Answer:
(value - mean)/standard deviation.
Now substitute 78 for the value, 84.5 for the mean, and 3.1 for the standard deviation.
Step-by-step explanation:
The weight of an orangutan is 64.5 kilograms
Solution:
It is stated that the orangutan's weight is 142 pounds
Additionally,
1 kilogram is approximately 2.2 pounds
We will now convert 142 pounds into kilograms
1 kilogram = 2.2 pounds
As a result, 142 pounds translates to
When rounding to the nearest tenth, we obtain 64.5 kilograms
Therefore the mass of the orangutan is 64.5 kilograms
In the ratio of miles traveled, Mr. Storey and Mrs. Storey's journey is represented as mr: mrs = 3: 1. The total of these ratio units equals 4 (3+1), which means each unit corresponds to 3200 divided by 4, resulting in 800 miles. Mr. Storey drove 3 multiplied by 800, equating to 2400 miles, while Mrs. Storey drove 1 multiplied by 800, totaling 800 miles.
Part A:
Considering the best possible outcome
The ideal case occurs if the two missing socks are from the same pair.
Consequently, there are 4 complete pairs remaining.
To choose 2 from the total of 10 socks (5 pairs), the number of combinations is given by 10C2 = 45.
Choosing 2 that are from the same pair means selecting one from 5 pairs, so the count is 5C1 = 5.
Thus, the probability for this best case is 5 / 45 = 1 / 9.
Part B:
Considering the worst-case outcome
This scenario occurs when the two missing socks are from different pairs.
As a result, we have 3 complete pairs left.
The total ways to select 2 socks from 10, again, is 10C2 = 45.
To select 2 that do not belong to the same pair, we calculate as follows: 10C2 - 5C1 = 45 - 5 = 40.
Therefore, the probability for the worst-case scenario is 40 / 45 = 8 / 9.
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!
