Answer:
F(x) = 0 ; x < 0
0.064 ; 0 ≤ x < 1
0.352 ; 1 ≤ x < 2
0.784 ; 2 ≤ x < 3
1 ; x ≥ 3
Explanation:
Each wafer is classified as pass or fail.
The wafers are independent.
Then, we can modelate X : ''Number of wafers that pass the test'' as a Binomial random variable.
X ~ Bi(n,p)
Where n = 3 and p = 0.6 is the success probability
The probatility function is given by :
Where 
is the combinatorial number
Let's calculate f(x) :
For the cumulative distribution function that we are looking for :
The cumulative distribution function for X is :
F(x) = 0 ; x < 0
0.064 ; 0 ≤ x < 1
0.352 ; 1 ≤ x < 2
0.784 ; 2 ≤ x < 3
1 ; x ≥ 3
Answer:
79 kW.
Explanation:
The equation for enthalpy is:
H2 = H1 + Q - L
Enthalpy is defined as:
H = G*(Cv*T + p*v)
This is specific volume.
The gas state equation is:
p*v = R*T (with specific volume)
The specific gas constant for air is:
287 K/(kg*K)
Then:
T1 = 60 + 273 = 333 K
T2 = 200 + 273 = 473 K
p1*v1 = 287 * 333 = 95.6 kJ/kg
p2*v2 = 287 * 473 = 135.7 kJ/kg
The Cv for air is:
Cv = 720 J/(kg*K)
So the enthalpies are:
H1 = 0.8*(0.72 * 333 + 95.6) = 268 kW
H2 = 0.8*(0.72 * 473 + 135.7) = 381 kW
Ang the heat is:
Q = 34 kW
Then:
H2 = H1 + Q - L
381 = 268 + 34 - L
L = 268 + 34 - 381 = -79 kW
This is the work from the point of view of the air, that's why it is negative.
From the point of view of the machine it is positive.
Answer:
False
Explanation:
According to the AISI (SAE) statements the last two numbers of the designation indicates the porcentage of carbon in the steel. And the firs numbers indicate the most important alloy alements.
In the case of the AISI 1045 indicates that the composition of this steel is 0,45% of carbon.
