You cant solve these if they are not equating up to something hence incomplete question.
Using statistical concepts, it is found that the number of outcomes that are possible for the complement of the union of Events J and K is of 43.
<h3>What is the union of events J and K?</h3>
It means that at least one of event J or event K is true, hence, it is composed by employees that are either considered support staff(less than 5 years of experience) or employees that have more than five years of experience, combining a total of 7 + 8 = 15 employees.
<h3>What is the complement?</h3>
The total number of outcomes of the union of J and K, plus the complement, add to the total number of 58, hence:
15 + x = 58
x = 43.
The number of outcomes that are possible for the complement of the union of Events J and K is of 43.
More can be learned about complementary events at brainly.com/question/9752956
A)46. Take 42 and divide it by 3. Then multiply 14 by 4.
b)2:1. There are 2 wings for every 1 beak on a bird.
c)15 minutes. multiply 50 times 60 and that's how many times it beats per minute. Then divide 45000 by that number and that's how many minutes it should take.
d) 28%. You have to add all the students together, and you get 200. Then you divide 56 by 200.
Answer:
| -3 | = 3
Step-by-step explanation:
minus is removed
Answer:
The probably genotype of individual #4 if 'Aa' and individual #6 is 'aa'.
Step-by-step explanation:
In a non sex-linked, dominant trait where both parents carry and show the trait and produce children that both have and don't have the trait, they would each have a genotype of 'Aa' which would produce a likelihood of 75% of children that carry the dominant traint and 25% that don't. Since the child of #1 and #2, #5, does not exhibit the trait, nor does the significant other (#6), then they both must have the 'aa' genotype. However, since #4 displays the dominant trait received from the parents, it is more likely they would have the 'Aa' genotype as by the punnet square of 'Aa' x 'Aa', 50% of their children would have the 'Aa' phenotype.
