With continuous data, it is possible to find the midpoint of any two distinct values. For instance, if h = height of tree, then its possible to find the middle height of h = 10 and h = 7 (which in this case is h = 8.5)
On the other hand, discrete data can't be treated the same way (eg: if n = number of people, then there is no midpoint between n = 3 and n = 4).
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With that in mind, we have the following answers
1) Continuous data. Time values are always continuous. Any two distinct time values can be averaged to find the midpoint
2) Continuous data. Like time values, temperatures can be averaged as well.
3) Discrete data. Place locations in a race or competition are finite and we can't have midpoints. We can't have a midpoint between 9th and 10th place for instance.
4) Continuous data. We can find the midpoint and it makes sense to do so when it comes to speeds.
5) Discrete data. This is a finite number and countable. We cannot have 20.5 freshman for instance.
I think it is 29%? I am not 100% sure though...
Substitute 
, so that
Then the resulting ODE in 
is separable, with
On the left, we can split into partial fractions:
Integrating both sides gives
Now solve for 
:
Answer:
E is not a subspace of 
Step-by-step explanation:
E is not a subspace of
In order to see this, we must find two points (a,b), (c,d) in E such that (a,b) + (c,d) is not in E.
Consider
(a,b) = (1,1)
(c,d) = (-1,-1)
It is easy to see that both (a,b) and (c,d) are in E since 1*1>0 and (1-)*(-1)>0.
But (a,b) + (c,d) = (1-1, 1-1) = (0,0)
and (0,0) is not in E.
By the way, it can be proved that in any vector space all sub spaces must have the vector zero.
