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Gre4nikov [31]
4 years ago
5

On the first statistics exam, the coefficient of determination between the hours studied and the grade earned was 80%. The stand

ard error of estimate was 10. There were 20 students in the class. Develop an ANOVA table for the regression analysis of hours studied as a predictor of the grade earned on the first statistics exam.
Engineering
1 answer:
True [87]4 years ago
4 0

Answer:

\left[\begin{array}{ccccc}&DF&SS&MS&F\\Regression&1&7200&7200&72\\Error&18&1800&100\\total&19&900\end{array}\right]

Explanation:

Sample size, n=20

Degrees of freedom is 1

Number of degrees of freedom for error is n-2 hence 20-2=18

Total number of degrees of freedom is 18+1=19

Standard error estimate is s_{y-x}=\sqrt {\frac {SSE}{n-2}}

Here, SSE=(n-2)s_{y-x}^{2}=(20-2)(10)^{2}=1800

Coefficient of determination r^{2}=\frac {SSE}{SS total}

Here, SSR=r^{2}(SSR+SSE)

SSR=\frac {r^{2}}{1-r^{2}} SSE=\frac {0.8}{1-0.8}(1800)=7200

The total sum of squares is

SS total=SSR+SSE=7200+1800=9000

MSR=SSR=7200

MSE=\frac {SSE}{n-2}=\frac {1800}{20-2}=100

F value is given by

F=\frac {MSR}{MSE}=\frac {7200}{100}=100

The ANOVA table is then  

\left[\begin{array}{ccccc}&DF&SS&MS&F\\Regression&1&7200&7200&72\\Error&18&1800&100\\total&19&900\end{array}\right]

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Water flows at a rate of 0.011 m3/s in a horizontal pipe whose diameter increases from 6 to 11 cm by an enlargement section. If
lakkis [162]

Answer:

Pressure change in pipe is 766.96 N/m²

Explanation:

The continuity equation is stated below,

AV = Q

A1V1 = A2V2

Where A is the cross-sectional area, V is the velocity and Q is the fluid flow rate.

To calculate the inlet velocity of the pipe,

V1 = Q/A1

V1 = Q/(π x d1²)

d1 is the inlet diameter of the pipe

Substituting values,

V1 = 0.011/(π x 1/4 x 0.06²)

V1 = 3.89 ms-¹

To determine the outlet velocity,

V2 = Q/A2

d2 is the outlet diameter of pipe

V2 = 0.011/(π x 1/4 x 0.11²)

V2 = 1.157 ms-²

Applying Bernoulli's equation for steady flow between the points,

P1/pg + a1(V1²/2g) + z1 + hp = P2/pg + a2(V2²/2g) + z2 + ht + hL

Collecting like terms,

The kinetic energy correction factor, a = a1 = a2

a((V1² - V2²)/2g) - hL = (P2 - P1)/pg

apg((V1² - V2²)/2g) - hLpg = P2 - P1

ap((V1² - V2²)/2) - hLpg = ∆P

p - density of water, g is the acceleration due to gravity and hL is the head loss due to friction in pipe.

Substituting values,

a = 1.05, p = 1000kg/m³, g = 9.81m/s², hL = 0.66m

∆P = (1000)(1.05)((3.89² - 1.157²)/2) - (0.66 x 1000 x 9.81)

∆P = 7241.56 - 6474.6

∆P = 766.96 N/m²

4 0
3 years ago
Question #4
inn [45]

Answer:

Deconstruction

Explanation:

8 0
3 years ago
What separates the work of technology transfer research from implementation of the products of such research?
tatuchka [14]

Answer:

The thing that separates the work of technology transfer research from implementation of the products of such research is:

Technology transfer research looks at how technology may be transferred but does not actually make the transfer.

Explanation:

This suggests that technology transfer research is different from the technology transfer (implementation) itself.  The first stops at making scientific investigations into technology transfer activities while the next step performs the actual transfer or implementation.  In other words, technology transfer conveys the results of scientific and technological research to the marketplace and to the wider society.  It is the bridge-builder between the research and the implementation.

4 0
3 years ago
For a bronze alloy, the stress at which plastic deformation begins is 274 MPa and the modulus of elasticity is 118 GPa. (a) What
likoan [24]

Answer:

86584N

132.306 mm

Explanation:

Q = 274

Modulus of elasticity = 118 gpa

1.

Area = 316mm² without plastic deformation

F = QA

= 274x10⁶x316x10^-6

= 274000000 x 0.000316

= 86584 N

This is the maximum load.

2.

Max length =

L = 132(1 + 274x10⁶/118x10⁹)

L = 132(1+274000000/118000000000)

L = 132(1+0.002322)

L = 132(1.002322)

L = 132.306

This is the maximum length to which it may be stretched without causing plastic deformation.

7 0
4 years ago
Chlorine is one of the important commodity chemicals for the global economy. Before the advent of large scale
artcher [175]

The composition of gas in the feed, the percentage conversion and the

theoretical yield are combined to give the product stream composition.

Response:

The composition of gas in the product stream are;

  • HCl: 0.4 kmol/h, Cl₂: 1.6 kmol/h, H₂O: 1.6 kmol/h, O₂: 0.5 kmol/h

<h3>How can percentage conversion give the contents of the product stream?</h3>

The amount of oxygen used = 30% exceeding the theoretical amount

Number of moles of hydrochloric acid = 4 kmol/h

Percentage conversion = 80%

Required:

The composition of the gas in the product feed.

Solution;

The given reaction is; 4HCl + O₂ \longrightarrow 2Cl₂ + 2H₂O

Percentage \ conversion = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{Moles \  of \ limiting \ reactant \ supplied \ in \ the \, feed}}

Which gives;

80 \% = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{4 \, kmol/h}}

Moles of limiting reactant reacted = 4 kmol/h × 0.80 = 3.6 kmol/h

Which gives;

Number of moles of HCl in the stream = 4 kmol/h - 3.6 kmol/h = 0.4 kmol/h

Number of moles of Cl₂ produced = 2 kmol/h × 0.8 = 1.6 kmol/h

Similarly;

Number of moles of H₂O produced = 2 kmol/h × 0.8 = 1.6 kmol/h

Number of moles of O₂ in the product stream = 30% × 1 kmol/h + 20% × 1 kmol/h = 0.5 kmol/h

The composition of the production stream is therefore;

  • <u>HCl: 0.4 kmol/h</u>
  • <u>Cl₂: 1.6 kmol/h</u>
  • <u>H₂O: 1.6 kmol/h</u>
  • <u>O₂: 0.5 kmol/h</u>

Learn more about theoretical and actual yield here:

brainly.com/question/14668990

brainly.com/question/82989

7 0
3 years ago
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