<span>He would click on the Test 1 column and press Sort, then click on the Test 2 column and press Sort.
It depends on what you learn because it says "shift"</span>
Scan the tables to make a digital copy of them
—-_-__-____- _—-_- -__-_-____-__
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Answer:
t= 8.7*10⁻⁴ sec.
Explanation:
If the signal were able to traverse this distance at an infinite speed, the propagation delay would be zero.
As this is not possible, (the maximum speed of interactions in the universe is equal to the speed of light), there will be a finite propagation delay.
Assuming that the signal propagates at a constant speed, which is equal to 2.3*10⁸ m/s (due to the characteristics of the cable, it is not the same as if it were propagating in vaccum, at 3.0*10⁸ m/s), the time taken to the signal to traverse the 200 km, which is equal to the propagation delay, can be found applying the average velocity definition:

If we choose x₀ = 0 and t₀ =0, and replace v= 2.3*10⁸ m/s, and xf=2*10⁵ m, we can solve for t:

⇒ t = 8.7*10⁻⁴ sec.
Answer:
Required memory size is 16k x 8
16k = 24 x 210 = 214
Hence, No. of address lines = 14
No. of data lines = 8
a) Size of IC 1024 x 1
Total number of ICs required = 16k x 8 / 1024 x 1 = 16 x 8 = 128
b) Size of IC 2k x 4
Total number of ICs required = 16k x 8 / 2k x 4 = 8 x 2 = 16
c) Size of IC 1k x 8
Total number of ICs required = 16k x 8 / 1k x 8 = 16 x 1 = 16
Explanation:
For a, 10 address lines from A0 to A9 are used to select any one of the memory location out of 1024 memory locations present in a IC.
For b, 11 address lines from A0 to A10 are used to select any one of the memory location out of 2k=2048 memory locations present in a IC.
For c, 10 address lines from A0 to A9 are used to select any one of the memory location out of 1k=1024 memory locations present in a IC.