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)
Given:
W(width) = (6L) - 9
L(length) = L
Equation:
2( [ 6L ] - 9) + 2 (L) = 150
= 12L - 18 + 2L = 150
= 12L + 2L = 150 + 18
=14L = 168
L = 168/14, so the length is 12. Let's check our work.
Width: 6(12) - 9 = 72 - 9 = 63
Length: 12
Since there are two lines of width and two lines of length:
2(12) + 2(63) = 24 + 126, which gives you a perimeter of 150 mm.
Hope this helped.
(2 × (6cm x 9cm)) + (2 × (6cm x 5.1cm)) + (2 × (9cm x 5.1cm)) = 261 square cm
9514 1404 393
Answer:
D. 9
Step-by-step explanation:
The segment lengths are equal, so ...
x^2 = 90-x
x^2 +x -90 = 0 . . . . . . add x-90 to put in standard form
(x +10)(x -9) = 0 . . . . . factor
The solutions are the values of x that make the factors zero: -10, 9.
The positive value of x is 9.
