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3 years ago

State 2 advantages of laptops over desktop computers

Computers and Technology

2 answers:

PSYCHO15rus [73]3 years ago

8 0

1. Laptops are portable
2. Laptops take less space

Tcecarenko [31]3 years ago

7 0

Answer:

portable ,and they have accessories such as a track pad, a keyboard, a microphone, and a web camera.

Explanation:

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Write a shell (text-based) program, called sum_second.py, that opens a text file called stuff.txt with 2 numbers per line separa

Ivanshal [37]

Answer:

See explaination for the program code

Explanation:

The Programming code:

inFile = open("stuff.txt", "r")

num = []

Sum = 0

for val in inFile.readlines():

value = val.split(",")

Sum = Sum + int(value[1])

print("Sum of second number on every line in the file is: ", Sum)

inFile.close()

Please kindly check attachment for output

6 0

3 years ago

Sketch f(x) = 5x2 - 20 labelling any intercepts.

Norma-Jean [14]

Answer:

The graph of the function is attached below.

The x-intercepts will be: (2, 0), (-2, 0)

The y-intercept will be: (-20, 0)

Explanation:

Given the function

$f\left(x\right)\:=\:5x^2-\:20$

As we know that the x-intercept(s) can be obtained by setting the value y=0

$y=\:5x^2-\:20$

switching sides

$5x^2-20=0$

Add 20 to both sides

$5x^2-20+20=0+20$

$5x^2=20$

Dividing both sides by 5

$\frac{5x^2}{5}=\frac{20}{5}$

$x^2=4$

$\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=\sqrt{4},\:x=-\sqrt{4}$

$x=2,\:x=-2$

so the x-intercepts will be: (2, 0), (-2, 0)

we also know that the y-intercept(s) can obtained by setting the value x=0

$y=\:5(0)^2-\:20$

$y=0-20$

$y=-20$

so the y-intercept will be: (-20, 0)

From the attached figure, all the intercepts are labeled.

8 0

2 years ago

Consider this binary search tree:______.

Nitella [24]

Answer:

The new root will be 2.

Explanation:

The binary tree is not properly presented (See attachment)

To answer this; first, we need to order the nodes of the tree in a pre-order traversal.

We use pre-order because the question says if something is removed from the left child.

So, the nodes in pre-order form is: 14, 2, 1, 5, 4, 16.

The root of the binary tree is 14 and if 14 is removed, the next is 2.

<em>Hence, the new root will be 2.</em>

7 0

3 years ago

Assume a TCP sender is continuously sending 1,090-byte segments. If a TCP receiver advertises a window size of 5,718 bytes, and

Arturiano [62]

Answer:

for the 5 segments, the utilization is 3.8%

Explanation:

Given the data in the question;

segment size = 1090 bytes

Receiver window size = 5,718 bytes

Link transmission rate or Bandwidth = 26 Mbps = 26 × 10⁶ bps

propagation delay = 22.1 ms

so,

Round trip time = 2 × propagation delay = 2 × 22.1 ms = 44.2 ms

we determine the total segments;

Total segments = Receiver window size / sender segment or segment size

we substitute

Total segments = 5718 bytes / 1090 bytes

Total segments = 5.24587 ≈ 5

Next is the throughput

Throughput = Segment / Round trip

Throughput = 1090 bytes / 44.2 ms

1byte = 8 bits and 1ms = 10⁻³ s

Throughput = ( 1090 × 8 )bits / ( 44.2 × 10⁻³ )s

Throughput = 8720 bits / ( 44.2 × 10⁻³ s )

Throughput = 197.285 × 10³ bps

Now Utilization will be;

Utilization = Throughput / Bandwidth

we substitute

Utilization = ( 197.285 × 10³ bps ) / ( 26 × 10⁶ bps )

Utilization = 0.0076

Utilization is percentage will be ( 0.0076 × 100)% = 0.76%

∴ Over all utilization for the 5 segments will be;

⇒ 5 × 0.76% = 3.8%

Therefore, for the 5 segments, the utilization is 3.8%

4 0

3 years ago

Consider the two computers A and B with the clock cycle times 100 ps and 150 ps respectively for some program. The number of cyc

Kipish [7]

Answer:

Option d) B is 1.33 times faster than A

Given:

Clock time, $t_{A} = 100 ps$

$t_{A} = 150 ps$

No. of cycles per instructions, $n_{A} = 2.0$

$n_{B} = 1.0$

Solution:

Let I be the no. of instructions for the program.

CPU clock cycle, $f_{A}$ = 2.0 I

CPU clock cycle, $f_{B}$ = 1.0 I

Now,

CPU time for each can be calculated as:

CPU time, T = $CPU clock cycle\times clock time$

$T_{A} = f_{A}\times t_{A} = 2.0 I\times 100 = 200 I ps$

$T_{B} = f_{B}\times t_{B} = 1.0 I\times 100 = 150 I ps$

Thus B is faster than A

Now,

$\frac{Performance of A}{Performance of B} = \frac{T_{A}}{T_{B}}$

$\frac{Performance of A}{Performance of B} = \frac{200}{150}$

Performance of B is 1.33 times that of A

7 0

3 years ago

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