Answer:
The statement can be written as
int result = cube(4);
Explanation:
A function is a block of reusable codes to perform some tasks. For example, the function in the question is to calculate the cube of a number.
A function can also operate on one or more input value (argument) and return a result. The <em>cube </em>function in the question accept one input value through its parameter <em>number </em>and the <em>number</em> will be multiplied by itself twice and return the result.
To call a function, just simply write the function name followed with parenthesis (e.g. <em>cube()</em>). Within the parenthesis, we can include zero or one or more than one values as argument(s) (e.g. <em>cube(4)</em>).
We can then use the "=" operator to assign the return output of the function to a variable (e.g. <em>int result = cube(4)</em>)
Answer:
The power developed in HP is 2702.7hp
Explanation:
Given details.
P1 = 150 lbf/in^2,
T1 = 1400°R
P2 = 14.8 lbf/in^2,
T2 = 700°R
Mass flow rate m1 = m2 = m = 11 lb/s Q = -65000 Btu/h
Using air table to obtain the values for h1 and h2 at T1 and T2
h1 at T1 = 1400°R = 342.9 Btu/h
h2 at T2 = 700°R = 167.6 Btu/h
Using;
Q - W + m(h1) - m(h2) = 0
W = Q - m (h2 -h1)
W = (-65000 Btu/h ) - 11 lb/s (167.6 - 342.9) Btu/h
W = (-65000 Btu/h ) - (-1928.3) Btu/s
W = (-65000 Btu/h ) * {1hr/(60*60)s} - (-1928.3) Btu/s
W = -18.06Btu/s + 1928.3 Btu/s
W = 1910.24Btu/s
Note; Btu/s = 1.4148532hp
W = 2702.7hp
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:
The heat input from the combustion phase is 2000 watts.
Explanation:
The energy efficiency of the heat engine (
), no unit, is defined by this formula:
(1)
Where:
- Heat input, in watts.
- Power output, in watts.
If we know that 
and 
, then the heat input from the combustion phase is:
The heat input from the combustion phase is 2000 watts.
Answer:
W= 8120 KJ
Explanation:
Given that
Process is isothermal ,it means that temperature of the gas will remain constant.
T₁=T₂ = 400 K
The change in the entropy given ΔS = 20.3 KJ/K
Lets take heat transfer is Q ,then entropy change can be written as
Now by putting the values
Q= 20.3 x 400 KJ
Q= 8120 KJ
The heat transfer ,Q= 8120 KJ
From first law of thermodynamics
Q = ΔU + W
ΔU =Change in the internal energy ,W=Work
Q=Heat transfer
For ideal gas ΔU = m Cv ΔT]
At constant temperature process ,ΔT= 0
That is why ΔU = 0
Q = ΔU + W
Q = 0+ W
Q=W= 8120 KJ
Work ,W= 8120 KJ
