Simplify: `(1/3+3/4(2/5-1/3))/(1 2/3\ of3/4-1/4\ of4/5)` `1/(63)` (b) `(23)/(40)` (c) `(23)/(55)` (d) `(23)/(63)`

(5 1/4-3 1/3)= (12)/(23) (b) 2 1(11)/(12) (d) (11)/(12)

((x+1)-(2x+4))/(3-5x)=(1)/(23)

|{:(1/3, 4),((-1)/6,3/2):}|=

(1)/(1.23)+(1)/(2.3.4)+(1)/(3.4.5)+(1)/(4.5.6)= a) (7)/(30) b) (11)/(30) c) (13)/(30) d (17)/(30)

((3(2)/(3))^(2)-(2(1)/(2))^(2))/((4(3)/(4))^(2)-(3(1)/(3))^(2))-:(3(2)/(3)-2(1)/(2))/(4(3)/(4)-3(1)/(3))=?(37)/(97) (b) (74)/(97)( c) 1(23)/(74) (d) None of these

Simplify: [0.9-{2.3-3.2-(7.1-5.4-3.5)}]

For teaching the concept of probability, Mrs. Verma decided to use two dice. Shet took a pair of die and write all the possible outcomes on the blackboard. All possible outcomes wave: (1,1), (1,2), (1,3), (1,4), (1,5), (1,6) (2,1), (2,2), (2,3), (2,4), (2,5), (2,6) (3,1), (3,2), (3,3), (3,4), (3,5), (3,6) (4,1), (4,2), (4,3), (4,4), (4,5), (4,6) (5,1), (5,2), (5,3), (5,4), (5,5), (5,6) (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) The probability that 4 will not come up on either of them is

For teaching the concept of probability, Mrs. Verma decided to use two dice. Shet took a pair of die and write all the possible outcomes on the blackboard. All possible outcomes wave: (1,1), (1,2), (1,3), (1,4), (1,5), (1,6) (2,1), (2,2), (2,3), (2,4), (2,5), (2,6) (3,1), (3,2), (3,3), (3,4), (3,5), (3,6) (4,1), (4,2), (4,3), (4,4), (4,5), (4,6) (5,1), (5,2), (5,3), (5,4), (5,5), (5,6) (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) The probability that 5 will come up at least once is: