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sergeinik [125]
2 years ago
14

Which is true regarding the system of equations? x minus 4 y = 1. 5 x minus 20 y = 4.

Mathematics
2 answers:
enot [183]2 years ago
7 0

Step-by-step explanation:

x - 4y = 1.5 \:  \:  \:  \: ...(1) \\ x - 20y = 4 \:  \:  \:  \: ...(2)

from equation (2) we get,

x = 20y + 4

now put the value of x = 20y + 4 in equation (1)

20y + 4 - 4y = 1.5 \\ 16y = 1.5 - 4 \\ 16y =  - 2.5 \\ y =  -  \frac{25}{16 \times 10}   \\ \\ y =  -  \frac{5}{32}

now put the value of y in equation (2) we get

x - 20( -  \frac{5}{32} ) = 4 \\  \\ x +  \frac{100}{32}  = 4 \\  \\ x +  \frac{25}{8}  = 4 \\  \\ x = 4 -  \frac{25}{8}  \\  \\ x =  \frac{32 - 25}{8}  \\  \\ x =  \frac{7}{8}

goldfiish [28.3K]2 years ago
3 0

Answer:

No solution

Step-by-step explanation:

x - 4y = 1 --> (1)

5x - 20y = 4 --> (2)

y = (¼)x - ¼ --> (1)

y = (¼)x - ⅕ --> (2)

Since these are 2 parallel lines with different y-intercepts, they will never meet

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40+5+0.6+0.07

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Step-by-step explanation:

6 0
2 years ago
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

<u>500         515             470        </u>

<u>∑2000     2060         1880</u>

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.

8 0
2 years ago
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