(a)
![f'(x) = \frac{d}{dx}[\frac{lnx}{x}]](https://tex.z-dn.net/?f=f%27%28x%29%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5B%5Cfrac%7Blnx%7D%7Bx%7D%5D)
Using the quotient rule:
For maximum, f'(x) = 0;
(b) <em>Deduce:
</em>
<em>
Soln:</em> Since x = e is the greatest value, then f(e) ≥ f(x) > f(0)
, since ln(e) is simply equal to 1
Now, since x > 0, then we don't have to worry about flipping the signs when multiplying by x.
Taking the exponential to both sides will cancel with the natural logarithmic function in the right hand side to produce:
, as required.
Answer:
Kindly check explanation
Step-by-step explanation:
Given the data :
Technician __Shutdown
Taylor, T___4
Rousche, R _ 3
Hurley, H__ 3
Huang, Hu___2
Gupta, ___ 5
The Numbe of samples of 2 possible from the 5 technicians :
We use combination :
nCr = n! ÷ (n-r)!r!
5C2 = 5!(3!)2!
5C2 = (5*4)/2 = 10
POSSIBLE COMBINATIONS :
TR, TH, THu, TG, RH, RHu, RG , HHu, HG, HuG
Sample means :
TR = (4+3)/2 = 3.5
TH = (4+3)/2 = 3.5
THu = (4+2) = 6/2 = 3
TG = (4 + 5) = 9/2 = 4.5
RH = (3+3) = 6/2 = 3
RHu = (3+2) /2 = 2.5
RG = (3 + 5) = 8/2 = 4
HHu = (3+2) = 2.5
HG = (3+5) = 8/2 = 4
HuG = (2+5) / 2 = 3.5
Mean of sample mean (3.5+3.5+3+4.5+3+2.5+4+2.5+4+3.5) / 10 = 3.4
Population mean :
(4 + 3 + 3 + 2 + 5) / 5 = 17 /5 = 3.4
Population Mean and mean of sample means are the same.
This distribution should be approximately normal.
If two events are independent, then P(A and B) = P(A) x P(B).
In your situation, you need to solve 0.185 = 0.25 x P(B).
Can you take it from there?
The perimeter is 16 inches.
