Answer:
Given that the frog jumps every 10 seconds
(using digits from a random number table)
- It requires 7 jumps with 2 in the reverse direction (either left or right) for the frog to get off the board in 60 seconds.
- Alternatively, 3 jumps in the same direction will also lead to the frog being off the board.
- Furthermore, it would take 5 jumps with one in the opposite direction within the time limit of 60 seconds to leave the board.
Step-by-step explanation:
A frog positioned right at the center of a 5ft long board is 2.5 ft away from either edge.
Every 10 seconds, the frog jumps left or right.
If the frog's jumps are LLRLRL, it will remain on the board at the leftmost square.
If it jumps as LLRLL, it will jump off the board after fifty seconds.
Given that the frog jumps every 10 seconds
(using digits from a random number table)
- It requires 7 jumps with 2 in reverse direction (either left or right) for the frog to get off the board in 60 seconds.
- Alternatively, 3 jumps in the same direction will also lead to the frog being off the board.
- Furthermore, it would take 5 jumps with one in the opposite direction within the time limit of 60 seconds to leave the board.
An arithmetic sequence is defined by the property that the difference between each pair of adjacent terms is consistently the same. When this requirement is met, that constant is recognized as the common difference.
This particular sequence is arithmetic, as the difference between each term and the preceding one is
Thus, the common difference is 3.4, considering that the adjacent terms differ by 3.4
Apply the Pythagorean theorem
x^2 + x^2 = 2x^2
sqrt(2x^2)=x sqrt(2)
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%.
