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gtnhenbr [62]
2 years ago
7

Suppose a manager of a certain mining company wants to determine the weekly food expenditure of the company’s employees. if ther

e are 2,549 employees and the manager decided to use only 500 employees as a sample, who will be included in the sample? to help the manager make his decision, suggest a sampling technique to be used and state the whole process of selecting the participants
Engineering
1 answer:
Nikolay [14]2 years ago
5 0

The sampling technique that can be used to selecting the participants will be a random sampling.

<h3>What is sampling?</h3>

It should be noted that sampling simply a method in statistical analysis that predetermined observations are taken from the population.

In this case, the sampling technique that can be used to selecting the participants will be a random sampling. This is need to give everyone an equal chance of being selected.

Learn more about sampling on:

brainly.com/question/17831271

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A metallic material with yield stress of 140 MPa and cross section of 300 mm x 100 mm, is subjected to a tensile force of 8.00 M
Readme [11.4K]

Answer:Yes,266.66 MPa

Explanation:

Given

Yield stress of material =140 MPa

Cross-section of 300\times 100 mm^2

Force(F)=8 MN

Therefore stress due to this Force(\sigma)

\sigma =\frac{F}{A}=\frac{8\times 10^6}{300\times 100\times 10^{-6}}

\sigma =266.66 \times 10^{6} Pa

\sigma =266.66 MPa

Since induced stress  is greater than Yield stress therefore Plastic deformation occurs

8 0
3 years ago
How to comment other people
Mekhanik [1.2K]
What you mean?? .!.!.!.!!.
4 0
2 years ago
Read 2 more answers
Air at a pressure of 6000 N/m^2 and a temperature of 300C flows with a velocity of 10 m/sec over a flat plate of length 0.5 m. E
White raven [17]

Answer:

Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J

Explanation:

To solve this problem we use the expression for the temperature film

T_{f}=\frac{T_{\inf}+T_{w}}{2}=\frac{300+27}{2}=163.5

Then, we have to compute the Reynolds number

Re=\frac{uL}{v}=\frac{10\frac{m}{s}*0.5m}{16.96*10^{-6}\rfac{m^{2}}{s}}=2.94*10^{5}

Re<5*10^{5}, hence, this case if about a laminar flow.

Then, we compute the Nusselt number

Nu_{x}=0.332(Re)^{\frac{1}{2}}(Pr)^{\frac{1}{3}}=0.332(2.94*10^{5})^{\frac{1}{2}}(0.699)^{\frac{1}{3}}=159.77

but we also now that

Nu_{x}=\frac{h_{x}L}{k}\\h_{x}=\frac{Nu_{x}k}{L}=\frac{159.77*26.56*10^{-3}}{0.5}=8.48\\

but the average heat transfer coefficient is h=2hx

h=2(8.48)=16.97W/m^{2}K

Finally we have that the heat transfer is

Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J

In this solution we took values for water properties of

v=16.96*10^{-6}m^{2}s

Pr=0.699

k=26.56*10^{-3}W/mK

A=1*0.5m^{2}

I hope this is useful for you

regards

8 0
3 years ago
Two Technicians are discussing ShopKey Pro. Technician
Leno4ka [110]

Answer:

Technician B

Explanation:

i took he test already

6 0
3 years ago
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valkas [14]
5 is the correct one to choose for this
6 0
3 years ago
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