-4x²+16x-24=-1(4x²-16x+24)=-4(x²-4x+6)
x²-4x+6 has no real roots. ☺☺☺☺
Step-by-step explanation: Remember that the parent graph f(x) = x², is a parabola that opens up and passes through the origin.
Notice that g(x) = x² + 2 has 2 added to the parent term.
This means that g(x) = x² + 2 is the graph of f(x) = x² translated two units up.
Answer:
Question incomplete.Please tell the functions.
Period For Interest = 9 months
Period taken = 54 months
Investment = 2000
Quarters for payment = 54 ÷ 9
=6 periods
2000×2 = 4000
4000×2 = 8000
8000×2 = 16000
16000×2 = 32000
32000×2 = 64000
64000×2 = 128000
Answer = 128000
Answer:
Cpk = Cpu = Cpl = 0.4785
Step-by-step explanation:
Solution:-
- The capability of a process is determined using the capability index ( Cpk ). The index helps in determining the output of the process lies within the specification limits.
- The key dimension on a product measure is:
105 ± 12 units
- Process is producing this product at a 6-sigma quality standards i.e ( process is producing this product has a standard deviation (σ) of three units.)
Where, Standard Error in measured value = 12 units.
- The process capability (Cpu) is defined based on the Upper Specification Limit of the process (USL):
Cpu = (USL - u) / 3σ
- The process capability (Cpl) is defined based on the Lower Specification Limit of the process (LSL):
Cpl = ( u - LSL) / 3σ
- Assume that the process is centered with respect to specifications, then the given absolute value of u = 105 units.
Where the specification limits are defined as:
LSL = u - SE = 105 - 12
LSL = 93 units
USL = u + SE = 105 + 12
USL = 117 units
- Now evaluate the Process capability on each of the two limits specification limits:
Cpu = (USL - u) / 3σ
Cpu = (117 - 105) / 3σ
Cpu = 4 / σ
Cpl = ( u - LSL) / 3σ
Cpl = ( 105 - 99 ) / 3σ
Cpl = 4 / σ
- The minimum of Cpu and Cpl is denoted as the Cpk value:
Cpk = min ( Cpu , Cpl )
Cpk = Cpu = Cpl = 4 / 8.36
Cpk = Cpu = Cpl = 4 / 8.36 = 0.4785
