Answer:
36.58% probability that one of the devices fail
Step-by-step explanation:
For each device, there are only two possible outcomes. Either it fails, or it does not fail. The probability of a device failling is independent of other devices. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which 
is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
A total of 15 devices will be used.
This means that 
Assume that each device has a probability of 0.05 of failure during the course of the monitoring period.
This means that 
What is the probability that one of the devices fail?
This is 
36.58% probability that one of the devices fail
Answer:
a_n = 2^(n - 1) 3^(3 - n)
Step-by-step explanation:
9,6,4,8/3,…
a1 = 3^2
a2 = 3 * 2
a3 = 2^2
As we can see, the 3 ^x is decreasing and the 2^ y is increasing
We need to play with the exponent in terms of n
Lets look at the exponent for the base of 2
a1 = 3^2 2^0
a2 = 3^1 2^1
a3 = 3^ 0 2^2
an = 3^ 2^(n-1)
I picked n-1 because that is where it starts 0
n = 1 (1-1) =0
n=2 (2-1) =1
n=3 (3-1) =2
Now we need to figure out the exponent for the 3 base
I will pick (3-n)
n =1 (3-1) =2
n =2 (3-2) =1
n=3 (3-3) =0
7 to 8 i think............................
Option A is the correct answer, y = x + 2
y - 3 = x-1
If you bring the 3 to the R.H.S, it becomes y = x + 2
If you substitute the values,
3 = 1 + 2
3 = 3
Thus, e equation is correct
