ai) P( pupils have read at least one of the magazines) = P(A) + P... pupils have read at least one of the magazines = 0.7 *400 =280 aii) P( pupils have read at least one of the magazines) = P(AB) + P...
1.a) Let $x be the sum of money and y be the number of pupils . x=18y-14 (x+14)/18=y------1 x=17y+4 (x-4)/17=y------2 Sub. 1 into...
...70 - 20 - 10 = 60 So there are 70 + 30 + 10 + 20 + 40 + 50 + 60 = 280 pupils have read at least one of magazines.
It is impossible for a P.3 pupil to solve the problem with equation. (一位小三學生沒法以方程式去解這問題...
...)+n(A∩ B∩ C)=280 (ii) Consider how many pupils only read A and B Number=n(A∩ B)-n(A...
...63-16﹞ =﹝458﹞﹝47﹞ = 21526 幫我轉反做中文 2.There are 40 pupils in P.6A and there are 10 girls more than boys...
The number of girls =800 x 37.5% =300 The number of boys =800-300 =500 The number of students after increasing =300 x (1+18%) + 500 x (1+14%) =354 + 570 =924 The percentage increase =[(924-800)/800] x 100% =15.5%