Answer:
the equation which can be simplified as
![\frac{1}{n {}^{18} }](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7Bn%20%7B%7D%5E%7B18%7D%20%7D%20)
is :-
![(n {}^{ - 3} ) {}^{6}](https://tex.z-dn.net/?f=%28n%20%7B%7D%5E%7B%20-%203%7D%20%29%20%7B%7D%5E%7B6%7D%20)
Step-by-step explanation:
here's the solution :-
=》
![\frac{1}{n {}^{18} }](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7Bn%20%7B%7D%5E%7B18%7D%20%7D%20)
=》
![n {}^{ - 18}](https://tex.z-dn.net/?f=n%20%7B%7D%5E%7B%20-%2018%7D%20)
=》
![(n {}^{ - 3}) {}^{6}](https://tex.z-dn.net/?f=%28n%20%7B%7D%5E%7B%20-%203%7D%29%20%20%7B%7D%5E%7B6%7D%20)
Answer:
4 students sing in the class.
Step-by-step explanation:
20×.2=4
The answer would be 6, just like the person who said rn
Answer:
The probability of a student receiving a C in the course is p=0.3.
Step-by-step explanation:
We have a absolute frequency for each of the grades (A to F), of a total of 300 course tests.
It is assumed that this sample gives a dood estimation of the distribution of the grades. Then, we can estimate the probability of obtaining a C in the course usign the relative frequency for C.
The relative frequency is calculated as the division between the absolute frequency (in this case, 90 for a C grade) and the size of the sample (in this example, 300).
![p_C=\dfrac{X_C}{N}=\dfrac{90}{300}=0.3](https://tex.z-dn.net/?f=p_C%3D%5Cdfrac%7BX_C%7D%7BN%7D%3D%5Cdfrac%7B90%7D%7B300%7D%3D0.3)