The GCF of 26 and 70 is 2.
Answer:

In order to satisfy this distribution we need that each observation on this case comes from a normal distribution, because since the sample size is not large enough we can't apply the central limit theorem.
Step-by-step explanation:
For this case we have that the sample size is n =6
The sample man is defined as :

And we want a normal distribution for the sample mean

In order to satisfy this distribution we need that each observation on this case comes from a normal distribution, because since the sample size is not large enough we can't apply the central limit theorem.
So for this case we need to satisfy the following condition:

Because if we find the parameters we got:


And the deviation would be:

And we satisfy the condition:

Answer:
<em>The function to represent this relationship will be:
</em>
Step-by-step explanation:
varies inversely with
. That means......
where
is a proportional constant.
Given that, when
, then 
Plugging these values into the above equation, we will get.....

Now plugging this
into equation (1).....

So, the function to represent this relationship will be: 
A, sub in a random number as x in all equations and solve, eg- x= 3
2-3= -1
3-2=1
-2 +3= 1
3-2=1
Answer:
-3x + 31
Step-by-step explanation:
First distribute
2x + 16 -5x +15
Collect like terms
-3x + 31