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Crank
3 years ago
11

What is the federal E-Rate program?

Engineering
1 answer:
Annette [7]3 years ago
8 0

Answer:

c. A program that offers discounts to libraries and schools ensuring they have affordable access to modern telecommunications and information services.

Explanation:

Federal E-Rate program refers to the Schools and Libraries Program of the Universal Service Fund managed by the Universal Service Administrative Company (USAC) and being directed by the Federal Communications Commission (FCC).

The program offers telecommunications and internet access to schools and libraries in the United States at discounts of between 20% and 90% in order to make the services affordable to them.

The discounts received by each of he beneficiary schools receive which is between the rage of 20% and 90% is determined by the degree of poverty and the urban/rural status of the population or students being served.

In the program, connectivity and maintenance services are provided by the Schools and Libraries Program, while school that applied to the program has to provide other items like software, hardware (e.g. computers), and among other items that will make then to use the connectivity provided.

I wish you the best.

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Cool I think u should do Marvel first

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3 years ago
Cast iron has about how much carbon content?
Aliun [14]

Answer:

c

Explanation:

7 0
3 years ago
What is 39483048^349374*3948048/3i4u4
Verizon [17]

Answer:

1.  3.81813506×10^2^9

2.  1.71479428×10^6^5

3.  9.38483383×10^2^6

4.  1.150847×10^2^9

Explanation:

Feel free to give brainliest

Have a great day!

3 0
3 years ago
A reservoir delivers water to a horizontal pipeline 39 long The first 15 m has a diameter of 50 mm, after which it suddenly beco
allsm [11]

Answer:

The difference of head in the level of reservoir is 0.23 m.

Explanation:

For pipe 1

d_1=50 mm,f_1=0.0048

For pipe 2

d_2=75 mm,f_2=0.0058

Q=2.8 l/s

Q=2.8\times 10^{-3]

We know that Q=AV

Q=A_1V_1=A_2V_2

A_1=1.95\times 10^{-3}m^2

A_2=4.38\times 10^{-3} m^2

So V_2=0.63 m/s,V_1=1.43 m/s

head loss (h)

h=\dfrac{f_1L_1V_1^2}{2gd_1}+\dfrac{f_2L_2V_2^2}{2gd_2}+0.5\dfrac{V_1^2}{2g}

Now putting the all values

h=\dfrac{0.0048\times 15\times 1.43^2}{2\times 9.81\times 0.05}+\dfrac{0.0058\times 24\times 0.63^2}{2\times 9.81\times 0.075}+0.5\dfrac{1.43^2}{2\times 9.81}

So h=0.23 m

So the difference of head in the level of reservoir is 0.23 m.

8 0
3 years ago
Technician A says the final drive assembly always has a gear ratio of 1:1. Technician B says the final drive assembly provides f
Olenka [21]

Answer:

Technician B only is correct

Explanation:

The last stage of gears found between the vehicle transmission system and the wheels is the final drive ratio. The function of the final drive gear assembly is to enable a gear reduction control stage to reduce the rotation per minute and increase the wheel torque, such that the vehicle performance can be adjusted and the final gear ratio can be between 3:1 and 4.5:1 not 1:1

Therefore, technician B only is correct

5 0
3 years ago
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