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Nadya [2.5K]
3 years ago
15

What are the important things to remember when arriving for an interview?

Engineering
1 answer:
ludmilkaskok [199]3 years ago
8 0

Answer: That you are dressed appropriately, to speak in a formal manner, and to be confident in your answers.

You might be interested in
Four of the minterms of the completely specified function f(a, b, c, d) are m0, m1, m4, and m5.
Sveta_85 [38]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a) The required additional minterms  for f so that f has eight primary implicants with two literals and no other prime implicant are m_{2},m_{3},m_{7},m_{8},m_{11},m_{12},m_{13},m_{14} and m_{15}

b) The essential prime implicant are c' d',a'b',ab and cd

c) The minimum sum-of-product expression for f are

                  a'b' +ab +c'd'+cd+a'c',\\ a'b'+ab+c'd'+cd+a'd,\\a'b'+ab+c'd'+cd+bc'  and  \\ a'b'+ab+c'd' +cd+bd

Explanation:

The explanation is shown on the second third and fourth image

8 0
3 years ago
Air (cp = 1.005 kJ/kg·°C) is to be preheated by hot exhaust gases in a cross-flow heat exchanger before it enters the furnace. A
uysha [10]

Answer:

Q=67.95 W

T=119.83°C

Explanation:

Given that

For air

Cp = 1.005 kJ/kg·°C

T= 20°C

V=0.6 m³/s

P= 95 KPa

We know that for air

P V = m' R T

95 x 0.6 = m x 0.287 x 293

m=0.677 kg/s

For gas

Cp = 1.10 kJ/kg·°C

m'=0.95 kg/s

Ti=160°C   ,To= 95°C

Heat loose by gas = Heat gain by air

[m Cp ΔT] for air =[m Cp ΔT] for gas

by putting the values

0.677 x 1.005 ( T - 20)= 0.95 x 1.1 x ( 160 -95 )

T=119.83°C

T is the exit temperature of the air.

Heat transfer

Q=[m Cp ΔT] for gas

Q=0.95 x 1.1 x ( 160 -95 )

Q=67.95 W

7 0
3 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
2 years ago
In Example 2-1, part c, the data were represented by the normal distribution function f(x)=0.178 exp(-0.100(x-451)2 Use this dis
valkas [14]

Answer:

P ( 2.5 < X < 7.5 ) = 0.7251

Explanation:

Given:

- The pmf for normal distribution for random variable x is given:

                                      f(x)=0.178 exp(-0.100(x-4.51)^2)

Find:

the fraction of individuals demonstrating a response in the range of 2.5 to 7.5.

Solution:

- The random variable X follows a normal distribution with mean u = 4.51, and standard deviation s.d as follows:

                               s.d = sqrt ( 1 / 0.1*2)

                               s.d = sqrt(5) =2.236067

- Hence, the normal distribution is as follows:

                               X ~ N(4.51 , 2.236)

- Compute the Z-score values of the end points 2.5 and 7.5:

              P ( (2.5 - 4.51) / 2.236 < Z < (7.5 - 4.51 ) / 2.236 )

              P ( -0.898899327 < Z < 1.337168651 )  

- Use the Z-Table for the probability required:

              P ( 2.5 < X < 7.5 ) = P ( -0.898899327 < Z < 1.337168651 ) = 0.7251            

6 0
3 years ago
(a) Design a lag compensation to meet the following specifications: The step response settling time is to be less than 5 sec, th
Pavlova-9 [17]

Answer:

Please see the attached file for the complete answer.

Explanation:

Download pdf
4 0
2 years ago
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