Commercial aircraft used for flying in instrument conditions are required to have two independent radios instead of one. assume that for a typical flight, the probability of a radio failure is 0.00260.0026. what is the probability that a particular flight will be safe with at least one working radio? why does the usual rounding rule of three significant digits not work here? is this probability high enough to ensure flight safety?

We are given that Commercial aircraft used for flying in instrument conditions are required to have two independent radios instead of one.

The probability of failure of one radio = 0.0026

So, the probability of failure of both radio =

Now we are supposed to find  the probability that a particular flight will be safe with at least one working radio

A) the probability that a particular flight will be safe with at least one working radio = 1- 0.00000676=0.99999324

B)The usual rounding rule of three significant digits not work here because usual rounding rule of three significant digits makes it 1 which means that both radios are working properly.

C)Yes,this probability is high enough to ensure flight safety because it is close to 1

-14

step-by-step explanation:

2x2 – 2(x+4)

=2x2-2（5+4）

=4-2×9

=4-18

=-14

c) yes, because of asa or aas