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Sliva [168]
3 years ago
7

What is the standard deviation of the following data set:

Engineering
1 answer:
EastWind [94]3 years ago
3 0

Answer:

See Explanation

Explanation:

The question is incomplete as the data set are missing. However, I'll use the following data to answer your question:

<em>5,12,3,18,6,8,2,10 </em>

Start by calculating the mean:

Mean = \frac{\sum x}{n}

Mean= \frac{5+12+3+18+6+8+2+10}{8}

Mean= \frac{64}{8}

Mean = 8

Standard deviation is calculated using:

SD = \sqrt{\frac{(x_i - Mean)^2}{n}}

This gives:

SD = \sqrt{\frac{(5 - 8)^2+(12 - 8)^2+(3 - 8)^2+(18 - 8)^2+(6 - 8)^2+(8 - 8)^2+(2 - 8)^2+(10 - 8)^2}{8}}

SD = \sqrt{\frac{(-3)^2+(4)^2+(- 5)^2+(10)^2+(-2)^2+(0)^2+( - 6)^2+(2)^2}{8}}

SD = \sqrt{\frac{9+16+25+100+4+0+36+4}{8}}

SD = \sqrt{\frac{194}{8}}

SD = \sqrt{24.25}

SD = 4.92

<em>Apply the above steps in the original question, then you will get your correct answer.</em>

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The solid cylinders AB and BC are bonded together at B and are attached to fixed supports at A and C. The modulus of rigidity is
romanna [79]

Answer:

a) 0.697*10³ lb.in

b) 6.352 ksi

Explanation:

a)

For cylinder AB:

Let Length of AB = 12 in

c=\frac{1}{2}d=\frac{1}{2} *1.1=0.55in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}0.55^4=0.1437\ in^4\\

\phi_B=\frac{T_{AB}L}{GJ}=\frac{T_{AB}*12}{3.3*10^6*0.1437}  =2.53*10^{-5}T_{AB}

For cylinder BC:

Let Length of BC = 18 in

c=\frac{1}{2}d=\frac{1}{2} *2.2=1.1in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}1.1^4=2.2998\ in^4\\

\phi_B=\frac{T_{BC}L}{GJ}=\frac{T_{BC}*18}{5.9*10^6*2.2998}  =1.3266*10^{-6}T_{BC}

2.53*10^{-5}T_{AB}=1.3266*10^{-6}T_{BC}\\T_{BC}=19.0717T_{AB}

T_{AB}+T_{BC}-T=0\\T_{AB}+T_{BC}=T\\T_{AB}+T_{BC}=14*10^3\ lb.in\\but\ T_{BC}=19.0717T_{AB}\\T_{AB}+19.0717T_{AB}=14*10^3\\20.0717T_{AB}=14*10^3\\T_{AB}=0.697*10^3\ lb.in\\T_{BC}=13.302*10^3\ lb.in

b) Maximum shear stress in BC

\tau_{BC}=\frac{T_{BC}}{J}c=13.302*10^3*1.1/2.2998=6.352\ ksi

Maximum shear stress in AB

\tau_{AB}=\frac{T_{AB}}{J}c=0.697*10^3*0.55/0.1437=2.667\ ksi

8 0
3 years ago
Vector A extends from the origin to a point having polar coordinates (7, 70ᵒ ) and vector B extends from the origin to a point h
yaroslaw [1]

Answer:

13.95

Explanation:

Given :

Vector A polar coordinates = ( 7, 70° )

Vector B polar coordinates = ( 4, 130° )

To find A . B we  will

A ( r , ∅ ) = ( 7, 70 )

A = rcos∅ + rsin∅

therefore ; A  = 2.394i + 6.57j

B ( r , ∅ ) = ( 4, 130° )

B = rcos∅ + rsin∅

therefore ;  B = -2.57i + 3.06j

Hence ; A .B

( 2.394 i + 6.57j ) . ( -2.57 + 3.06j ) = 13.95

8 0
3 years ago
A heat engine operates between 2 reservoirs at TH and 18oC. The heat engine receives 17,000 kJ/h from the high temperature reser
lisabon 2012 [21]

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem

4 0
3 years ago
For welding the most important reason to use jigs and fixtures in a welding shop is to
diamong [38]

Answer:

Reduce manufacturing costs.

Explanation:

Hope This Helps

Have A Great Day

7 0
2 years ago
Describe how you could engineer the situation to produce even more friction and heat
lana66690 [7]
True the use many abstract power
6 0
3 years ago
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