A basketball is filled with atmospheric air. What percent of air mass of the same temperature must be added to the ball to incre
1 answer:
You might be interested in
Answer:
a. 0.11
b. 110 students
c. 50 students
d. 0.46
e. 460 students
f. 540 students
g. 0.96
Explanation:
(See attachment below)
a. Probability that a student got an A
To get an A, the student needs to get 9 or 10 questions right.
That means we want P(X≥9);
P(X>9) = P(9)+P(10)
= 0.06+0.05=0.11
b. How many students got an A on the quiz
Total students = 1000
Probability of getting A = 0.11 ---- Calculated from (a)
Number of students = 0.11 * 1000
Number of students = 110 students
So,the number of students that got A is 110
c. How many students did not miss a single question
For a student not to miss a single question, then that student scores a total of 10 out of possible 10
P(10) = 0.05
Total Students = 1000
Number of Students = 0.05 * 1000
Number of Students = 50 students
We see that 5
d. Probability that a student pass the quiz
To pass, a student needed to get at least 6 questions right.
So we want P(X>=6);
P(X>=) =P(6)+P(7)+P(8)+P(9)+P(10)
=0.08+0.12+0.15+0.06+0.05=0.46
So, the probability of a student passing the quiz is 0.46
e. Number of students that pass the quiz
Total students = 1000
Probability of passing the quiz = 0.46 ----- Calculated from (d)
Number of students = 0.46 * 1000
Number of students = 460 students
So,the number of students that passed the test is 460
f. Number of students that failed the quiz
Total students = 1000
Total students that passed = 460 ----- Calculated from (e)
Number of students that failed = 1000 - 460
Number of students that failed = 540
So,the number of students that failed is 540
g. Probability that a student got at least one question right
This means that we want to solve for P(X>=1)
Using the complement rule,
P(X>=1) = 1 - P(X<1)
P(X>=1) = 1 - P(X=0)
P(X>=1) = 1 - 0.04
P(X>=1) = 0.96
Answer:
2285kw
Explanation:
since it is an isentropic process, we can conclude that it is a reversible adiabatic process. Hence the energy must be conserve i.e the total inflow of energy must be equal to the total outflow of energy.
Mathematically,
Note: from the question we have only one source of inflow and two source of outflow (the exhaust at a pressure of 50kpa and the feedwater at a pressure of 5ookpa). Also the power produce is another source of outgoing energy .
Where are the mass flow rate and the enthalpies at the inlet at a pressure of 3Mpa
,
are the mass flow rate and the enthalpies at the outlet 2 where we have a pressure of 500kpa respectively.
,
and are the mass flow rate and the enthalpies at the outlet 3 where we have a pressure of 50kpa respectively.
,
We can now express write out the required equation by substituting the new expression for the energies
from the above equation, the unknown are the enthalpy values and the mass flow rate.
first let us determine the enthalpy values at the inlet and the out let using the Superheated water table.
It is more convenient to start from outlet 3 were we have a temperature and pressure value of (50kpa or 0.05Mpa ). using double interpolation method on the superheated water table to determine the enthalpy value with careful calculation we have
= 2682.4 KJ/KG , at this point also from the table the entropy value ,
value is 7.6953 KJ/Kg.K.
Next we determine the enthalphy value at outlet 2. But in this case, we don't have a temperature value, hence we use the entrophy value since the entropy is constant at all inlet and outlet.
So, from the superheated water table again, at a pressure of 500kpa (0.5Mpa) and entropy value of 7.6953 KJ/Kg.K with careful interpolation we arrive at a enthalpy value of 3206.5KJ/Kg.
Finally for inlet one at a pressure of 3Mpa, interpolting with an entropy value of 7.6953KJ/Kg.K we arrive at enthalpy value of 3851.2KJ/Kg.
Now we determine the mass flow rate at each inlet and outlet. since mass must also be balance, i.e
From the question the, the mass flow rate at the inlet is 2Kg/s
Since 5% flow is delivered into the feedwater heating,
Also for the outlet 3 the remaining 95% will flow out. Hence
Now, from
we substitute values
.
Hence the power produced is 2285kW