The AAA system contains 3 independent components in parallel, each with probability of failure 0.4. Find the expected number of
1 answer:
Answer:
Step-by-step explanation:
For each component, there are only two possible outcomes. Either it fails, or it does not. The components are independent. We want to know how many outcomes until r failures. The expected value is given by
In which r is the number of failures we want and p is the probability of a failure.
In this problem, we have that:
r = 1 because we want the first failed unit.
The expected number of systems inspected until the first failed unit is 2.5
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Answer:
The answer: The factors are (2x-9) and (x+6).
The Problem:
Factor
Step-by-step explanation:
So I'm going to do trial factors using the choices to aid me.
Factored form for this problem if it exist will be in the form:
.
In general this is what it would look like if we factored any quadratic in terms of (given the quadratic is not prime but technically you could factor even over the complex numbers).
Let's look at:
We want to choose such that when you multiply them you get 2. Well those would have to be 2 and 1.
we want to choose such that when you multiply them you get -54. Based on the choices we want to get with -9 and 6, or 9 and -6. We don't know the order we want to choose it in either.
For example which of these would work:
We are going to consider only the outer and inner of FOIL since we already know the first times the first is and the last times the last is
.
Let's test the first one:
Outer: 2x(9)=18x
Inner: -6(x)=-6x
------------------------ADD!
18x-6x=12x
The first choice did not give us the middle term 3x.
Trying the second one would give us the opposite since they are in the same form as previous just the + and - are switched.
Let's look at the third one:
Outer: 2x(6)=12x
Inner: -9(x)=-9x
--------------------------ADD!
12x-9x=3x
This is the winner.
The answer: The factors are (2x-9) and (x+6).
Answer:
Step-by-step explanation:
PART 1:
Possible roots by noticing the coefficient of first term:
x = 1, -1
PART 2:
1 | 1 -1 -4 4
<u> 1 0 -4</u>
1 0 -4 0
The remainder is zero, hence one factor is (x-1)
PART 3:
PART 4: