Answer
given,
work = 5 J
spring stretch form 30 cm to 39 cm
x = 0.39 - 0.30 = 0.09 m
k = 1234.568 N/m
a) work when spring is stretched from 32 cm to 34 cm
x₂= 0.34 -0.30 = 0.04 m
x₁ = 0.32 - 0.30 = 0.02
W = 0.741 J
b) F = k x
25 = 1234.568 × x
x = 0.0205 m
x = 2.05 cm
Answer:
(a) The probability that a randomly selected parcel arrived late is 0.026.
(b) The probability that a parcel was late was being shipped through the overnight mail service A₁ is 0.615.
(c) The probability that a parcel was late was being shipped through the overnight mail service A₂ is 0.192.
(d) The probability that a parcel was late was being shipped through the overnight mail service A₃ is 0.192.
Step-by-step explanation:
Consider the tree diagram below.
(a)
The law of total probability sates that:
Use the law of total probability to determine the probability of a parcel being late.
Thus, the probability that a randomly selected parcel arrived late is 0.026.
(b)
The conditional probability of an event A provided that another event B has already occurred is:
Compute the probability that a parcel was late was being shipped through the overnight mail service A₁ as follows:
Thus, the probability that a parcel was late was being shipped through the overnight mail service A₁ is 0.615.
(c)
Compute the probability that a parcel was late was being shipped through the overnight mail service A₂ as follows:
Thus, the probability that a parcel was late was being shipped through the overnight mail service A₂ is 0.192.
(d)
Compute the probability that a parcel was late was being shipped through the overnight mail service A₂ as follows:
Thus, the probability that a parcel was late was being shipped through the overnight mail service A₃ is 0.192.
