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photoshop1234 [79]
3 years ago
7

When using a line graph, why is it inportant to only graph 1 - 3 series of data?. A. A line graph in Microsoft Excel will not al

low you to graph more than 3 series of data.. . B. It will not show changes over time of you use more than 3 series of data.. . C. The lines are too thick to use more than 3 series.. . D. More than 3 series of data causes too many lines on the graph, which makes ti confusing to read.
Computers and Technology
1 answer:
ElenaW [278]3 years ago
8 0
The most necessary point you need to follow when using a line graph, is to make your graph readable. So, the reason why is it inportant to only graph 1 - 3 series of data is that more than 3 series of data causes too many lines on the graph, which makes ti confusing to read. The last option is the correct one.
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Jacob is a teacher and wants to sort his grades based on Test 1 and then on Test 2.
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4 0
3 years ago
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When you are given a set of tables and asked to create a database to store their data, the first step is to ________?
Vlad1618 [11]
Scan   the tables  to make a digital copy of them
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3 years ago
2. When a business practices offensive behavior, you have many options. The option with the loudest voice is
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4 0
2 years ago
Find the propagation delay for a signal traversing the in a metropolitan area through 200 km, at the speed of light in cable (2.
Ede4ka [16]

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:

v = \frac{(xf-xo)}{(t-to)}

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 =\frac{xf}{v}  =\frac{2e5 m}{2.3e8 m/s} =8.7e-4 sec.

⇒ t = 8.7*10⁻⁴ sec.

4 0
3 years ago
Design a 16K x 8 memory using the following ICs:
Montano1993 [528]

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.

3 0
3 years ago
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