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iren [92.7K]
2 years ago
12

The distribution of SAT scores of all college-bound seniors taking the SAT in 2014 was approximately normal with mean μ=1497 and

standard deviation σ=322. A certain test-retake preparation course is designed for students whose SAT scores are in the lower 25%, percent of those who take the test in a given year. What is the maximum SAT score in 2014 that meets the course requirements?
Engineering
1 answer:
Sav [38]2 years ago
4 0

Answer:

1279

Explanation:

We have the mean u = 1497

Standard deviation sd = 322

We find the x distribution using 25%

P(Z<z) = 0.25

Z = -0.675

From here we use the formula for z score

X = z(sd) + u

X = -0.675*322 + 1497

X = -217.35 + 1497

X = 1279.6

Which is approximately 1279

So we conclude that the maximum sat scores in year 2014that meets with the requirements of this course is 1279

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Rooftop gardens have several significant benefits which includes

Reduction of the surrounding temperatures and the Urban heat Island temperatures.

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2 years ago
WILL GIVE BRAINLIEST!Technician A says it takes two revolutions of the crankshaft to fit all eight cylinders on a V8. Technician
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3 years ago
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Select all that apply: Contaminated sharps should<br><br> not be<br><br> ----
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Answer:

Contaminated sharps should not be bent, recapped or removed.

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4 0
3 years ago
A welding rod with κ = 30 (Btu/hr)/(ft ⋅ °F) is 20 cm long and has a diameter of 4 mm. The two ends of the rod are held at 500 °
SOVA2 [1]

Answer:

In Btu:

Q=0.001390 Btu.

In Joule:

Q=1.467 J

Part B:

Temperature at midpoint=274.866 C

Explanation:

Thermal Conductivity=k=30  (Btu/hr)/(ft ⋅ °F)= \frac{30}{3600} (Btu/s)/(ft.F)=8.33*10^{-3}  (Btu/s)/(ft.F)

Thermal Conductivity is SI units:

k=30(Btu/hr)/(ft.F) * \frac{1055.06}{3600*0.3048*0.556} \\k=51.88 W/m.K

Length=20 cm=0.2 m= (20*0.0328) ft=0.656 ft

Radius=4/2=2 mm =0.002 m=(0.002*3.28)ft=0.00656 ft

T_1=500 C=932 F

T_2=50 C= 122 F

Part A:

In Joules (J)

A=\pi *r^2\\A=\pi *(0.002)^2\\A=0.00001256 m^2

Heat Q is:

Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{51.88*0.000012566*(500-50}{0.2}\\ Q=1.467 J

In Btu:

A=\pi *r^2\\A=\pi *(0.00656)^2\\A=0.00013519 m^2

Heat Q is:

Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{8.33*10^{-3}*0.00013519*(932-122}{0.656}\\ Q=0.001390 Btu

PArt B:

At midpoint Length=L/2=0.1 m

Q=\frac{k*A*(T_1-T_2)}{L}

On rearranging:

T_2=T_1-\frac{Q*L}{KA}

T_2=500-\frac{1.467*0.1}{51.88*0.00001256} \\T_2=274.866\ C

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<h3>What are the Characteristics of Code of Ethics?</h3>

The code of ethics are known to be a kind of a universal moral values, that is one that state that what a person expect of any given employee such as been trustworthy, respectful, responsible, and others.

Note that Rules of Practice, Professional Obligations and Codes of Ethics. are known to be put in place to avoid issues that may lead to conflict.

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Therefore, The professional ethics for computer engineers are:

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Learn more about Engineering rules from

brainly.com/question/17169621

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