Acarton of 12 rechargable batteries contains one that is defective. in how many ways can the inspector choose 3 of the batteries and (a) get the one that is defective, (b)not get the one that is defective?

For part "b":  you have 1 defective and 11 no defective, so you have a combination 3 by 3 with 11 elements wich are not defective, so 11C2= 165.
For part "a":you have to choose the only that is defective so you have to order just 2 no defective, i mean from 11 no defective we have 11C2 = 55 

1. 0.3 •4 is the same as 4 groups of 0.32. 7•0.8 is the same as 7 groups of 0.83. it increases by 0.5 every time, so 0.5x4 it increases by 0.9 every time, so 0.9x5. 0.9•12=10.86. 0.47•5=2.35i can finish the next ones in a separate post