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.
#1.
Here what you do is cross multiply.
#2.
Same as #1, just cross multiply.
You get the gist?
Answer:
It's 4
Step-by-step explanation:
It asks how many 3/8 cups are in 1 1/2 cups. So, we divide.
1 1/2 ÷ 3/8. We have to change the 1 1/2 into a improper fraction. 1 1/2 = 3/2. 3/2 ÷ 3/8 = 3/2 ÷ 8/3 = 24/6. Simplify, and you get 4. :)
Step-by-step explanation:
from the point -10 to -4 thre is a difference of 6 uints . Base 6
from the x-axis to furthest point,
there is a difference of 6 units. Height 6
