The answer to that question would be 14
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:
It should be 20minutes at a easy pace
Step-by-step explanation:
I think the answer is 52 cm. If you add all the numbers and find out the two missing numbers you will get 52. ( if I’m sorry so how, I’m sorry :))
Answer:
x=0.69897009
Step-by-step explanation:
expand by logarithms
10^x=5
=log(10^x)=log(5)
=x log(10)=log (5)
log base 10 of 10 is 1
x.l =log(5)
multiple bothside by 1
x=log(5)
x=0.6989709
