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Airida [17]
2 years ago
15

Imagine you compare the effectiveness of four different types of stimulant to keep you awake while revising statistics using a o

ne-way ANOVA. The null hypothesis would be that all four treatments have the same effect on the mean time kept awake. How would you interpret the alternative hypothesis?
At least two of the stimulants will have different effects on the mean time spent awake.
All four stimulants have different effects on the mean time spent awake.
Two of the four stimulants have the same effect on the mean time spent awake.
None of the above
Engineering
1 answer:
Arlecino [84]2 years ago
4 0

Answer:

The correct answer is At least two of the stimulants will have different effects on the mean time spent awake.

Explanation:

The null hypothesis exposed leads to consider an alternative hypothesis that differs from the behavior of the 4 stimulants. In this case it is related to the effects on patients, since in practice at least one will present differences compared to the others, influencing the sleep time of consumers.

You might be interested in
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
Is an isothermal process necessarily internally reversible? Explain your answer with an example
torisob [31]

Answer:

please give me brainlist and follow

Explanation:

Example of an irreverseble isothermal process is mixing of two fluids on the same temperature - it requires a lot of energy to unmix Jack and coke. ... Example of an reversible process with changing temperature is isentropic expansion.

5 0
1 year ago
In which of the following branches of engineering is the practice not restricted?
fgiga [73]

Answer:

a) civil engineering.

Explanation:

Civil engineering is a professional engineering program that deals with the construction, design, and maintenance of all the natural and man-made environments including dams, buildings, railways, and roads.

Civil engineering is the branch of engineering that is the practice not restricted because civil engineer is not restricted to academic profession but practice in designing and construction can make someone a professional civil engineer.

Hence, the correct answer is "a)."

6 0
2 years ago
Read 2 more answers
A horizontal, cylindrical, tank, with hemispherical ends, is used to store liquid chlorine at 10 bar. The vessel is 4 m internal
sp2606 [1]

Answer: 0.021818m =2.18x10^-²m

Explanation: Using Laplace principle

Design pressure is =12 bar= 1200000N/m²

T = Wall thickness

Design stress=110Mn/m²= 110,000,000N/m²

Radius= diameter/2 =4/2=2m

Design stress=Hoop stress =Pr/t where p=internal pressure or internal pressure, r=radius and t= wall thickness.

As Laplace equation stated.

110,000,000= 1,2000,00 x 2/t

t= 2,400,000/110,000,000.

t= 0.021818m

t=2.18x10^-2m.

3 0
2 years ago
Laminar flow normally persists on a smooth flat plate until a critical Reynolds number value is reached. However, the flow can b
grandymaker [24]

Answer:

At L = 0.1 m

h⁻_lam = 11.004K   W/m^1.5

h⁻_turb = 7.8848K   W/m^1.8

At L = 1 m

h⁻_lam = 3.48K   W/m^1.5

h⁻_turb = 4.975K   W/m^1.8

Explanation:

Given that;

h_lam(x)= 1.74 W/m^1.5. Kx^-0.5

h_turb(x)= 3.98 W/m^1.8 Kx^-0.2

conditions for plates of length L = 0.1 m and 1 m

Now

Average heat transfer coefficient is expressed as;

h⁻ = 1/L ₀∫^L hxdx

so for Laminar flow

h_lam(x)= 1.74 . Kx^-0.5  W/m^1.5

from the expression

h⁻_lam = 1/L ₀∫^L 1.74 . Kx^-0.5   dx

= 1.74k / L { [x^(-0.5+1)] / [-0.5 + 1 ]}₀^L

= 1.74k/L = [ (x^0.5)/0.5)]⁰^L

= 1.74K × L^0.5 / L × 0.5

h⁻_lam= 3.48KL^-0.5

For turbulent flow

h_turb(x)= 3.98. Kx^-0.2 W/m^1.8

form the expression

1/L ₀∫^L 3.98 . Kx^-0.2   dx

= 3.98k / L { [x^(-0.2+1)] / [-0.2 + 1 ]}₀^L

= (3.98K/L) × (L^0.8 / 0.8)

h⁻_turb = 4.975KL^-0.2

Now at L = 0.1 m

h⁻_lam = 3.48KL^-0.5  =  3.48K(0.1)^-0.5  W/m^1.5

h⁻_lam = 11.004K   W/m^1.5

h⁻_turb = 4.975KL^-0.2 = 4.975K(0.1)^-0.2

h⁻_turb = 7.8848K   W/m^1.8

At L = 1 m

h⁻_lam = 3.48KL^-0.5  =  3.48K(1)^-0.5  W/m^1.5

h⁻_lam = 3.48K   W/m^1.5

h⁻_turb = 4.975KL^-0.2 = 4.975K(1)^-0.2

h⁻_turb = 4.975K   W/m^1.8

Therefore

At L = 0.1 m

h⁻_lam = 11.004K   W/m^1.5

h⁻_turb = 7.8848K   W/m^1.8

At L = 1 m

h⁻_lam = 3.48K   W/m^1.5

h⁻_turb = 4.975K   W/m^1.8

3 0
2 years ago
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