answer:Let's start by calculating the number of pages in each textbook. History textbook: 160 pages Geography textbook: 160 pages + 70 pages = 230 pages Now, let's find the number of pages in the math textbook, which is half of the sum of the pages in the history and geography textbooks: Math textbook: (160 pages + 230 pages) / 2 = 390 pages / 2 = 195 pages Now we know the number of pages in the history, geography, and math textbooks. Let's denote the number of pages in the science textbook as S pages. According to the information given, the total number of pages when all textbooks are stacked is 905 pages. So we can write the following equation: 160 pages (history) + 230 pages (geography) + 195 pages (math) + S pages (science) = 905 pages Now, let's solve for S: 160 + 230 + 195 + S = 905 585 + S = 905 S = 905 - 585 S = 320 pages Now we have the number of pages in the science textbook, which is 320 pages. To find the ratio of the number of pages in the science textbook to the number of pages in the history textbook, we divide the number of pages in the science textbook by the number of pages in the history textbook: Ratio = Number of pages in science textbook / Number of pages in history textbook Ratio = 320 pages / 160 pages Ratio = 2 / 1 Therefore, the ratio of the number of pages in the science textbook to the number of pages in the history textbook is boxed{2:1} .

answer:Let the income be represented by 5x and the expenditure by 4x, where x is a common multiplier. According to the problem, the income (5x) is Rs. 19000. So we can write: 5x = 19000 To find the value of x, we divide both sides of the equation by 5: x = 19000 / 5 x = 3800 Now that we have the value of x, we can calculate the expenditure (4x): Expenditure = 4x = 4 * 3800 = 15200 The savings of the person would be the difference between his income and his expenditure: Savings = Income - Expenditure Savings = 19000 - 15200 Savings = Rs. 3800 So, the person's savings are Rs. boxed{3800} .

answer:(1) When m=1, f(x)=|x+1|+|2x-1|, then f(x)= begin{cases} -3x,x < -1 2-x,-1leqslant xleqslant frac {1}{2} 3x,x > frac {1}{2}end{cases}, by f(x)geqslant 3 we get xleqslant -1 or xgeqslant 1, thus the solution set of the original inequality is (-infty,-1]cup[1,+infty); (2) From frac {1}{2}f(x)leqslant |x+1|, we have frac {1}{2}|x+m|+ frac {1}{2}|2x-1|leqslant |x+1|, given xin[m,2m] and m < frac {1}{4}, so 0 < m < frac {1}{4}, and x > 0 thus frac {1}{2}x+ frac {m}{2}leqslant |x+1|- frac {1}{2}|2x-1|, i.e., mleqslant x+2-|2x-1|; let t(x)=x+2-|2x-1|, then t(x)= begin{cases} 3x+1,0 < x < frac {1}{2} 3-x,xgeqslant frac {1}{2}end{cases}, so when xin[m,2m], t(x)_{min}=t(m)=3m+1, thus mleqslant 3m+1, solving for m we get mgeqslant - frac {1}{2}, therefore the range of values for the real number m is boxed{(0, frac {1}{4})}.

answer:To find the probability that A and B are not both at the ends of the row simultaneously, we need to determine the total number of arrangements and then subtract the number of arrangements where they are both at the ends. First, the total number of ways that 5 people can be arranged in a row is 5! = 120. Next, we consider the arrangements where A and B are both at the ends. Since there are two possible positions for A (either end) and once A is placed, there are two positions for B (the other end), this gives us 2! arrangements for A and B. The remaining 3 people can be arranged in 3! ways. Therefore, the number of arrangements where A and B are both at the ends is 2! times 3! = 12. Now, the number of arrangements where A and B are not both at the ends is 120 - 12 = 108. Finally, the probability that A and B do not stand at the two ends at the same time is frac{108}{120}. Simplifying this gives us: frac{108}{120} = frac{9}{10} = 0.9 So the probability is boxed{0.9}.