Answer:
Follows are the solution to this question:
Step-by-step explanation:
In the given question some of the data is missing so, its correct question is defined in the attached file please find it.
Let
A is quality score of A
B is quality score of B
C is quality score of C
![\to P[A] =0.55\\\\\to P[B] =0.28\\\\\to P[C] =0.17\\](https://tex.z-dn.net/?f=%5Cto%20P%5BA%5D%20%3D0.55%5C%5C%5C%5C%5Cto%20P%5BB%5D%20%3D0.28%5C%5C%5C%5C%5Cto%20P%5BC%5D%20%3D0.17%5C%5C)
Let F is a value of the content so, the value is:
![\to P[\frac{F}{A}] =0.15\\\\\to P[\frac{F}{B}] =0.12\\\\\to P[\frac{F}{C}] =0.14\\](https://tex.z-dn.net/?f=%5Cto%20P%5B%5Cfrac%7BF%7D%7BA%7D%5D%20%3D0.15%5C%5C%5C%5C%5Cto%20P%5B%5Cfrac%7BF%7D%7BB%7D%5D%20%3D0.12%5C%5C%5C%5C%5Cto%20P%5B%5Cfrac%7BF%7D%7BC%7D%5D%20%3D0.14%5C%5C)
Now, we calculate the tooling value:
![\to p[\frac{C}{F}]](https://tex.z-dn.net/?f=%5Cto%20p%5B%5Cfrac%7BC%7D%7BF%7D%5D)
using the baues therom:
![\to p[\frac{C}{F}] = \frac{p[C] \times p[\frac{F}{C} ]}{p[A] \times p[\frac{F}{A}] + p[B] \times p[\frac{F}{B}]+p[C] \times p[\frac{F}{C}] }](https://tex.z-dn.net/?f=%5Cto%20p%5B%5Cfrac%7BC%7D%7BF%7D%5D%20%3D%20%20%5Cfrac%7Bp%5BC%5D%20%5Ctimes%20p%5B%5Cfrac%7BF%7D%7BC%7D%20%5D%7D%7Bp%5BA%5D%20%5Ctimes%20p%5B%5Cfrac%7BF%7D%7BA%7D%5D%20%2B%20p%5BB%5D%20%5Ctimes%20p%5B%5Cfrac%7BF%7D%7BB%7D%5D%2Bp%5BC%5D%20%5Ctimes%20p%5B%5Cfrac%7BF%7D%7BC%7D%5D%20%7D)
Answer:
i think it is 6 but not sure
Step-by-step explanation:
12•1
13•6
12/78 divide both numbers by 6 to simplify
Answer- 2/13
Answer:
4(n-16)
Step-by-step explanation:
4 Four
( ) Times
n number
16 16
- difference
Answer:
x + 3
Step-by-step explanation:
A 'sum' means addition, 'a number' is represented by a variable, in this case, I chose 'x'. I could have picked any variable I wanted.
The sum of three and a number means: three plus some number, which is
3 + x in math language. We write variables before constants, so it gets
rewritten as x + 3
