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Nonamiya [84]
3 years ago
7

Suppose that you have the following declaration:

Computers and Technology
1 answer:
masha68 [24]3 years ago
6 0

Answer:

Explanation:

The code is written in C++:

#include <iostream>

#include <stack>

#include <math.h>

#include <iomanip>

using namespace std;

/*

* Supporting method to print contents of a stack.

*/

void print(stack<double> &s)

{

if(s.empty())

{

cout << endl;

return;

}

double x= s.top();

s.pop();

print(s);

s.push(x);

cout << x << " ";

}

int main(){

// Declaration of the stack variable

stack<double> stack;

//rray with input values

double inputs[] = {25,64,-3,6.25,36,-4.5,86,14,-12,9};

/*

* For each element in the input, if it is positive push the square root into stack

* otherwise push the square into the stack

*/

for(int i=0;i<10;i++){

if(inputs[i]>=0){

stack.push(sqrt(inputs[i]));

}else{

stack.push(pow(inputs[i],2));

}

}

//Print thye content of the stack

print(stack);

}

OUTPUT:

5     8     9     2.5     6     20.25     9.27362     3.74166     144     3

-------------------------------------------------------------------------

Process exited after 0.01643 seconds with return value 0

Press any key to continue . . . -

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5 0
3 years ago
Jim, the IT director, is able to complete IT management task very well but is usually two weeks late in submitting results compa
oee [108]

Answer:Effective but not efficient

Explanation:

Jim is effective because he was able to complete the IT tasks well but he is not efficient because he didn't submit the result on time because being efficient includes management of time.

5 0
4 years ago
A computer retail store has 15 personal computers in stock. A buyer wants to purchase 3 of them. Unknown to either the retail st
Dima020 [189]

Answer:

a. 1365 ways

b. Probability = 0.4096

c. Probability = 0.5904

Explanation:

Given

PCs = 15

Purchase = 3

Solving (a): Ways to select 4 computers out of 15, we make use of Combination formula as follows;

^nC_r = \frac{n!}{(n-r)!r!}

Where n = 15\ and\ r = 4

^{15}C_4 = \frac{15!}{(15-4)!4!}

^{15}C_4 = \frac{15!}{11!4!}

^{15}C_4 = \frac{15 * 14 * 13 * 12 * 11!}{11! * 4 * 3 * 2 * 1}

^{15}C_4 = \frac{15 * 14 * 13 * 12}{4 * 3 * 2 * 1}

^{15}C_4 = \frac{32760}{24}

^{15}C_4 = 1365

<em>Hence, there are 1365 ways </em>

Solving (b): The probability that exactly 1 will be defective (from the selected 4)

First, we calculate the probability of a PC being defective (p) and probability of a PC not being defective (q)

<em>From the given parameters; 3 out of 15 is detective;</em>

So;

p = 3/15

p = 0.2

q = 1 - p

q = 1 - 0.2

q = 0.8

Solving further using binomial;

(p + q)^n = p^n + ^nC_1p^{n-1}q + ^nC_2p^{n-2}q^2 + .....+q^n

Where n = 4

For the probability that exactly 1 out of 4 will be defective, we make use of

Probability =  ^nC_3pq^3

Substitute 4 for n, 0.2 for p and 0.8 for q

Probability =  ^4C_3 * 0.2 * 0.8^3

Probability =  \frac{4!}{3!1!} * 0.2 * 0.8^3

Probability = 4 * 0.2 * 0.8^3

Probability = 0.4096

Solving (c): Probability that at least one is defective;

In probability, opposite probability sums to 1;

Hence;

<em>Probability that at least one is defective + Probability that at none is defective = 1</em>

Probability that none is defective is calculated as thus;

Probability =  q^n

Substitute 4 for n and 0.8 for q

Probability =  0.8^4

Probability = 0.4096

Substitute 0.4096 for Probability that at none is defective

Probability that at least one is defective + 0.4096= 1

Collect Like Terms

Probability = 1 - 0.4096

Probability = 0.5904

8 0
3 years ago
Is this statement true or false?
quester [9]

Answer:

NOPE

Explanation:

sometimes presentations can be for one person only. For instance if you work in a company sometimes you present for your boss only etc.

Hope this helped :)

6 0
3 years ago
Read 2 more answers
Identify the operation which is NOT one of the parts of the fivebasic set operations in relational algebra?
wariber [46]

Answer:

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Explanation:

Five basic set operators in relational algebra are as follows:

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  • Set difference - minus operation on two relations

As we can see, Join is not part of the basic set operations but it is implemented using the Cartesian Product operator.

5 0
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