Answer:
mean = 1 power failure
variance = 1 (power failure)²
Step-by-step explanation:
Since the mean is computed as
mean = E(x) = ∑ x * p(x) for all x
then for the random variable x=power failures , we have
mean = ∑ x * p(x) = 0 * 0.4 + 1* 0.3 + 2*0.2 + 3* 0.1 = 1 power failure
since the variance can be calculated through
variance = ∑[x-E(x)]² * p(x) for all x
but easily in this way
variance = E(x²) - [E(x)]² , then
E(x²) = ∑ x² * p(x) = 0² * 0.4 + 1²* 0.3 + 2²*0.2 + 3²* 0.1 = 2 power failure²
then
variance = 2 power failure² - (1 power failure)² = 1 power failure²
therefore
mean = 1 power failure
variance = 1 power failure²
Answer:
Is there any other information?
Step-by-step explanation:
I cannot reach a meaningful solution from the given information. To prove that S was always true, you would have to prove that N was always false. To prove that N was always false you would have to prove that L was always false. For the statement (L ^ T) -> K to be true, you only need K to be true, so L can be either true or false.
Therefore, because of the aforementioned knowledge, I do not believe that you can prove S to be true.
Answer:
The roots of the equation 2m²+3=m are non-real roots.
Step-by-step explanation:
Given equation:
2m²+3=m
2m²-m+3=0
Here, from the equation we can obtain the following values:
a = 2, b = -1, c = 3
Discriminant of an equation is given as:
D = b²-4ac
= (-1)²-4(2)(3)
= 1 - 21
= -20
Discriminant can tell what kinds of roots the equation have.
In our case, the discriminant is less than 0.
When D < 0, the roots of the equation are complex conjugates.(non-real)
Answer:
1
Step-by-step explanation:
