Answer: 2 2/5 is 4/5 of 3
Step-by-step explanation:
We can view that as what do we multiply 3 by to get 2 2/5 and call that value x
3x = 2 2/5
x = (2 2/5)/3
Let's turn 2 2/5 into an improper fraction... 2 2/5 = 12/5
So we have
x = (12/5)/3 = (12/5)(1/3) = 4/5
Thus 2 2/5 is 4/5 of 3
Verify: 3(4/5) = 12/5 = 2 2/5
I think the answer is -1=x-3<4
To find the perimeter you do length + width so you do 12.4 + 5.9 to get the perimeter.
If 3 =
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, then we know 9 =

because 9 is 3 squared.

can be simplified to

So now we know

=

·

Since

and

have the same base, we can express it as a single term by adding the exponents together.
So the answer is C)
Answer:
(c)approximately Normal, mean 112, standard deviation 1.414.
Step-by-step explanation:
To solve this problem, we have to understand the Central Limit Theorem
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a random variable X, with mean
and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
.
In this problem, we have that:

Using the Central Limit Theorem
The distribution of the sample mean IQ is approximately Normal.
With mean 112
With standard deviation 
So the correct answer is:
(c)approximately Normal, mean 112, standard deviation 1.414.