means to say that for any given 
, we can find 
such that anytime 
(i.e. the whenever 
is "close enough" to 5), we can guarantee that 
(i.e. the value of 
is "close enough" to the limit value).
What we want to end up with is
Dividing both sides by 3 gives
which suggests 
is a sufficient threshold.
The proof itself is essentially the reverse of this analysis: Let be given. Then if
and so the limit is 7. QED
No,
is not a subspace of
. A simple counter-example to the contrary: let
with
. However, scaling by -1 gives the vector
and
.
180-(180-49-48)
180-(83)
97
180-((180-141)+(180-76))
180-(39+104)
180-143
137
180-((180-101)+60)
180-(79+60)
180-139
41
360-147-79
213-79
139
180-(360-109*2)
180-(360-218)
180-142
38
(180-118)/2
62/2
31
Hope this helps :)
Rounded to the nearest tenth would be one since the four isn't higher. So the answer would be 10,386.1. You did mean the tenths place right?? because could have also meant tenth(8)
Indirect material cost: y
explained by units produced: x
Linear regression. Cost estimation equation: y=mx+b
Constant: b=$15,640
Standard error of y estimate=$3,600
r^2=0.7704
Number of observations: n=22
x coeffient: m=11.25
Standard error of x coefficient=2.19
m=11.25, b=15,640 → y=11.25x+15,640
Answer: The cost estimation equation is y=11.25x+15,640
