Answer:
a) 0.03
b) 0.68
c) 0.32
Step-by-step explanation:
We are given the following in the question:
B: companies in the area of biotechnology
I: companies in the area of information technology
P(B) = 0.2
P(I) = 0.15
The two events are given to be independent.
a) P(both companies become profitable)
0.03 is the probability that both companies become profitable
b) P(neither company becomes profitable)
0.68 is the probability that neither company becomes profitable.
c) P(at least one of the two companies become profitable)
0.32 is the probability that at least one of the two companies become profitable
Answer:
5/9 of a chance
Step-by-step explanation:
or a 55.55- chance
<span>0.316 in expanded form is= 0.3 + 0.01 + 0.006. </span>
Answer:
This problem is incomplete, we do not know the fraction of the students that have a dog and also have a cat. Suppose we write the problem as:
"In Mrs.Hu's classroom, 4/5 of the students have a dog as a pet. X of the students who have a dog as a pet also have cat as a pet. If there are 45 students in her class, how many have both a dog and a cat as pets?"
Where X must be a positive number smaller than one, now we can solve it:
we know that in the class we have 45 students, and 4/5 of those students have dogs, so the number of students that have a dog as a pet is:
N = 45*(4/5) = 36
And we know that X of those 36 students also have a cat, so the number of students that have a dog and a cat is:
M = 36*X
now, we do not have, suppose that the value of X is 1/2 ("1/2 of the students who have a dog also have a cat")
M = 36*(1/2) = 18
So you can replace the value of X in the equation and find the number of students that have a dog and a cat as pets.
