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2 years ago

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The length of the service period expected from an asset for example in years or machine hours is the?

1 answer:

VARVARA [1.3K]2 years ago

8 0

Answer:

The length of the service period expected from an asset for example in years or machine hours is the?

Step-by-step explanation:

The length of the service period expected from an asset for example in years or machine hours is the?

You might be interested in

It is believed that as many as 23% of adults over 50 never graduated from high school. We wish to see if this percentage is the

JulijaS [17]

Answer:

1) n=48

2) n=298

3) n=426

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$p$ represent the real population proportion of interest

$\hat p$ represent the estimated proportion for the sample

n is the sample size required (variable of interest)

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

Part 1

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.90=0.10$ and $\alpha/2 =0.05$ . And the critical value would be given by:

$z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.1$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

We can assume that the estimated proportion is 0.23 for the 25 to 30 group. And replacing into equation (b) the values from part a we got:

$n=\frac{0.23(1-0.23)}{(\frac{0.1}{1.64})^2}=47.63$

And rounded up we have that n=48

Part 2

The margin of error on this case changes to 0.04 so if we use the same formula but changing the value for ME we got:

$n=\frac{0.23(1-0.23)}{(\frac{0.04}{1.64})^2}=297.7$

And rounded up we have that n=298

Part 3

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.04$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

We can assume that the estimated proportion is 0.23 for the 25 to 30 group. And replacing into equation (b) the values from part a we got:

$n=\frac{0.23(1-0.23)}{(\frac{0.04}{1.96})^2}=425.22$

And rounded up we have that n=426

3 0

2 years ago

Amad was curious if triangles \triangle ABC△ABCtriangle, A, B, C, and \triangle EDF△EDFtriangle, E, D, F were congruent. He was

11111nata11111 [884]

Answer:

There is no error, Amad is correct.

Step-by-step explanation:

Khan Academy Checked.

3 0

2 years ago

What is the balance on an amortized loan of $110,000 after the first payment if the interest rate is 5.5% with a monthly P&I

marishachu [46]

The interest due on the first payment is
.. I = Prt
.. I = 110,000*.055*(1/12)
.. I = 504.17

Then the decrease in principal resulting from the first payment is
.. 568.00 -504.17 = 63.83
and the new balance is
.. $110,000.00 -63.83 = $109,936.17

4 0

2 years ago

Tiesha enjoys reading in her spare time. She reads 4 pages every 1/10 of an hour.

dangina [55]

Answer:

p = 40h

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

The average sea level in 1900 at London bridge was 33 feet. In 1990 it was 33.08 feet. Use linear interpolation or extrapolation

eduard

Answer:

In 1956, the height will be 33.05 ft

Step-by-step explanation:

Let +x+ = number of years after 1900

Calll 1900 +x+=+0+

Then 1990 is +x+=+90+

Let +y+ = height in feet

--------------------------

You are gven the points

( 0, 33 )

( 90, 33.08 )

Use general point-slope formula

+%28+y+-+33+%29+%2F+%28+x+-+0+%29+=+%28+33.08+-+33+%29+%2F+%28+90+-+0+%29+

Multiply both sides by +90x+

+90%2A%28+y+-+33+%29+=+.08x+

+90y+-+2970+=+.08x+

+90y+=+.08x+%2B+2970+

+y+=+.000889x+%2B+33+ ( equation to use )

--------------------------

Check:

does it go through ( 90, 33.08 ) ?

+33.08+=+.000889%2A90+%2B+33+

+33.08+=+.08001+%2B+33+

+33.08+=+33.08001+

close enough

-------------------------------

(a)

+x+=+61%0D%0A%7B%7B%7B+y+=+.000889x+%2B+33+

+y+=+.000889%2A61+%2B+33+

+y+=+.0542+%2B+33+

+y+=+33.0542+

In 1961, the height will be 33.05 ft

8 0

2 years ago