Answer:
Answer for the question:
"Show that for a linearly separable dataset, the maximum likelihood solution for the logisitic regression model is obtained by finding a weight vector w whose decision boundary wx.
"
is explained in the attachment.
Explanation:
Answer:
80.7lbft/hr
Explanation:
Flow rate of water in the system = 3.6x10^-6
The height h = 100
1s = 1/3600h
This implies that
Q = 3.6x10^-6/[1/3600]
Q = 0.0000036/0.000278
Q = 0.01295
Then the power is given as
P = rQh
The specific weight of water = 62.3 lb/ft³
P = 62.3 x 0.01295 x 100
P = 80.675lbft/h
When approximated
P = 80.7 lbft/h
This is the average power that could be generated in a year.
This answers the question and also corresponds with the answer in the question.
Answer:
the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi
Explanation:
Given that;
depth 1 = 71 ft
depth 2 = 10 ft
pressure p = 17 psi = 2448 lb/ft²
depth h = 71 ft - 10 ft = 61 ft
we know that;
p = P_air + yh
where y is the specific weight of ethyl alcohol ( 49.3 lb/ft³ )
so we substitute;
p = 2448 + ( 49.3 × 61 )
= 2448 + 3007.3
= 5455.3 lb/ft³
= 37.88 psi
Therefore, the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi
Answer:
The theoretical maximum specific gravity at 6.5% binder content is 2.44.
Explanation:
Given the specific gravity at 5.0 % binder content 2.495
Therefore
95 % mix + 5 % binder gives S.G. = 2.495
Where the binder is S.G. = 1, Therefore
Per 100 mass unit we have (Mx + 5)/(Vx + 5) = 2.495
(95 +5)/(Vx +5) = 2.495
2.495 × (Vx + 5) = 100
Vx =35.08 to 95
Or density of mix = Mx/Vx = 95/35.08 = 2.7081
Therefore when we have 6.5 % binder content, we get
Per 100 mass unit
93.5 Mass unit of Mx has a volume of
Mass/Density = 93.5/2.7081 = 34.526 volume units
Therefore we have
At 6.5 % binder content.
(100 mass unit)/(34.526 + 6.5) = 2.44
The theoretical maximum specific gravity at 6.5% binder content = 2.44.
Answer:
Thermal resistance for a wall depends on the material, the thickness of the wall and the cross-section area.
Explanation:
Current flow and heat flow are very similar when we are talking about 1-dimensional energy transfer. Attached you can see a picture we can use to describe the heat flow between the ends of the wall. First of all, a temperature difference is required to flow heat from one side to the other, just like voltage is required for current flow. You can also see that 
represents the thermal resistance. The next image explains more about the parameters which define the value of the thermal resistances which are the following:
- Wall Thickness. More thickness, more thermal resistance.
- Material thermal conductivity (unique value for each material). More conductivity, less thermal resistance.
- Cross-section Area. More cross-section area, less thermal resistance.
A expression to define the thermal resistance for the wall is as follows: 
, where l is the distance between the tow sides of the wall, that is to say the wall thickness; A is the cross-section area and k is the material conducitivity.
