We have to determine which value is equivalent to | f ( i ) | if the function is: f ( x ) = 1 - x. We know that for the complex number: z = a + b i , the absolute value is: | z | = sqrt( a^2 + b^2 ). In this case: | f ( i )| = | 1 - i |. So: a = 1, b = - 1. | f ( i ) | = sqrt ( 1^2 + ( - 1 )^2) = sqrt ( 1 + 1 ) = sqrt ( 2 ). Answer: <span>C. sqrt( 2 )</span>
Answer:
c. A two-tailed test should be performed since the alternative hypothesis states that the parameter is not equal to the hypothesized value.
Step-by-step explanation:
Let p1 be the average score on a final exam who texted on a regular basis during the lectures for a particular class
And p2 be the average score on a final exam who did not texted at all during the lectures for a particular class
According to the Cameron's point of interest, null and alternative hypotheses are:
p1 = p2
p1 ≠ p2
Two tailed test should be performed since the alternative hypothesis states that the parameter is not equal to the hypothesized value.
Answer:
the ablity to choose one is the right now
Step-by-step explanation:
Answer:
Step-by-step explanation:
Suppose the time required for an auto shop to do a tune-up is normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - u)/s
Where
x = points scored by students
u = mean time
s = standard deviation
From the information given,
u = 102 minutes
s = 18 minutes
1) We want to find the probability that a tune-up will take more than 2hrs. It is expressed as
P(x > 120 minutes) = 1 - P(x ≤ 120)
For x = 120
z = (120 - 102)/18 = 1
Looking at the normal distribution table, the probability corresponding to the z score is 0.8413
P(x > 120) = 1 - 0.8413 = 0.1587
2) We want to find the probability that a tune-up will take lesser than 66 minutes. It is expressed as
P(x < 66 minutes)
For x = 66
z = (66 - 102)/18 = - 2
Looking at the normal distribution table, the probability corresponding to the z score is 0.02275
P(x < 66 minutes) = 0.02275
Answer:
a) 23.76%
b) 7.8%
Step-by-step explanation:
a) probability that a failure is due to loose keys.
loose key failure (27%) comes under mechanical failure(88%)
hence, probability that a failure is due to loose keys= 0.27×0.88= 0.2376= 23.76%
b) probability that a failure is due to improperly connected wire which comes under electrical failure = 0.12×0.13
probability that a failure is due to poorly welded wires which comes under electrical failure= 0.52×0.12
now, the probability that a failure is due to improperly connected or poorly welded wires. = 0.12(0.52+0.13)= 0.078= 7.8%
