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1 year ago

A) Your friend is faced with a situation i n whi ch t here is a difficulty in

Engineering

1 answer:

Arlecino [84]1 year ago

7 0

The Steps my friend need to make an ethical decision are:

Step One: He or she needs to define the Problem.
Step Two: He or she needs to find out possible Resources to solve the problem and then Brainstorm on the List for the Potential Solutions to the problem.
Step Three : Evaluate and examine those Alternatives. ...
Step Four : Make his or her Decision, Implement It and then examine your Decision.

<h3>What is ethical decision model?</h3>

An ethical decision-making model is known to be a kind of a tool that can be used by a person to help create the ability to think in regards to an ethical dilemma and come to an ethical decision.

Note that Ethical decision-making is one that is dependent on the key character values such as trustworthiness, respect, role, fairness, and others.

Therefore, The Steps my friend need to make an ethical decision are:

Step One: He or she needs to define the Problem.
Step Two: He or she needs to find out possible Resources to solve the problem and then Brainstorm on the List for the Potential Solutions to the problem.
Step Three : Evaluate and examine those Alternatives. ...
Step Four : Make his or her Decision, Implement It and then examine your Decision.

Learn more about ethical decision from

brainly.com/question/21105288

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If it is desired to lay off a distance of 10,000' with a total error of no more than Â± 0.30 ft. If a 100' tape is used and the

Ira Lisetskai [31]

Answer:

± 0.003 ft

Explanation:

Since our distance is 10,000 ft and we need to use a full tape measure of 100 ft. We find that 10,000 = 100 × 100.

Let L' = our distance and L = our tape measure

So, L' = 100L

Now by error determination ΔL' = 100ΔL

Now ΔL' = ± 0.30 ft

ΔL = ΔL'/100

= ± 0.30 ft/100

= ± 0.003 ft

So, the maxim error per tape is ± 0.003 ft

3 0

3 years ago

PLZZ HELP

9966 [12]

Answer:

Could ask a family member to help

Explanation:

5 0

3 years ago

Read 2 more answers

A reversible power cycle R and an irreversible power cycle I operate between the same hot and cold thermal reservoirs. Cycle I h

anygoal [31]

Answer: Attached below is the missing diagram

answer :

A) 1) Wr > WI, 2) Qc' > Qc

B) 1) QH' > QH, 2) Qc' > Qc

Explanation:

л = w / QH = 1 - Qc / QH and QH = w + Qc

A) each cycle receives same amount of energy by heat transfer

( Given that ; Л1 = 1/3 ЛR )

1) develops greater bet work 

WR develops greater work ( i.e. Wr > WI )

2) discharges greater energy by heat transfer

Qc' > Qc

solution attached below

B) If Each cycle develops the same net work

1) Receives greater net energy by heat transfer from hot reservoir

QH' > QH ( solution is attached below )

2) discharges greater energy by heat transfer to the cold reservoir

Qc' > Qc

solution attached below

4 0

2 years ago

Kam

jolli1 [7]

I think D i’m not sure

7 0

3 years ago

Assume a program requires the execution of 50 x 106 FP instructions, 110 x 106 INT instructions, 80 x 106 L/S instructions, and

Pavlova-9 [17]

Answer:

Part A:

$1.3568*10^{-5}=\frac{5300* New\ CPI_1+11660*1+8480*4+1696*2}{2*10^9\ Hz} \\ New\ CPI_1=-4.12$

CPI cannot be negative so it is not possible to for program to run two times faster.

Part B:

$1.3568*10^{-5}=\frac{5300*1+11660*1+8480*New\ CPI_3+1696*2}{2*10^9\ Hz} \\ New\ CPI_3=0.8$

CPI reduced by $1-\frac{0.8}{4} = 0.80$ =80%

Part C:

New Execution Time= $\frac{5300*0.6+11660*0.6+8480*2.8+1696*1.4}{2*10^9\ Hz}=1.81472*10^{-5}\ s$

Increase in speed= $1-\frac{1.81472*10^{-5}}{2.7136*10^{-5}} =0.33125= 33.125\%$

Explanation:

FP Instructions=50*106=5300

INT Instructions=110*106=11660

L/S Instructions=80*106=8480

Branch Instructions=16*106=1696

Calculating Execution Time:

Execution Time= $\frac{\sum^4_{i=1} Number\ of\ Instruction*\ CPI_{i}}{Clock\ Rate}$

Execution Time= $\frac{5300*1+11660*1+8480*4+1696*2}{2*10^9\ Hz}$

Execution Time= $2.7136*10^{-5}\ s$

Part A:

For Program to run two times faster,Execution Time (Calculated above) is reduced to half.

New Execution Time= $\frac{2.7136*10^{-5}}{2}=1.3568*10^{-5}\ s$

$1.3568*10^{-5}=\frac{5300* New\ CPI_1+11660*1+8480*4+1696*2}{2*10^9\ Hz} \\ New\ CPI_1=-4.12$

CPI cannot be negative so it is not possible to for program to run two times faster.

Part B:

For Program to run two times faster,Execution Time (Calculated above) is reduced to half.

New Execution Time= $\frac{2.7136*10^{-5}}{2}=1.3568*10^{-5}\ s$

$1.3568*10^{-5}=\frac{5300*1+11660*1+8480*New\ CPI_3+1696*2}{2*10^9\ Hz} \\ New\ CPI_3=0.8$

CPI reduced by $1-\frac{0.8}{4} = 0.80$ =80%

Part C:

$New\ CPI_1=0.6*Old\ CPI_1=0.6*1=0.6\\New\ CPI_2=0.6*Old\ CPI_2=0.6*1=0.6\\New\ CPI_3=0.7*Old\ CPI_3=0.7*4=2.8\\New\ CPI_4=0.7*Old\ CPI_4=0.7*2=1.4$

New Execution Time= $\frac{\sum^4_{i=1} Number\ of\ Instruction*\ CPI_{i}}{Clock\ Rate}$

New Execution Time= $\frac{5300*0.6+11660*0.6+8480*2.8+1696*1.4}{2*10^9\ Hz}=1.81472*10^{-5}\ s$

Increase in speed= $1-\frac{1.81472*10^{-5}}{2.7136*10^{-5}} =0.33125= 33.125\%$

8 0

3 years ago