Ask question

Ask question

All categories

4 years ago

11

A Sampling of 409 motor Shafts from a very large production Batch shows a sample standard deviation in Diameter of 0.021 mm with

a sample mean of 9.251 mm Estimate the mean diameter of the entire batch at a %95 level of confidences?

1 answer:

andrezito [222]4 years ago

7 0

Answer:

9.248 < $\mu$ < 9.253 mm

Explanation:

Given data:

standard deviation = 0.021 mm

sample mean = 9.251 mm

total sample = 409 m

confidence interval level = 95%

mean diameter of entire batch $\mu = \bar X \pm z \frac{\sigma}{\sqrt{n}}$

where

$\bar X$ - sample mean = 9.251 mm

z = critical value

plugginf all value in the above relation to get the mean diameter

$\mu = 9.251 \pm 1.96 \times \frac{0.021}{\sqrt{409}}$

9.248 < $\mu$ < 9.253 mm

You might be interested in

What is the first step necessary for initiating the visual transduction cascade in rods?.

irinina [24]

'Capturing a photon by<em> isomerization and rhodopsin </em>of retinal' is the first step necessary for initiating the visual transduction cascade in rods.

Visual transduction refers to the process in the eye where absorption of light in the 'retina' is translated into electrical signals that then reach the brain. It is correct to state that visual transduction is the photochemical reaction that takes place when light or photon is converted to an electric signal in the retina. The visual pigment in the rods, called rhodopsin, is a membrane protein placed in the outer segments of the rods.

When initiating the visual transduction cascade in rods the first vital step is to capture a photon by<em> isomerization and rhodopsin</em> of retinal'.

You can leran more about visual transduction at

brainly.com/question/13798113

#SPJ4

4 0

1 year ago

Drag the tiles to the correct boxes to complete the pairs.

Jobisdone [24]

Answer:

Check image, right on PLATO

3 0

3 years ago

Water flows through a horizontal 60 mm diameter galvanized iron pipe at a rate of 0.02 m3/s. If the pressure drop is 135 kPa per

maksim [4K]

Answer:

pipe is old one with increased roughness

Explanation:

discharge is given as

$V =\frac{Q}{A} = \frac{ 0.02}{\pi \4 \times (60\times 10^{-3})^2}$

V = 7.07 m/s

from bernou;ii's theorem we have

$\frac{p_1}{\gamma} +\frac{V_1^2}{2g} + z_1 = \frac{p_2}{\gamma} +\frac{V_2^2}{2g} + z_2 + h_l$

as we know pipe is horizontal and with constant velocity so we have

$\frac{P_1}{\gamma } + \frac{P_2 {\gamma } + \frac{flv^2}{2gD}$

$P_1 -P_2 = \frac{flv^2}{2gD} \times \gamma$

$135 \times 10^3 = \frac{f \times 10\times 7.07^2}{2\times 9.81 \times 60 \times 10^{-5}} \times 1000 \times 9.81$

solving for friction factor f

f = 0.0324

fro galvanized iron pipe we have $\epsilon = 0.15 mm$

$\frac{\epsilon}{d} = \frac{0.15}{60} = 0.0025$

reynold number is

$Re =\frac{Vd}{\nu} = \frac{7.07 \times 60\times 10^{-3}}{1.12\times 10^{-6}}$

Re = 378750

from moody chart

$For Re = 378750 and \frac{\epsilon}{d} = 0.0025$

$f_{new} = 0.025$

therefore new friction factor is less than old friction factoer hence pipe is not new one

now for Re = 378750 and f = 0.0324

from moody chart

we have $\frac{\epsilon}{d} =0.006$

$\epsilon = 0.006 \times 60$

$\epsilon = 0.36 mm$

thus pipe is old one with increased roughness

5 0

3 years ago

Nec ________ covers selection of time-delay fuses for motor- overload protection.

Murljashka [212]

Nec Article 430 covers selection of time-delay fuses for motor- overload protection.

<h3>What article in the NEC covers motor overloads?</h3>

Article 430 that is found in National Electrical Code (NEC) is known to be state as “Motors, Motor Circuits and Controllers.” .

Note that the article tells that it covers areas such as motors, motor branch-circuit as well as feeder conductors, motor branch-circuit and others.

Therefore, Nec Article 430 covers selection of time-delay fuses for motor- overload protection.

Learn more about motor- overload from

brainly.com/question/20738481

#SPJ1

6 0

1 year 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