Answer:Hence, x = 5, y = 0 is the required solution.
Step-by-step explanation:
Answer:
P=0.147
Step-by-step explanation:
As we know 80% of the trucks have good brakes. That means that probability the 1 randomly selected truck has good brakes is P(good brakes)=0.8 . So the probability that 1 randomly selected truck has bad brakes Q(bad brakes)=1-0.8-0.2
We have to find the probability, that at least 9 trucks from 16 have good brakes, however fewer than 12 trucks from 16 have good brakes. That actually means the the number of trucks with good brakes has to be 9, 10 or 11 trucks from 16.
We have to find the probability of each event (9, 10 or 11 trucks from 16 will pass the inspection) . To find the required probability 3 mentioned probabilitie have to be summarized.
So P(9/16 )= C16 9 * P(good brakes)^9*Q(bad brakes)^7
P(9/16 )= 16!/9!/7!*0.8^9*0.2^7= 11*13*5*16*0.8^9*0.2^7=approx 0.02
P(10/16)=16!/10!/6!*0.8^10*0.2^6=11*13*7*0.8^10*0.2^6=approx 0.007
P(11/16)=16!/11!/5!*0.8^11*0.2^5=13*21*16*0.8^11*0.2^5=approx 0.12
P(9≤x<12)=P(9/16)+P(10/16)+P(11/16)=0.02+0.007+0.12=0.147
Answer: The proportion of employees who either have MBAs or are managers are 0.58.
Step-by-step explanation:
Since we have given that
Probability of employees having managerial positions = 67%
Probability of employees having MBA degrees = 58%
Probability of managers having MBA degrees = 67%
So, using probability formulas, we get that
Hence, the proportion of employees who either have MBAs or are managers are 0.58.
