(1) R+G+B=50
(2) R=B+6
(3) G=B-4 so substitute G in (1)
R+B-4+B=50 substitute (2) in here
B+6+B-4+B=50 group like terms and solve
3B+2=50
3B=50-2
B=48/3=16 substitute in (2) and (3)
R=16+6=22
G=16-4=12
Check
22+12+16=50✅
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.
You can't multiply anything to get a prime number ?
The 1st one is true
the 2nd one is false
the 3rd one is false
the 4th one is true
the 5th one is true
Answer:
the answer is the first one. im sorry if im wrong. good luck though
Step-by-step explanation:
