Answer:
a) 0.00019923%
b) 47.28%
Step-by-step explanation:
a) To determine the likelihood that all sockets in the sample are defective, we can use the following approach:
The first socket is among a group that has 5 defective out of 38, leading to a probability of 5/38.
The second socket is then taken from a group of 4 defective out of 37, following the selection of the first defective socket, resulting in a probability of 4/37.
Extending this logic, the chance of having all 5 defective sockets is computed as: (5/38)*(4/37)*(3/36)*(2/35)*(1/34) = 0.0000019923 = 0.00019923%.
b) Using similar reasoning as in part a, the first socket has a probability of 33/38 of not being defective as it's chosen from a set where 33 sockets are functionally sound. The next socket has a proportion of 32/37, and this continues onward.
The overall probability calculates to (33/38)*(32/37)*(31/36)*(30/35)*(29/34) = 0.4728 = 47.28%.
Answer:
5 ft
Step-by-step explanation:
Denote the height of the previous jump as j. Therefore, it follows that 1.1j equals 5.5 ft.
By dividing both sides by 1.1, we derive j = 5 ft. This indicates the height from the prior jump.
Response:

Detailed explanation:
We are aware that

She made the trip between her residence and the airport twice.
Therefore,
Multiply the distance from her house to the airport by 4.

Response:
(C) Fourth
Detailed reasoning:
We will create a list based on the given constraints.
Chris is last in the order, positioned right after Andy.
Frank holds the first position in line and is one spot ahead of Betty.
Mark stands in between Betty and Mary.
- Frank
- Betty
- Mark
- Mary
- Andy
- Chris