Answer:
So if a value is less than 24.86 would be considered significantly low and a value higher than 29.38 would be considered as significantly high.
The value for the analysis is 29.8 and as we can see 29.8>29.38 so then we can consider 29.8 as a value significantly high.
Step-by-step explanation:
For this case we have the mean given and the deviation
The Range Rule of Thumb says "that the range is about four times the standard deviation"
So then we will ave approximately most of the value within 2 deviations from the mean, so we can find the limits considered normally like this:
So if a value is less than 24.86 would be considered significantly low and a value higher than 29.38 would be considered as significantly high.
The value for the analysis is 29.8 and as we can see 29.8>29.38 so then we can consider 29.8 as a value significantly high.
Answer:
the answer is table 2
Step-by-step explanation:
and you don't suck you just have to try harder
hope this helped
Answer:
=-3 Stable
=-1 Instable
=1 Semistable
=2.5 Instable
Step-by-step explanation:
Equilibrium constant solutions (or critical points) occur whenever x′ = f(x) = 0
x′=−(x+3)(x+1)3(x−1)2(x−2.5)=0
=-3
=-1
=1
=2.5
we can see the plot of x′ = f(x) in the figure annexed. In order to analyse the stability of the constant solutions, we must see how the function changes around the constant solutions X's:
(i.) If f(x) < 0 on the left of X, and f(y) > 0 on the right of
X, then the equilibrium solution is unstable.
(ii.) If f(X) > 0 on the left of X, and f(y) < 0 on the right of
X, then the equilibrium solution y = c is stable.
(iii.) If f(x) > 0 on both sides of X, or f(x) < 0 on both
sides of c, then the equilibrium solution y = X is
semistable.
Then:
=-3 Stable
=-1 Instable
=1 Semistable
=2.5 Instable
Answer:
I would say 3
Step-by-step explanation:
Most reasonable
Y=7
It's a horizontal line at height 7