Properly identifying and assigning the proper variables to each part of the problem before attempting to solve it
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
Hi,
A is an arithmetic sequence
U(n)=U(n-1)+3
4t + 75 => 475, t = 100
Is this what you were looking for?
a) 32.25* 40 = 1290
b) 32.25 (1.5) *12 =580.50
c) 1290+580.50=1870.50
