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Elina [12.6K]
3 years ago
11

Type the correct answer in the box. Spell all words correctly.

Engineering
1 answer:
Alik [6]3 years ago
8 0

Answer:

management engineer

Explanation:

A Management Engineer is someone with a commercial engineering understanding and marketing techniques.

The role of a management engineer covers overseeing the activities of all the engineering departments in an engineering firm particularly in the area of product development, implementation, and production.

This is to ensure the products meet and satisfy the business methods and techniques specifically in the area of cost efficiency and quality product.

Under Management Engineering there are subdivisions of Manufacturing and Industrial Engineer, which both fit the role of Malcolm.

Hence, in this case, Malcolm work as a Management Engineer

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social entrepreneurship

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How wold you classify the earliest examples of S.T.E.M discoveries provided in this lesson?
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the answer is C

Explanation:

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Question 9.1 from the textbook. Consider the following workload: Process Burst Time Priority Arrival Time P1 50 4 0 P2 20 1 20 P
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Explanation:

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5 0
3 years ago
Casein, a dairy product used in making cheese, contains 25% moisture when wet. A dairy sells this product for $40/100 kg. If req
Nataly_w [17]

Based on the percent moisture content of the dried product, the mass of dried casein produced os 852.3 kg.

<h3>What is the mass of casein in wet casein?</h3>

The mass of casein in 1000 Kg of wet casein is 75% 1000 kg = 750 Kg

Mass of water 250 kg

The mass of casein is constant while the moisture content can be changed.

At 12% moisture content;

750 kg = 88%%

100 % = 100 ×750/88 = 852.27 kg

Therefore, the mass of dried casein produced os 852.3 kg.

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3 0
2 years ago
A cylinder with a frictionless piston contains 0.05 m3 of air at 60kPa. The linear spring holding the piston is in tension. The
AleksAgata [21]

Answer:

18 kJ

Explanation:

Given:

Initial volume of air = 0.05 m³

Initial pressure = 60 kPa

Final volume = 0.2 m³

Final pressure = 180 kPa

Now,

the Work done by air will be calculated as:

Work Done = Average pressure × Change in volume

thus,

Average pressure = \frac{60+180}{2}  = 120 kPa

and,

Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³

Therefore,

the work done = 120 × 0.15 = 18 kJ

4 0
3 years ago
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