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:
The choice that best describes the given sentence above is, it is a complete and correct sentence. What makes this sentence complete is having both the subject and the verb in a simple form, and still expresses a complete thought. The verb and simple predicate "volunteered" is enough to describe the subject "Cassidy".
Answer:
2 13/14 feet
Step-by-step explanation:
The remaining length is the difference between the starting length and the amount cut off:
5 3/7 - 2 1/2 = (5 -2) +(3/7 -1/2) = 3 +((3·2-7·1)/(7·2)) = 3 -1/14
= 2 13/14 . . . feet remaining
_____
<em>Additional comment</em>
Addition and subtraction of fractions can be accomplished using a formula that does not rely on finding a least common denominator. In this case, the denominators have no common factors, so their product is, in fact, the least common denominator.

Answer:
(x+6)^2+3
Step-by-step explanation:
translate 6 left
left is negative
so (x+6)^2. it is + because the x goes in the opposite direction causing (x-6)^2 to go right
and 3 units up so add x by 3 to get (x+6)^2+3
Answer:
a

b

Ca
Cb
Explanation:
From the question we are told that
The sample size is n = 100
The upper limit of the 95% confidence interval is b = 47.2 years
The lower limit of the 95% confidence interval is a = 34.5 years
Generally the sample mean is mathematically represented as

=> 
=> 
Generally the margin of error is mathematically represented as

=> 
=> 
Considering question C a
From the question we are told the confidence level is 90% , hence the level of significance is
=>
The sample size is n = 22
Given that the sample size is not sufficient enough i.e
we will make use of the student t distribution table
Generally the degree of freedom is mathematically represented as

=> 
=> 
Generally from the student t distribution table the critical value of
at a degree of freedom of 21 is
Considering question C b
From the question we are told the confidence level is 80% , hence the level of significance is
=>
Generally from the normal distribution table the critical value of
is