Answer:
0.057
0.6140
0.3158
0.0701
Step-by-step explanation:
Given that:
Let :
P(L) = Number transported by land = half = 50% = 0.5
P(S) = number transported by sea = 30% = 0.3
P(A) = Number transported by air = (100 - (50 + 30))% = 20% = 0.2
P(H) = highway transport = 40% of land transport = 0.4 * 0.5 = 0.2
P(R) = Rail shipment =(100- 40)% = 60% of land transport = 0.6 * 0.5 = 0.3
Percentage of damaged cargo :
Let probability of damage = P(d)
P(d | H) = 0.1
P(d | R) = 0.05
P(d | S) = 0.06
P(d | A) = 0.02
1) What percentage of all cargoes may be expected to be damaged
[P(d | H)*p(H)] + [P(d | R)*p(R)] + [P(d | S)*p(S)] + [P(d | A)*p(A)]
(0.1*0.2) + (0.05*0.3) + (0.06*0.3) + (0.02*0.2) = 0.057
(2) If a damaged cargo is received, what is the probability that it was shipped by ;
land?
([P(d | H)*p(H)] + [P(d | R)*p(R)]) / [P(d | H)*p(H)] + [P(d | R)*p(R)] + [P(d | S)*p(S)] + [P(d | A)*p(A)]
((0.1*0.2) + (0.05*0.3)) / (0.1*0.2) + (0.05*0.3) + (0.06*0.3) + (0.02*0.2)
= 0.035 / 0.057
= 0.6140
By sea?
[P(d | S)*p(S)] / [P(d | H)*p(H)] + [P(d | R)*p(R)] + [P(d | S)*p(S)] + [P(d | A)*p(A)]
(0.06 * 0.3) / 0.057
= 0.3158
By air?
[P(d | A)*p(A)] / [P(d | H)*p(H)] + [P(d | R)*p(R)] + [P(d | S)*p(S)] + [P(d | A)*p(A)]
(0.02 * 0.2) / 0.057
= 0.0701
- Vector distance between points A and Q.
- Vector force exerted at point A.