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8_murik_8 [283]

2 years ago

14

In a print statement, you can set the __________ argument to a space or empty string to stop the output from advancing to a new

line.

Computers and Technology

1 answer:

jeka942 years ago

6 0

In a print statement, you can set the End argument to a space or empty string to stop the output from advancing to a new line.

<h3>What is a Print statement?</h3>

A print "statement" is known to be a kind of statement that looks like a call to the make (print) or println process of the System.

Note that In a print statement, you can set the End argument to a space or empty string to stop the output from advancing to a new line so that it can work properly.

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Why is it important to know much time you spend on task

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The answer is a) <span>so you can provide enough time studying. </span>

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What does a professional programmer usually do first to gain an understanding of a problem?

ElenaW [278]

The correct answer to your question is:

Professional programmers work directly with, and interview customers to find out what the problem is.

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2 years ago

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How does Accenture help companies harness the power of data to achieve optimal business outcomes?

EleoNora [17]

Accenture help companies harness the power of data to achieve optimal business outcomes as they develop data governance frameworks that go on to form the basis of the company's data analytics strategy.

<h3>What is Accenture known for?</h3>

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See full question below

How does Accenture help companies harness the power of data to achieve optimal business outcomes?

- They help companies think about and define the "art of the possible" with their data assets.

- They develop data governance frameworks that go on to form the basis of the company's data analytics strategy.

- They utilize only their Data and Analytics Platform to enable companies.

- They provide an interactive front end for a database solution that changes the data culture.

5 0

1 year ago

Suppose that you have the following definitions:

spin [16.1K]

Suppose that you have the following definitions:struct time Type struct tourTypen the variable destination of type your type are tourType input(){cout<<"Enter the city, distance and time in hr and min"<>destination.city name;

<h3>What is C++ statements?</h3>

Expression statements. These statements examine an expression for its aspect outcomes or for its go-back value. Null statements. These statements may be supplied in which a declaration is needed with the aid of using the C++ syntax however in which no motion is to be taken.

#include
#include
the use of namespace std;
struct timeType
;
struct tourType
;
tourType destination;
tourType input(){
cout<<"Enter the city, distance and time in hr and min"<>destination.cityName;
cin>>destination.distance;
cin>>destination.travelTime.hr;
cin>>destination.travelTime.min;
go back destination;
}
void display()cityName;
d.distance = t->distance;
d.travelTime.hr = t->travelTime.hr;
d.travelTime.min = t->travelTime.min;">
int main().

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6 0

1 year 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

2 years ago