125.638 rounded to the nearest hundredth equal 125.64
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:
4x + 10
Step-by-step explanation:
The answer is 0.9902 please give 5 stars if helpful
Answer:
all real numbers
Step-by-step explanation:
The graphs of positive and negative x^2 parabolas will always have a domain of all real numbers. Even though you only have a portion of the graph and see a "restriction" on your domain values, it is incorrect to assume that the domain is limited to what you can see. As the branches of the parabola keep going up and up and up, the values of x keep getting bigger and bigger and bigger. Again, this is true for all + or - parabolas. ....... Can consider brainlist*:)
