Fill `4/7....... 5/14`

Solve : (3x +4)/(7) - (x +5)/(14) = (x)/(28) + (x +1)/(14) The following steps are involved in solving the above problem. Arrange them in sequential order (A) 5x + 3 = (3x + 2)/(2) rArr 10 x + 6 = 3x + 2 (B) rArr 7x = -4 (C) Given (3x + 4)/(7) - (x + 5)/(14) = (x)/(28) + (x +1)/(14) rArr (6x + 8 -x -5)/(14) = (x + 2x + 2)/(28) (D) rArr x = -(4)/(7)

4(3)/(7)+2(5)/(14)

(1/4 xx 2/7)- ( 5/(14) xx 2/3)

Divide: (i) 5/9 "by"2/3 (ii) 5 4/7"by"13/14 (iii) 4 2/7"by"2 2/5

Fill in the blanks: 5/7 div 5/7 = ___

Which of the following cannot be valid assignmentof probabilities for outcomes of sample space S={omega_(1),omega_(2),omega_(3),omega_(4),omega_(5),omega_(6),omega_(7)} ? Assignement omega_(1),omega_(2),omega_(3),omega_(4),omega_(5),omega_(6),omega^(7) a. 0.1,0.01,0.05,0.03,0.01,0.3,0.6 b. 1/7,1/7,1/7,1/7,1/7,1/7,1/7 c. 0.1,0.2,0.3,0.4,0.5,0.6,0.7 d. -0.1,0.2,0.3,0.4,-0.2,0.1,0.3 e. 1/14,2/14,3/14,4/14,5/14,6/14,15/14

Which of the following cannot be valid assignment of probability for elementary events or outcomes of samples space S={w_1, w_2, w_3, w_4, w_5, w_6, w_7}: Elementary events w_1 w_2 w_3 w_4 w_5 w_6 w_7 i. 0.1 0.01 0.05 0.03 0.01 0.2 0.6 ii. 1/7 1/7 1/7 1/7 1/7 1/7 1/7 iii. 0.7 0.6 0.5 0.4 0.3 0.2 0.1 iv. 1/(14) 2/(14) 3/(14) 4/(14) 5/(14) 6/(14) (15)/(14)