Suppose that, in flight, airplane engines will fail with probability 1 − p, independently from engine to engine. If an airplane needs a majority of its engines operative to complete a successful flight, for what values of p is a 5-engine plane preferable to a 3-engine plane?

The answer is p ≥  

Step-by-step explanation:

The probability of failure of an engine is 1 - p.

The probability that the engine operates successfully  is  = 1 - (1 - p) = p

For a successful flight a majority of its engine have to operate.

So for a 5 engine plane to have a successful flight at least 3 of the engines have to operate successfully.

For a 3 engine plane at least 2 of its engines  have to operate for the total flight for the flight to be successful.

Therefore using the binomial distribution we find the probability for a successful flight for a 5 plane engine is given by

=

=

=

We find the probability for a successful flight for a 3 plane engine is given by

=

=

A 5 engine plane will be preferable to a 3 engine plane

Solving the above inequality we get p ≥    because  p cannot be greater than 1 and p greater than 0 does not really signify much.

The answer is p ≥  

b.-4 units

all you   do is count how many times you move

i believe that the answer is y+6=1/3(x+2)