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liraira [26]
3 years ago
12

What are two constraints that continuous-media files have that conventional data files generally do not have?

Computers and Technology
1 answer:
musickatia [10]3 years ago
4 0

Answer:dATA space and gb

Explanation:

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Write a function that will sum all of the numbers in a list, ignoring the non-numbers. The function should takes one parameter:
olasank [31]

In python 3.8:

def func(value_list):

   lst = [x for x in value_list if type(x) == int or type(x) == float]

   return sum(lst)

print(func(["h", "w", 32, 342.23, 'j']))

This is one solution using list comprehensions. I prefer this route because the code is concise.

def func(value_list):

   total = 0

   for x in value_list:

       if type(x) == int or type(x) == float:

           total += x

   return total

print(func(["h", "w", 32, 342.23, 'j']))

This is the way as described in your problem.

3 0
3 years ago
There are three types of value for money. Which of the following is not a method of value?
AlladinOne [14]
The answer to this question is c
4 0
3 years ago
When you ____ an exception, you send a message that an error has occurred to another part of the program.
jekas [21]
The correct answer is most definitely B
8 0
3 years ago
Read 2 more answers
Putting commands in correct order so computers can read the commands
lana66690 [7]

Answer:

its b I think I'm pretty sure

8 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
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