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3 years ago

The average starting salary for this year's graduates at a large university (LU) is $20,000 with

Engineering

1 answer:

storchak [24]3 years ago

8 0

Answer:

(a) 0.0968 (b) the probability that a randomly selected LU graduate will have a salary of exactly $30,400 is 0.0000 (c) 29.12% (d) 3000 students graduates this year from this university

Explanation:

Solution

For the problem given,

The average salary starting for this year's graduates at a large university (LU) is = 20,000

So,

The mean μ = $ 20,000

Standard deviation is б = $ 8000

Note: kindly find the complete steps taken to get the solution to this questions attached below.

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WHICH TASK BEST FITS THE ROLE OF A DESIGN ENGINEER ?

pshichka [43]

Answer: drafting blueprints

Explanation:

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2 years ago

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bogdanovich [222]

Answer:

Hi there

Your answer is:

Explanation:

The "metal contraption", as the text goes on to say, is treated like a friend to Bonnie. Her mom comments on this by contrasting the metal contraption to a puppy.

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3 years ago

Read 2 more answers

A slab-milling operation is performed on a 0.7 m long, 30 mm-wide cast-iron block with a feed of 0.25 mm/tooth and depth of cut

denis23 [38]

Answer:

a) $T_m=1.787min$

b) $MRR=35259.7mm^3/min$

Explanation:

From the question we are told that:

Cast-iron block Dimension:

Length $l=0.7m=>700mm$

Width $w=30mm$

Feed $F=0.25mm/tooth$

Depth $dp=3mm$

Diameter $d=75mm$

Number of cutting teeth $n=8$

Rotation speed $N=200rpm$

Generally the equation for Approach is mathematically given by

$x=\sqrt{Dd-d^2}$

$X=\sqrt{75*3-3^2}$

$X=14.69mm$

Therefore

Effective length is given as

$L_e=Approach +object Length$

$L_e=700+14.69$

$L_e=714.69mm$

Generally the equation for Machine Time is mathematically given by

$T_m=\frac{L_e}{F_m}$

Where

$F_m=F*n*N$

$F_m=0.25*8*200$

$F_m=400$

Therefore

$T_m=\frac{714.69}{400}$

$T_m=1.787min$

Generally the equation for Material Removal Rate. is mathematically given by

$MRR=\frac{L*B*d}{t_m}$

$MRR=\frac{700*30*3}{1.787}$

$MRR=35259.7mm^3/min$

3 0

3 years ago

Heating of Oil by Air. A flow of 2200 lbm/h of hydrocarbon oil at 100°F enters a heat exchanger, where it is heated to 150°F by

irakobra [83]

Answer:

2062 lbm/h

Explanation:

The air will lose heat and the oil will gain heat.

These heats will be equal in magnitude.

qo = -qa

They will be of different signs because one is entering iits system and the other is exiting.

The heat exchanged by oil is:

qo = Gp * Cpo * (tof - toi)

The heat exchanged by air is:

qa = Ga * Cpa * (taf - tai)

The specific heat capacity of air at constant pressure is:

Cpa = 0.24 BTU/(lbm*F)

Therefore:

Gp * Cpo * (tof - toi) = Ga * Cpa * (taf - tai)

Ga = (Gp * Cpo * (tof - toi)) / (Cpa * (taf - tai))

Ga = (2200 * 0.45 * (150 - 100)) / (0.24 * (300 - 200)) = 2062 lbm/h

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3 years ago

A 1 m wide continuous footing is designed to support an axial column load of 250 kN per meter of wall length. The footing is pla

creativ13 [48]

Answer:

correct option is (A) 0.5

Explanation:

given data

axial column load = 250 kN per meter

footing placed = 0.5 m

cohesion = 25 kPa

internal friction angle = 5°

solution

we know angle of internal friction is 5° that is near to 0°

so it means the soil is almost cohesive soil.

and for a pure cohesive soil

$N_{\gamma }$ = 0

and we know formula for $N_{\gamma }$ is

$N_{\gamma }$ = (Nq - 1 ) × tan(Ф) ..................1

so here Ф is very less $N_{\gamma }$ should be nearest to zero

and its value can be 0.5

so correct option is (A) 0.5

7 0

3 years ago