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.
Answer:
(C) They have the same coefficient of variation
Step-by-step explanation:
The coefficient of variation (CV) is calculated using the formula:
Where 
represents standard deviation and 
represents the mean.
Bob's average weight is 200 pounds with a standard deviation of 16 pounds
This indicates that 
.
Thus, his coefficient of variation is
Mary's average weight is 125 pounds, with a standard deviation of 10 pounds.
This implies 
Therefore, her coefficient of variation is
Since both have the same coefficient of variation, the accurate response is.
(C) They have the same coefficient of variation
Since m∠abe = 2b, and angle abe consists of angles abf and ebf, we can write:
m∠abe = m∠abf + m∠ebf
To find m∠ebf, rearrange:
m∠ebf = m∠abe - m∠abf
Substitute the given expressions:
m∠ebf = 2b - (7b - 24)
Simplify:
m∠ebf = 2b - 7b + 24
m∠ebf = -5b + 24.
Step-by-step explanation:
a) 7!
In absence of any restrictions, the answer is 7! as it represents the permutations of all animals.
b) 4! x 3!
Considering there are 6 cats and 5 dogs, the first and last slots must be occupied by cats to ensure alternate arrangements. The only options available then are based on the arrangement of the cats among themselves and the dogs among themselves, yielding 4! permutations for the cats and 3! for the dogs, thus leading to a total of 4! x 3! arrangements.
c) 3! x 5!
Here, the arrangement of the dogs among themselves can occur in 3! ways. Considering the dogs as a singular “object,” we can arrange this unit with the 4 cats, providing 5! total arrangements possible, leading to 3! · 5! arrangement possibilities.
d) 2 x 4! x 3!
In this scenario, both cats and dogs must be grouped together, allowing positions where all cats come before the dogs or vice versa. As there are two configurations, the resultant count is 2 multiplied by both arrangements, resulting in 2 x 4! x 3!
