are relatively prime are ((1, 104)), ((104, 1)), ((8, 13)), and ((13, 8)). Thus, the corresponding values of ( m ) and ( n ) are: [ m = 1 cdot 1, quad n = 1 cdot 104 quad text{or} quad m = 1 cdot 104, quad n = 1 cdot 1, ] [ m = 1 cdot 8, quad n = 1 cdot 13 quad text{or} quad m = 1 cdot 13, quad n = 1 cdot 8. ] Therefore, the possible values of ( m + n ) are: [ 1 + 104 = 105 quad text{and} quad 8 + 13 = 21. ]Combining all the possible values from both cases, we get:[boxed{21, 105, 309}.]

answer:Some misunderstood the concept of plane section, while others were satisfied with recognizing that the section is a parallelogram, but the answer is: rectangle.