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Alexeev081 [22]
3 years ago
14

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

uces. He knows from numerous previous samples that this service life is normally distributed with a mean of 500 hours and a standard deviation of 20 hours. On three recent production batches, he tested service life on random samples of four headlamps, with these results: Sample Service Life (hours) 1 495 500 505 500 2 525 515 505 515 3 470 480 460 470 What is the standard deviation of the sampling distribution of sample means for whenever service life is in control? Multiple Choice 5 hours 6.67 hours 10 hours 11.55 hours 20 hours
Mathematics
1 answer:
olchik [2.2K]3 years ago
3 0

Answer:

The standard deviation of the sampling distribution of sample means for whenever service life is in control 10 hours.

Step-by-step explanation:

The standard error (\sigma_{M}) of the mean is the standard deviation of the sampling distribution of the mean.

The formula to compute the standard error is:

\sigma_{M}=\frac{\sigma}{\sqrt{n}}

The information provided is:

\sigma = 20\ \text{hours}\\n=4

Compute the standard deviation of the sampling distribution of sample means for whenever service life is in control as follows:

\sigma_{M}=\frac{\sigma}{\sqrt{n}}

      =\frac{20}{\sqrt{4}}\\\\=\frac{20}{2}\\\\=10\ \text{hours}

Thus, the standard deviation of the sampling distribution of sample means for whenever service life is in control 10 hours.

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3 years ago
List ALL of the common factors of 25 and 10
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Step-by-step explanation:

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3 years ago
Xavier worked 10 hours on Monday and 15 hours on Wednesday. His total pay was $280.00. What is his rate per hour?
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Number of hours Xavier worked on Monday = 10 hours
Number of hours Xavier worked on Tuesday = 15 hours
Total pay received by Xavier = $280
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4 years ago
A person on tour has dollar 360 for his daily expenses. If he extends his tour for 4 days, he has to cut down his daily expenses
mash [69]

Answer:

The original duration of the tour = 20 days

Step-by-step explanation:

Solution:

Total expenses for the tour = $360

Let the original tour duration be for x days.

So, for x days the total expense = $360

<em>Thus the daily expense in dollars can be given by</em> = \frac{360}{x}

Tour extension and effect on daily expenses.

The tour is extended by 4 days.

<em>Tour duration now</em> = (x+4) days

On extension, his daily expense is cut by $3

<em>New daily expense in dollars </em>= (\frac{360}{x}-3)

Total expense in dollars can now be given as:  (x+4)(\frac{360}{x}-3)

Simplifying by using distribution (FOIL).

(x.\frac{360}{x})+(x(-3)+(4.\frac{360}{x})+(4(-3))

360-3x+\frac{1440}{x}-12

348-3x+\frac{1440}{x}

We know total expense remains the same which is = $360.

So, we have the equation as:

348-3x+\frac{1440}{x}=360

Multiplying each term with x to remove fractions.

348x-3x^2+1440=360x

Subtracting 348x both sides

348x-348x-3x^2+1440=360x-348x

-3x^2+1440=12x

Dividing each term with -3.

\frac{-3x^2}{-3}+\frac{1440}{-3}=\frac{12x}{-3}

x^2-480=-4x

Adding 4x both sides.

x^2+4x-480=-4x+4x

x^2+4x-480=0

Solving using quadratic formula.

For a quadratic equation: ax^2+bx+c=0

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

Plugging in values from the equation we got.

x=\frac{-4\pm\sqrt{(4)^2-4(1)(-480)}}{2(1)}

x=\frac{-4\pm\sqrt{16+1920}}{2}

x=\frac{-4\pm\sqrt{1936}}{2}

x=\frac{-4\pm44}{2}

So, we have

x=\frac{-4+44}{2}   and x=\frac{-4-44}{2}

x=\frac{40}{2}   and x=\frac{-48}{2}

∴ x=20           and x=-24

Since number of days cannot be negative, so we take x=20 as the solution for the equation.

Thus, the original duration of the tour = 20 days

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3 years ago
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