Marcus should have substituted -10for b, not 10
Marcus should have subtracted 4(1)(25) in the square root
This equation, when solved correctly, only has one real number solution
So A, C, D
Answer:
5x-2
Step-by-step explanation:
Since f(x)-g(x) you take 3x+2 and subtract -2x-4.
3x+2+2x-4
5x-2
Answer:
see picture with work shown
Answer:
B. (1,5) and (5.25, 3.94)
Step-by-step explanation:
The answer is where the 2 equations intersect.
We need to solve the following system of equations:
y = -x^2 + 6x
4y = 21 - x
From the second equation:
x = 21 - 4y
Plug this into the first equation:
y = -(21 - 4y)^2 + 6(21 - 4y)
y = -(441 - 168y + 16y^2)+ 126 - 24y
y = -441 + 168y - 16y^2 + 126 - 24y
16y^2 + y - 168y + 24y + 441 - 126 = 0
16y^2 - 143y + 315 = 0
y = [-(-143) +/- sqrt ((-143)^2 - 4 * 16 * 315)]/ (2*16)
y = 5, 3.938
When y = 5:
x = 21 - 4(5) = 1
When y = 3.938
x = 21 - 4(3.938) = 5.25.
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)
