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Answer:
6.5 × 10¹⁵/ cm³
Explanation:
Thinking process:
The relation
With the expression Ef - Ei = 0.36 × 1.6 × 10⁻¹⁹
and ni = 1.5 × 10¹⁰
Temperature, T = 300 K
K = 1.38 × 10⁻²³
This generates N₀ = 1.654 × 10¹⁶ per cube
Now, there are 10¹⁶ per cubic centimeter
Hence,
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:
See explaination
Explanation:
Lets first consider the term Isentropic efficiency. The isentropic efficiency of a compressor or pump is defined as the ratio of the work input to an isentropic process, to the work input to the actual process between the same inlet and exit pressures. IN practice, compressors are intentionally cooled to minimize the work input.
Please kindly check attachment for the step by step solution of the given problem.
