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zmey [24]
2 years ago
9

Jen is developing the positioning statement for a new line of sunglasses. In a meeting, the marketing team tells Jen that she ha

s succinctly and clearly expressed the competitive advantage of the new sunglasses brand. The team approves of the way Jen wants to express the brand's competitive advantage, and tells her to continue her work on fully developing the positioning statement. Which element of crafting the positioning statement has Jen's team just approved?
Engineering
1 answer:
Allushta [10]2 years ago
8 0

Answer:

Unique Selling Proposition

Explanation:

Unique Selling Proposition (USP) is the distinguishing feature that makes one product or business better than its competitor in the market. Jen has developed a competitive advantage of the new sunglasses brand making the company to sell their products. This is an example of USP.

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Discuss your interpretation of the confidence-precision trade-off, and provide a few examples of how you might make a choice in
LenKa [72]

The given question is incomplete, the complete question is as follows:

Our text describes a trade-off that we must make as engineers between our confidence in the value of a parameter versus the precision with which we know the value of that parameter. That trade-off might be affected by whether we are looking at a two-sided or bounded (one-sided) interval.

Question: Discuss your interpretation of the confidence-precision trade-off, and provide a few examples of how you might make a choice in one direction or the other in an engineering situation.

Answer: A balancing point is required to be reached to obtain a better confidence level in the predicted values.

Explanation:

The confidence interval and precision are the two terms that aims at providing the accurate estimation of the measurability of an object. If the precision increases, we can compromise on the confidence level and if the confidence level increases, then the precision of the predicted value also dilutes.

Thus a balance point is required to be reached between these two variables so that we get better confidence in the values being predicted without losing the correct estimation on precision. Ensuring that both the confidence and precision are maintained.

8 0
2 years ago
thermal energy is being added to steam at 475.8 kPa and 75% quality. determine the amount of thermal energy to be added to creat
eduard

Answer:

q_{in} = 528.6\,\frac{kJ}{kg}

Explanation:

Let assume that heating process occurs at constant pressure, the phenomenon is modelled by the use of the First Law of Thermodynamics:

q_{in} = h_{g} - h_{mix}

The specific enthalpies are:

Liquid-Vapor Mixture:

h_{mix} = 2217.2\,\frac{kJ}{kg}

Saturated Vapor:

h_{g} = 2745.8\,\frac{kJ}{kg}

The thermal energy per unit mass required to heat the steam is:

q_{in} = 2745.8\,\frac{kJ}{kg} - 2217.2\,\frac{kJ}{kg}

q_{in} = 528.6\,\frac{kJ}{kg}

7 0
2 years ago
The uniform 8–ft board is pushed down into the water so it makes an angle of θ = 30o with the water surface. If the cross sectio
lana66690 [7]

Answer:

The vertical force, F = 19.9 lb.

Explanation:

3 0
2 years ago
Water at 70 kPa and 1008C is compressed isentropically in a closed system to 4 MPa. Determine the final temperature of the water
Nikitich [7]

Answer:

The answer is "909.3928  KJ".

Explanation:

70 \ kPa  \ \ and \ \ 100^{\circ}C \\\\s_i= 7.56162\ \frac{kJ}{kgK}\\\\u_i= 2509.39 \ \frac{kJ}{kg}\\\\

The method is isentropic since the cylinders are shielded.

Calculating the work:

w= u_2-u_i \\\\

   = 3418.7728-2509.38 \\\\=909.3928 \ KJ

3 0
2 years ago
Small droplets of carbon tetrachloride at 68 °F are formed with a spray nozzle. If the average diameter of the droplets is 200 u
Licemer1 [7]

Answer:

the difference in pressure between the inside and outside of the droplets is 538 Pa

Explanation:

given data

temperature = 68 °F

average diameter = 200 µm

to find out

what is the difference in pressure between the inside and outside of the droplets

solution

we know here surface tension of carbon tetra chloride at 68 °F is get from table 1.6 physical properties of liquid that is

σ = 2.69 × 10^{-2} N/m

so average radius = \frac{diameter}{2} =  100 µm = 100 ×10^{-6} m

now here we know relation between pressure difference and surface tension

so we can derive difference pressure as

2π×σ×r = Δp×π×r²    .....................1

here r is radius and  Δp pressure difference and σ surface tension

Δp = \frac{2 \sigma }{r}    

put here value

Δp = \frac{2*2.69*10^{-2}}{100*10^{-6}}  

Δp = 538

so the difference in pressure between the inside and outside of the droplets is 538 Pa

7 0
2 years ago
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