A calculator requires a keystroke assembly and a logic circuit. Assume that 83% of the keystroke assemblies and 88% of the logic circuits are satisfactory. Assuming that defects in keystroke assemblies are independent of defects in logic circuits, find the probability that a finished calculator will be satisfactory. Group of answer choices

0.7304

Step-by-step explanation:

According to the Question,

We have,

P(keystroke satisfactory) =0.83 ,  P(logic satisfactory)= 0.88

Since the events are independent. So, the probability that a finished calculator will be satisfactory

⇒ 0.83×0.88

⇒0.7304

it could 14.28 but if its not that one then it has to be the 17 one. hope this ! : )

step-by-step explanation:

answer: there are x+184 pages in olivia's book.

step-by-step explanation:

let the total number of pages in the book which olive has already read be x.

then if olive has 184 pages left to read, then the total number of pages are in olive's book= pages she has read+pages she has left to read

therefore, the total number of pages in olivia's book=

hence, there are x+184 pages in olivia's book.