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2 years ago

Pls help will give brainly (again)

Mathematics

1 answer:

Elan Coil [88]2 years ago

7 0

Answer:

Step-by-step explanation:

First, simplify the exponents:

2*(3^6*3*-5)

2*((3*3*3*3*3*3)*(1/3*3*3*3*3))

2*(729*0.00411522633)

Then solve from there:

2*2.99999999457

5.99999998914

Since there is only whole numbers and a fraction present, im assuming you have to round:

5.99999998914

6 is your answer

Hope this helps!

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Answer:

C) 515 hours.

D) 500 hours

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Step-by-step explanation:

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

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x1`= ∑x1/n1= 2000/4= 500 hours

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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

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

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

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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.

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Step-by-step explanation:

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