Match the polynomial expression on the left with the simplified version on the right.

Mathematics

1 answer:

Vitek1552 [10]1 year ago

3 0

Given the following question:

First expression:

$\begin{gathered} \frac{12x^3-14x^2+16x-8}{3x-2} \\ \text{ Factor the expression:} \\ 12x^3-14x^2+16x-8=2(6x^3-7x^2+8x-4) \\ \frac{2\left(6x^3-7x^2+8x-4\right)}{3x-2} \\ \text{ Factor:} \\ 2\left(6x^3-7x^2+8x-4\right)=(3x-2)(2x^2-x+2) \\ =(3x-2)(2x^2-x+2)=2\left(3x-2\right)\left(2x^2-x+2\right) \\ \frac{2\left(3x-2\right)\left(2x^2-x+2\right)}{3x-2} \\ \text{ Cancel the common factor:} \\ -(3x-2) \\ 4x^2-2x+4 \end{gathered}$

Second expression:

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A design engineer wants to construct a sample mean chart for controlling the service life of a halogen headlamp his company prod

tekilochka [14]

Answer:

C) 515 hours.

D) 500 hours

c) sample 3

Step-by-step explanation:

1. Sample 2 mean = x2`= ∑x2/n2= 2060/4= 515 hours

Sample Service Life (hours)

1 2 3

495 525 470

500 515 480

505 505 460

500 515 470

∑2000 2060 1880

x1`= ∑x1/n1= 2000/4= 500 hours

x2`= ∑x2/n2= 2060/4= 515 hours

x3`= ∑x3/n3= 1880/4= 470 hours

2. The mean of the sampling distribution of sample means for whenever service life is in control is 500 hours . It is the given mean in the question and the limits are determined by using μ ± σ , μ±2 σ or μ ± 3 σ.

In this question the limits are determined by using μ ± σ .

3. Upper control limit = UCL = 520 hours

Lower Control Limit= LCL = 480 Hours

Sample 1 mean = x1`= ∑x1/n1= 2000/4= 500 hours

Sample 2 mean = x2`= ∑x2/n2= 2060/4= 515 hours

Sample 3 mean = x3`= ∑x3/n3= 1880/4= 470 hours

This means that the sample mean must lie within the range 480-520 hours but sample 3 has a mean of 470 which is out of the given limit.

We see that the sample 3 mean is lower than the LCL. The other two means are within the given UCL and LCL.

This can be shown by the diagram.