Check the attached file for the solution.
Answer:
she would need 7 ounces
Step-by-step explanation:
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:
Simple
Rise/Run
for one week they run 10 miles
2 weeks 20 miles
3 weeks 30 miles
4 weeks 40 miles
the rate of change would be 10/1 (Why? Because you rise +10 each time per week.)
divide that it would be 10 (:
gOoD lUCk
second answer copied me so.
