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Katyanochek1 [597]

2 years ago

15

To complete this quest, answer the following questions in the submission box below. Make x an array going from 0 to 100 by steps

of 2 units. (Array indexing) Make y an array going from 0 to 50 by steps of 1 unit. (Array indexing) Make z an array that is equal to [x1y1, x2y2, ..., xnyn]. (Arithmetic operations) Find the first 5 values of the z array. (Array indexing)

Computers and Technology

1 answer:

Law Incorporation [45]2 years ago

5 0

Answer:

C++.

Explanation:

#include <iostream.h>

void main(int argc,char* arg[]) {

// Arrays

int x[100];

int y[50];

int z[50];

////////////////////////////////////////////////////////////////////////////

int count = 0;

for (int i = 0; i < 100; i+=2) {

z[count] = x[i] * y[count];

count++;

}

for (int i =0; i < 5; i++) {

cout<<z[i]<<endl;

}

getche();

}

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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

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2 years ago

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SashulF [63]

Answer:

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Explanation:

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Different type of selectors are used in CSS.

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Hence,

The right answer is: Option A. h3

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3 years ago

Two cars A and B leave an intersection at the same time. Car A travels west at an average speed of x miles per hour and car B tr

drek231 [11]

Answer:

Here is the C++ program:

#include <iostream> // to use input output functions

#include <cmath> // to use math functions like sqrt()

#include <iomanip> //to use setprecision method

using namespace std; //to access objects like cin cout

int main () { //start of main function

double speedA; //double type variable to store average speed of car A

double speedB; //double type variable to store average speed of car B

int hour; //int type variable to hold hour part of elapsed time

int minutes; //int type variable to hold minutes part of elapsed time

double shortDistance; // double type variable to store the result of shortest distance between car A and B

double distanceA; //stores the distance of carA

double distanceB; //stores the distance of carB

double mins,hours; //used to convert the elapsed time

cout << "Enter average speed of car A: " << endl; //prompt user to enter the average speed of car A

cin >> speedA; //reads the input value of average speed of car A from user

cout << "Enter average speed of car B: " << endl ; //prompt user to enter the average speed of car B

cin >> speedB; //reads the input value of average speed of car A from user

cout << "Enter elapsed time (in hours and minutes, separated by a space): " << endl; //prompts user to enter elapsed time

cin>> hour >> minutes; //reads elapsed time in hours and minutes

mins = hour * 60; //computes the minutes using value of hour

hours = (minutes+mins)/60; //computes hours using minutes and mins

distanceA = speedA * (hours); // computes distance of car A

distanceB = speedB * (hours); //computes distance of car B

shortDistance =sqrt((distanceA * distanceA) + (distanceB * distanceB)); //computes shortest distance using formula √[(distanceA)² + (distanceB)²)]

cout << "The (shortest) distance between the cars is: "<<fixed<<setprecision(2)<<shortDistance;

//display the resultant value of shortDistance up to 2 decimal places

Explanation:

I will explain the program with an examples:

Let us suppose that the average speeds of cars are:

speedA = 70

speedB = 55

Elapsed time in hours and minutes:

hour = 2

minutes = 30

After taking these input values the program control moves to the statement:

mins = hour * 60;

This becomes

mins = 2 * 60

mins = 120

Next

hours = (minutes+mins)/60;

hours = (30 + 120) / 60

= 150/60

hours = 2.5

Now the next two statements compute distance of the cars:

distanceA = speedA * (hours);

this becomes

distanceA = 70 * (2.5)

distanceA = 175

distanceB = speedB * (hours);

distanceB = 55 * (2.5)

distanceB = 137.5

Next the shortest distance between car A and car B is computed:

shortDistance = sqrt((distanceA * distanceA) + (distanceB * distanceB));

shortDistance = sqrt((175 * 175) + (137.5 * 137.5))

= sqrt(30625 + 18906.25)

= sqrt(49531.25)

= 222.556173

shortDistance = 222.56

Hence the output is:

The (shortest) distance between the cars is: 222.56

3 0

3 years ago