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konstantin123 [22]
3 years ago
6

Wasserman (1989) studied a process for the manufacturing of steel bolts. Historically, these bolts have a mean thickness of 10.0

mm and a standard deviation of 1.6 mm. In a quality check, the engineer has a sample of 5 randomly selected and measured.
Assuming a near-normal distribution, what are the mean and standard deviation of the sample mean of these quality checks?
Mathematics
1 answer:
zalisa [80]3 years ago
7 0

Answer:

The mean of of the sample mean of these quality checks is 10 and the standard deviation is 0.7155.

Step-by-step explanation:

To solve this question, we use the central limit theorem.

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 10, \sigma = 1.6, n = 5, s = \frac{1.6}{\sqrt{5}} = 0.7155

The mean of of the sample mean of these quality checks is 10 and the standard deviation is 0.7155.

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hope this helps
3 0
3 years ago
Write the decimal in fractional notation: 0.097 A 97/1000 B 97/10,000 C 970 /1000 D 97/10
jeyben [28]
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3 years ago
Law of cosines.Anyone good in Trigonometry?​
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Answer:

c=11.8\ units

Step-by-step explanation:

we know that

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C=120^o

substitute the given values

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2 years ago
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Answer:

Step-by-step explanation:

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The equation of motion is f(x) = -16x^2 + 42x + 12.  We need to determine the maximum of this function.  To do this, find the x-coordinate of the vertex, which is x = -b/(2a), or x = -42/(2*-16), or 1.31 sec.

Evaluating f(x) = -16x^2 + 42x + 12 at x = 1.31 sec, we get f(1.31) = 39.6.

So it appears that f(x) has a higher max than does g(x); the difference is approx. 39.6 - 33, or 6.6

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Answer:

x=17

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3 years ago
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