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.
Answer:
30
Step-by-step explanation:
V=lwh
V=2*3*5
V=6*5
V=30
Step-by-step explanation:
Given:
x = 26214.47
s = 5969.25
n = 15
t = 2.046
The confidence interval is:
CI = x ± t (s / √n)
CI = 26214.47 ± 2.046 (5969.25 / √15)
CI = (23061.06, 29367.88)
0.000334 is the scientific notation
