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iren [92.7K]
3 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]3 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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When trying to solve a frame problem it will typically be necessary to draw many free body diagrams. a)-True b)-False
klasskru [66]

Answer:

True

Explanation:

When trying to solve a frame problem in Engineering or Physics, it will typically be necessary to draw more than one body diagram.

When we have several parts of the frame or a set of frames, we have the anchor point, as well as the intersections of frames. Besides that, usually, there is a particle or rigid body together with the frame system. In this sense, usually, it is required to analyze a body diagram for the particle or rigid body suspended, as well as the intersections of the frames. So, usually, it will be required a minimum of two body diagrams.

If the system is more complex, or there are many intersections points, it will be required more than two body diagrams.

Finally, indeed, it will typically be necessary to draw many-body diagrams. 

6 0
3 years ago
Which option identifies why Ethan’s skills are valuable to his team in the following scenario?
larisa [96]

Answer:

Explanation:

The options are:

- In an isometric drawing, multiple angles and axes can be shown in one sketch.

- There is no room for detail in an isometric drawing, so the detail is shown in the orthographic projection.

- Only one sketch will be needed since all other previous designs will no longer be necessary.

- Computer programs will not be necessary to create the exact dimensions of the design.

Orthographic projections are in either the First or Third Angles but the angles are fixed and do not provide perspective view.  Isometric drawings are perspective views from different angles.

So Ethan's skill is valuable because "In an isometric drawing, multiple angles and axes can be shown in one sketch."

4 0
3 years ago
g In the above water treatment facility, chemical concentration (mg/gal) within the tank can be considered uniform. The initial
vlada-n [284]

Answer:

0.05 mg / gallon

Explanation:

mass of chemecila coming in per minute = 50*10 = 500 mg/min

at a time t min , M = mass of chemical = 500*t mg

conecntartion of chemecal = 500t/10000 = 0.05 mg / gallon

8 0
3 years ago
What are the important factors needed to be considered while selecting a brake or clutch?
AysviL [449]

Answer:

The correct answer is: the following factors are needed to properly consider while selecting a brake or clutch:

-Engagement

-Friction

-Electromagnetic

-Mechanical

-Actuation

-Electric

-Fluid power

-Self-actuation

-Key concepts

-Application notes

-Selection criteria

Explanation:

Clutches and brakes are important devices in many rotating drive systems, it is very important to guarantee the security and the proper function of them accomplishing a high quality parameters in those factors.

4 0
3 years ago
In the circuit given below, R1 = 17 kΩ, R2 = 74 kΩ, and R3 = 5 MΩ. Calculate the gain 1formula58.mml when the switch is in posit
Elenna [48]

Answer:a

a) Vo/Vi = - 3.4

b) Vo/Vi = - 14.8

c) Vo/Vi = - 1000

Explanation:

a)

R1 = 17kΩ

for ideal op-amp

Va≈Vb=0 so Va=0

(Va - Vi)/5kΩ + (Va -Vo)/17kΩ = 0

sin we know Va≈Vb=0

so

-Vi/5kΩ + -Vo/17kΩ = 0

Vo/Vi = - 17k/5k

Vo/Vi = -3.4

║Vo/Vi ║ = 3.4    ( negative sign phase inversion)

b)

R2 = 74kΩ

for ideal op-amp

Va≈Vb=0 so Va=0

so

(Va-Vi)/5kΩ + (Va-Vo)74kΩ = 0

-Vi/5kΩ + -Vo/74kΩ = 0

Vo/Vi = - 74kΩ/5kΩ

Vo/Vi = - 14.8

║Vo/Vi ║ = 14.8  ( negative sign phase inversion)

c)

Also for ideal op-amp

Va≈Vb=0 so Va=0

Now for position 3 we apply nodal analysis we got at position 1

(Va - Vi)/5kΩ + (Va - Vo)/5000kΩ = 0           ( 5MΩ = 5000kΩ )

so

-Vi/5kΩ + -Vo/5000kΩ = 0

Vo/Vi = - 5000kΩ/5kΩ

Vo/Vi = - 1000

║Vo/Vi ║ = 1000  ( negative sign phase inversion)

3 0
3 years ago
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