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Alexandra [31]
4 years ago
6

• Suppose that a particular algorithm has time complexity T(n) = 10 ∗ 2n, and that execution of the algorithm on a particular ma

chine takes T seconds for n inputs. Now, suppose you are presented with a machine that is 64 times as fast as your current machine. How many inputs can you process on you new machine in T seconds?
Engineering
1 answer:
elena-s [515]4 years ago
8 0

Answer:

The number of inputs processed by the new machine is 64

Solution:

As per the question:

The time complexity is given by:

T(n) = 10\times 2n

where

n = number of inputs

T = Time taken by the machine for 'n' inputs

Also

The new machine is 65 times faster than the one currently in use.

Let us assume that the new machine takes the same time to solve k operations.

Then

T(k) = 64 T(n)

\frac{T(k)}{T(n)} = 64

\frac{20k}{20n} = 64

k = 64n

Thus the new machine will process 64 inputs in the time duration T

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If there are 16 signal combinations (states) and a baud rate (number of signals/second) of 8000/second, how many bps could I sen
Mice21 [21]

Answer:

32000 bits/seconds

Explanation:

Given that :

there are 16  signal combinations (states) = 2⁴

bits  n = 4

and a baud rate (number of signals/second) = 8000/second

Therefore; the number of bits per seconds can be calculated as follows:

Number of bits per seconds = bits  n × number of signal per seconds

Number of bits per seconds =  4 × 8000/second

Number of bits per seconds = 32000 bits/seconds

6 0
3 years ago
Which of the following conditions were present in over 80% of paddling fatalities from 1995-2000?
Minchanka [31]

Answer:

80% of the people that were killed weren't wearing a safety flotation device ( in correct terminology Personal Flotation Device, or PFD )

Explanation:

Hence they drowned due to the lack of safety.

3 0
3 years ago
What is the value of the work interaction in this process?
Cloud [144]

Answer:

The answer is "-121\  \frac{KJ}{Kg}".

Explanation:

Please find the correct question in the attachment file.

using formula:

\to W=-P_1V_1+P_2V_2 \\\\When \\\\\to W= \frac{P_1V_1-P_2V_2}{n-1}\ \   or \ \  \frac{RT_1 -RT_2}{n-1}\\\\

W =\frac{R(T_1 -T_2)}{n-1}\\\\

    =\frac{0.287(25 -237)}{1.5-1}\\\\=\frac{0.287(-212)}{0.5}\\\\=\frac{-60.844}{0.5}\\\\=-121.688 \frac{KJ}{Kg}\\\\=-121 \frac{KJ}{Kg}\\\\

7 0
3 years ago
In a lab, scientists grew several generations of offspring of a plant using the method shown. What conclusion can you make about
exis [7]
Answer: c) they have low genetic variability among them.


When a plant is grown for several generations of offspring of a plant, then there are some common things which are to be noted which are found similar in the offspring and in the parent of the offspring. The flowers and fruits and the time or season they come in are absolutely the same.
6 0
1 year ago
(a) For a given material, would you expect the surface energy to be greater than, the same as, or less than the grain boundary e
aksik [14]

Answer:

(a) Surface energy is greater than grain boundary energy due to the fact that the bonds of the atoms on the surface are lower than those of the atoms at the grain boundary. The energy is also directly proportional to the number of bonds created.

(b) The energy of a high-angle grain boundary is higher than that of a small-angle grain boundary because the high-angle grain boundary has a higher misalignment and smaller number of bonds than a small-angle grain boundary.

Explanation:

(a) Surface energy is greater than grain boundary energy due to the fact that the bonds of the atoms on the surface are lower than those of the atoms at the grain boundary. The energy is also directly proportional to the number of bonds created.

(b) The energy of a high-angle grain boundary is higher than that of a small-angle grain boundary because the high-angle grain boundary has a higher misalignment and smaller number of bonds than a small-angle grain boundary.

5 0
3 years ago
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