answer:Given the significant differences in income situations, stratified sampling should be adopted. The process is as follows: 1. Divide the 150 employees into three strata: senior titles, intermediate titles, and general staff. 2. Calculate the number of people to be sampled from each stratum based on the ratio of the sample size to the total population, resulting in 3 people, 9 people, and 18 people, respectively. 3. Conduct simple random sampling within each stratum to select the corresponding number of people. 4. Combine the selected individuals from each stratum to form a single sample. Therefore, the appropriate method of selection is boxed{text{stratified sampling}}.

answer:1. **Initial Setup**: - At the beginning: 40% liked, 60% disliked. - At the end: 80% liked, 20% disliked. 2. **Change in Responses**: - Increase in "like" responses from 40% to 80%, an increase of 40%. - Decrease in "dislike" responses from 60% to 20%, a decrease of 40%. 3. **Minimum Change Scenario**: - At least 40% of students must have switched from "dislike" to "like" to account for the increase. - Minimum possible change = 40%. 4. **Maximum Change Scenario**: - Consider extremes where all 60% initially disliking switch to liking, and to balance, 40% from initial liking switch to disliking. - Maximum possible change: 60% initially disliking + 40% initially liking = 100%. 5. **Difference Between Maximum and Minimum**: - Difference: 100% - 40% = 60%. The maximum possible number of students changing their opinion is when every student changes their opinion, matching the scenario where all initially disliking students turn to liking and an adequate proportion of liking students turn to disliking to account for final percentages. The minimum change is when only a necessary proportion switches sides just enough to account for the observed increases. The difference between maximum and minimum possible values of students changing opinions is 60%. **Conclusion:** Based on the situation described, the calculations illustrate that the computed difference is logical and consistent with the observed changes in student opinions over the school year. The final answer is boxed{textbf{(C)} 60%}

answer:1. Montrer que (n-2) et (n^{2}-n-1) sont premiers entre eux Par l'absurde, supposons qu'il existe un nombre premier (p) qui divise à la fois (n-2) et (n^2 - n - 1). Puisque (p) divise (n-2), il existe un entier (k) tel que: [ n = 2 + kp ] Nous devons montrer que ce même (p) doit aussi diviser (n^2 - n - 1). Substituons (n) dans (n^2 - n - 1): [ n^2 - n - 1 = (2 + kp)^2 - (2 + kp) - 1 ] [ = 4 + 4kp + k^2p^2 - 2 - kp - 1 ] [ = k^2p^2 + 3kp + 1 ] Puisque (p) divise (k^2p^2) et (3kp), pour que (p) divise (n^2 - n - 1), il doit diviser le dernier terme (1), ce qui est absurde car un nombre premier (p > 1) ne peut diviser 1. Ainsi, (n-2) et (n^2 - n - 1) sont premiers entre eux par contradiction. (blacksquare) 2. Trouver tous les entiers (m, n geq 0) tels que (n^3 - 3n^2 + n + 2 = 5^m) Observons que: [ (n-2)(n^2-n-1) = n^3 - 3n^2 + n + 2 ] D'après la première question, (n-2) et (n^2 - n - 1) sont premiers entre eux. Par conséquent, au moins une des deux expressions doit être égale à (pm 1) afin que leur produit soit une puissance de (5). Considérons les cas suivants pour que leur produit soit égale à (5^m): 1. (n-2 = 1) [ n = 3 ] Substituons (n = 3) dans (n^3 - 3n^2 + n + 2): [ 3^3 - 3 cdot 3^2 + 3 + 2 = 27 - 27 + 3 + 2 = 5 = 5^1 ] Donc, (m = 1). 2. (n^2 - n - 1 = 1) [ n^2 - n - 2 = 0 ] [ (n-2)(n+1) = 0 ] Donc, (n = 2) ou (n = -1). Si (n = 2): [ n^3 - 3n^2 + n + 2 = 8 - 12 + 2 + 2 = 0 neq 5^m ] Si (n = -1), (n) négatif, ce qui est exclu car (n geq 0). 3. (n-2 = -1) [ n = 1 ] Substituons (n = 1) dans (n^3 - 3n^2 + n + 2): [ 1^3 - 3 cdot 1^2 + 1 + 2 = 1 - 3 + 1 + 2 = 1 = 5^0 ] Donc, (m = 0). 4. (n^2 - n - 1 = -1) [ n^2 - n = 0 ] [ n(n-1) = 0 ] Donc, (n = 0) ou (n = 1). Si (n = 1), alors: [ n^3 - 3n^2 + n + 2 = 1 - 3 + 1 + 2 = 1 = 5^0 ] Donc, (m = 0). Donc les solutions sont ( (n, m) = (3, 1) ) et ( (n, m) = (1, 0) ). [ boxed{(3, 1), (1, 0)} ] 3. Trouver tous les entiers (m, n geq 0) tels que (2n^3 - n^2 + 2n + 1 = 3^m) Étant donné l'expression, nous regardons modulo (3). Considérons (n equiv 0, 1, 2 pmod{3}): 1. Si (n equiv 0 pmod{3}): [ n = 3k quad text{pour un entier } k, ] [ 2(3k)^3 - (3k)^2 + 2(3k) + 1 = 2 cdot 27k^3 - 9k^2 + 6k + 1 ] [ = 54k^3 - 9k^2 + 6k + 1 equiv 1 pmod{3} ] Donc, (2n^3 - n^2 + 2n + 1 equiv 1 pmod{3}). 2. Si (n equiv 1 pmod{3}): [ n = 3k + 1 quad text{pour un entier } k, ] [ 2(3k+1)^3 - (3k+1)^2 + 2(3k+1) + 1 ] [ = 2 cdot (27k^3 + 27k^2 + 9k + 1) - (9k^2 + 6k + 1) + (6k + 2) + 1 ] [ = 54k^3 + 54k^2 + 18k + 2 - 9k^2 - 6k - 1 + 6k + 2 + 1 ] [ = 54k^3 + 45k^2 + 18k + 4 equiv 1 pmod{3} ] 3. Si (n equiv 2 pmod{3}): [ n = 3k + 2 quad text{pour un entier } k, ] [ 2(3k+2)^3 - (3k+2)^2 + 2(3k+2) + 1 ] [ = 2 cdot (27k^3 + 54k^2 + 36k + 8) - (9k^2 + 12k + 4) + (6k + 4) + 1 ] [ = 54k^3 + 108k^2 + 72k + 16 - 9k^2 - 12k - 4 + 6k + 4 + 1 ] [ = 54k^3 + 99k^2 + 66k + 17 equiv 2 pmod{3} ] Ainsi, si (2n^3 - n^2 + 2n + 1 = 3^m) pour (m geq 1), alors (3) divise (n) et: [ 2n^3 - n^2 + 2n + 1 equiv 1 pmod{3} ] Ce qui n'est pas divisible par (3), il doit donc être que (m = 0). En conséquence, nous avons: [ 2n^3 - n^2 + 2n + 1 = 1 ] [ 2n^3 - n^2 + 2n = 0 ] [ n(2n^2 - n + 2) = 0 ] L'équation (2n^2 - n + 2 = 0) n'a pas de solutions réelles car son discriminant est: [ Delta = (-1)^2 - 4 cdot 2 cdot 2 = 1 - 16 = -15 ] Donc, la seule solution est (n = 0). Pour cette valeur: [ 2 cdot 0^3 - 0^2 + 2 cdot 0 + 1 = 1 = 3^0 ] La seule solution est donc: [ n = 0 quad text{et} quad m = 0 ] [ boxed{(0, 0)} ]

answer:The equation of the right latus rectum of the hyperbola C: frac{x^{2}}{a^{2}} - frac{y^{2}}{b^{2}} = 1 is x = frac{a^{2}}{c}, The equations of the two asymptotes are y = pm frac{b}{a}x, hence we get A(frac{a^{2}}{c}, frac{ab}{c}) and B(frac{a^{2}}{c}, -frac{ab}{c}), The side length of the equilateral triangle ABF is |AB| = frac{2ab}{c}, The distance from F(c, 0) to the right latus rectum is c - frac{a^{2}}{c}, Hence, c - frac{a^{2}}{c} = frac{sqrt{3}}{2} cdot frac{2ab}{c}, This implies b^{2} = c^{2} - a^{2} = sqrt{3}ab, Thus, b = sqrt{3}a, And b^{2} = c^{2} - a^{2} = 3a^{2}, So, c = 2a, and e = frac{c}{a} = 2, Therefore, the answer is: boxed{2}. We find the equation of the right latus rectum and the asymptotes to get the coordinates and distance of A and B. We then find the distance from F to the latus rectum. Using the relationship between the height and side length of an equilateral triangle, along with the relationships between a, b, c, and e of the hyperbola, we can calculate the required value. This problem tests the method of finding the eccentricity of a hyperbola. Pay attention to the use of the equations of the latus rectum and asymptotes. It also tests knowledge of the properties of an equilateral triangle and computational skills. This is a moderately difficult problem.