Not sure which one....

In manufacturing a particular set of motor shafts, only shaft diameters of between 38.00 and 37.50 mm are usable. If the process

Masteriza [31]

Answer:

$P(37.5$

And we can find this probability with this difference:

$P(-2.5$

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

$P(-2.5$

So then the percentage usable are 94,6%

Explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the diameters of a population, and for this case we know the distribution for X is given by:

$X \sim N(37.8,0.12)$

Where $\mu=37.8$ and $\sigma=0.12$

We are interested on this probability

$P(37.5$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(37.5$

And we can find this probability with this difference:

$P(-2.5$

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

$P(-2.5$

So then the percentage usable are 94,6%

4 0

3 years ago

An aluminum part will be subjected to cyclic loading where the maximum stress will be 300 MPa and the minimum stress will be-100

Dominik [7]

Answer:

a) The mean stress experimented by the aluminium part is 100 megapascals, b) The stress amplitude of the aluminium part is 400 megapascals, c) The stress ratio of the aluminium part is 4.

Explanation:

a) The mean stress is determined by this expression:

$\sigma_{m} = \frac{\sigma_{min}+\sigma_{max}}{2}$

Where:

$\sigma_{m}$ - Mean stress, measured in megapascals.

$\sigma_{min}$ - Minimum stress, measured in megapascals.

$\sigma_{max}$ - Maximum stress, measured in megapascals.

If we know that $\sigma_{min} = -100\,MPa$ and $\sigma_{max} = 300\,MPa$ , the mean stress is:

$\sigma_{m} = \frac{-100\,MPa+300\,MPa}{2}$

$\sigma_{m} = 100\,MPa$

The mean stress experimented by the aluminium part is 100 megapascals.

b) The stress amplitude is given by the following difference:

$\sigma_{a} = |\sigma_{max}-\sigma_{min}|$

Where $\sigma_{a}$ is the stress amplitude, measured in megapascals.

If we know that $\sigma_{min} = -100\,MPa$ and $\sigma_{max} = 300\,MPa$ , the stress amplitude is:

$\sigma_{a} = |300\,MPa-(-100\,MPa)|$

$\sigma_{a} = 400\,MPa$

The stress amplitude of the aluminium part is 400 megapascals.

c) The stress ratio ( $R$ ) is the ratio of the stress amplitude to mean stress. That is:

$R = \frac{\sigma_{a}}{\sigma_{m}}$

If we know that $\sigma_{m} = 100\,MPa$ and $\sigma_{a} = 400\,MPa$ , the stress ratio is:

$R = \frac{400\,MPa}{100\,MPa}$

$R = 4$

The stress ratio of the aluminium part is 4.

3 0

3 years ago

For a Cu-Ni alloy containing 53 wt.% Ni and 47 wt.% Cu at 1300°C, calculate the wt.% of the alloy that is solid and wt.% of allo

Zina [86]

Answer:

Hello your question is incomplete attached below is the complete question

answer: wt.% of alloy that is solid = 61.5%

wt.% of allot that is liquid = 38.5%

Explanation:

To determine the wt.% of the alloy that is solid

= $\frac{R}{R +S } * 100$

= $\frac{53-45}{58-45} * 100$ = 61.5%

To determine the wt.% of the alloy that is liquid

= $\frac{S}{S+R} * 100\\$

= $\frac{58-53}{58-45} *100$ = 38.5%

attached below is a free hand sketch as well

7 0

3 years ago

9. A vehicle is having routine maintenance performed.

dexar [7]

Answer: b

Explanation:

8 0

3 years ago

Explain biometric senser.

Contact [7]

Biometric sensors are used to collect measurable biological characteristics from a human being, which can then be used in conjunction with biometric recognition algorithms to perform automated person identification.

8 0

3 years ago

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