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4 years ago

Can someone please help me in answering this!?! The language is C++

Computers and Technology

1 answer:

Llana [10]4 years ago

5 0

Here is a somewhat cryptic solution that works:

#include <algorithm>

#include <cstdlib>

using namespace std;

void q(char c, int count)

{

for (int i = 0; i < count; i++) {

putchar(c);

}

void p(int b1, int plusses)

{

q(' ', b1);

q('+', plusses);

}

int main()

{

for (int i = -3; i <= 3; i++)

{

int pl = min(6, (3 - abs(i)) * 2 + 1);

p(6-pl, pl);

i == 0 ? p(0, 6) : p(6, 0);

p(0, pl);

putchar('\n');

}

getchar();

}

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How did mechanical clocks assist in completing work? How was work done previously?

lozanna [386]

Answer

Timekeeping has been a part of society since Ancient Egypt. The use of spring-powered mechanisms allowed clocks to be made smaller ... Essentially, the church bells and the mechanical clock now became the monitor of the working day.

Explanation:

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3 years ago

Imagine that the United States is characterized by a political structure in which the state government and the national governme

Alenkinab [10]

Answer:

The answer is " Co-operative federalism".

Explanation:

It is also known as Marble-cake governance, this federalism provides an idea, in which both governments (state and local), works together and productively resolves the common issues, that are distinct, but mostly identical.

It provides disagreement, over a policy in a national system, that is governed by the federal state.
It supports fitness, that's why the Co-operative federalism is the correct answer.

6 0

3 years ago

A datagram network allows routers to drop packets whenever they need to. The probability of a router discarding a packetis p. Co

tresset_1 [31]

Answer:

a.) k² - 3k + 3

b.) 1/(1 - k)²

c.) $k^{2} - 3k + 3 * \frac{1}{(1 - k)^{2} }\\\\= \frac{k^{2} - 3k + 3 }{(1-k)^{2} }$

Explanation:

a.) A packet can make 1,2 or 3 hops

probability of 1 hop = k ...(1)

probability of 2 hops = k(1-k) ...(2)

probability of 3 hops = (1-k)²...(3)

Average number of probabilities = (1 x prob. of 1 hop) + (2 x prob. of 2 hops) + (3 x prob. of 3 hops)

= (1 × k) + (2 × k × (1 - k)) + (3 × (1-k)²)

= k + 2k - 2k² + 3(1 + k² - 2k)

∴mean number of hops = k² - 3k + 3

b.) from (a) above, the mean number of hops when transmitting a packet is k² - 3k + 3

if k = 0 then number of hops is 3

if k = 1 then number of hops is (1 - 3 + 3) = 1

multiple transmissions can be needed if K is between 0 and 1

The probability of successful transmissions through the entire path is (1 - k)²

for one transmission, the probility of success is (1 - k)²

for two transmissions, the probility of success is 2(1 - k)²(1 - (1-k)²)

for three transmissions, the probility of success is 3(1 - k)²(1 - (1-k)²)² and so on

∴ for transmitting a single packet, it makes:

∞ n-1

T = ∑ n(1 - k)²(1 - (1 - k)²)

n-1

= 1/(1 - k)²

c.) Mean number of required packet = ( mean number of hops when transmitting a packet × mean number of transmissions by a packet)

from (a) above, mean number of hops when transmitting a packet = k² - 3k + 3

from (b) above, mean number of transmissions by a packet = 1/(1 - k)²

substituting: mean number of required packet = $k^{2} - 3k + 3 * \frac{1}{(1 - k)^{2} }\\\\= \frac{k^{2} - 3k + 3 }{(1-k)^{2} }$

6 0

3 years ago

Choose the term so achieve the output shown.

Aneli [31]

Answer:

sqrt

Explanation:

4 0

3 years ago

Consider a computer system, where each processor can independently perform a separate task. given that the response time on a co

vekshin1

First four tasks = 8 seconds
Next four tasks = 8 seconds
Last two tasks = 8 seconds

The total number of seconds to perform all tasks is 24 seconds.

The throughput, which would be the number of tasks per second, is calculated by taking the number of tasks and dividing by the number of seconds:

10 tasks / 24 seconds = 5/12 or 0.4167 tasks per second

8 0

3 years ago