Im pretty sure the answer is d
Answer:
28.6, that is, about 29 are expected to be defective
Step-by-step explanation:
For each battery, there are only two possible outcomes. Either it is defective, or it is not. The probability of a battery being defective is independent of other betteries. So the binomial probability distribution is used to solve this question.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
The expected value of the binomial distribution is:

The probability that a battery is defective is 1/14.
This means that 
400 batteries.
This means that 
How many are expected to be defective?

28.6, that is, about 29 are expected to be defective
The answer to that is: -12f+6g
Hope this helps ;-)
<span>4% is the same as 0.04 </span>
<span>so 0.04 x n = 56 where "n" is some number </span>
<span>0.04n = 56 Divide both sides by 0.04 </span>
<span>n = 1400</span>