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2 years ago

9

Unscramble the following letters to form a word whose meaning is "cooperative". What is the last letter of this word? Y C R N T

G S S E I I

Computers and Technology

1 answer:

AveGali [126]2 years ago

8 0

Synergistic answer should be c

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Security measures are sometimes described as a combination of physical, technical, and administrative (PTA) safeguards. Which of

attashe74 [19]

Answer:

Measures including device data encryption, anti-malware software, and communications encryption.

6 0

2 years ago

Write a while loop that prints usernum divided by 2 until user_num is less than 1. The value of user_num changes inside of the l

zaharov [31]

Answer:

The program to this question can be describes as follows:

Program:

#include <iostream> //defining header file

using namespace std;

int main() //defining main method

{

float user_num ;//defining float variable

cout<<"Enter any number: "; //message

cin>>user_num; //input value from the user

while (user_num >= 1) //defining loop to calculate value

{

user_num =user_num/ 2; //diving the value

cout<<user_num<<endl; //print value

}

return 0;

}

Output:

Enter any number: 20

10

5

2.5

1.25

0.625

Explanation:

In the above program, a float variable user_num is declared in which we store input value from the user end, in the next step, a while loop is declared, which calculates, the given value.

In the loop a condition is defined, that user_num value is greater than equal to 1, inside the loop it will divide the value of the user_num and store in this variable.
In this print, the method is used, which prints its variable values.

5 0

2 years ago

Which data type stores images and audio visual clips?

Xelga [282]

It could be a
<span> common settings</span>

4 0

2 years ago

A blood bank maintains two tables - DONOR, with information about people who are willing to donate blood and ACCEPTOR, with info

Kipish [7]

Answer:

The sql query is given below.

Since we need to count of males and females for a particular blood group, we put xxx for the blood group.

SELECT COUNT(SELECT DID FROM DONOR WHERE GENDER LIKE "M%") as Male_Donors,

COUNT(SELECT DID FROM DONOR WHERE GENDER LIKE "F%") as Female_Donors

FROM DONOR

WHERE BG = xxx;

Explanation:

The clauses in the query are as follows.

1. SELECT: all the columns required in the output are put in this clause.

2. FROM JOIN ON: the table(s) from which the above columns are taken are put in this clause.

3. WHERE: any condition required to filter the output is put in this clause.

The query is explained below.

1. Find the number of male donors. Number of anything can be found using COUNT() function. A query is required since gender is included in deciding the type of donor.

2. The query is defined to find number of male donors as follows.

COUNT( SELECT DID FROM DONOR WHERE GENDER LIKE "M%"; )

3. In the previous query, LIKE operator is used since it is not defined what value is stored for male donors.

4. Similarly, the query to find the number of female donors is formed.

COUNT( SELECT DID FROM DONOR WHERE GENDER LIKE "F%"; )

5. Next, the final query is formed as follows.

SELECT: both COUNT() functions will come here.

FROM: table name

WHERE: specific blood group will be put here

GROUP BY: this clause is optional and is not used in this query.

HAVING: this clause is optional and is not used in this query.

6. The query after putting all clauses is shown below.

SELECT COUNT(SELECT DID FROM DONOR WHERE GENDER LIKE "M%"),

COUNT(SELECT DID FROM DONOR WHERE GENDER LIKE "F%")

FROM DONOR

WHERE BG = xxx;

7. Alias is used in the above query for each column to get the final query.

SELECT COUNT(SELECT DID FROM DONOR WHERE GENDER LIKE "M%") as Male_Donors, COUNT(SELECT DID FROM DONOR WHERE GENDER LIKE "F%") as Female_Donors

FROM DONOR

WHERE BG = xxx;

7 0

2 years ago

Suppose a host has a 1-MB file that is to be sent to another host. The file takes 1 second of CPU time to compress 50%, or 2 sec

kozerog [31]

Answer: bandwidth = 0.10 MB/s

Explanation:

Given

Total Time = Compression Time + Transmission Time

Transmission Time = RTT + (1 / Bandwidth) xTransferSize

Transmission Time = RTT + (0.50 MB / Bandwidth)

Transfer Size = 0.50 MB

Total Time = Compression Time + RTT + (0.50 MB /Bandwidth)

Total Time = 1 s + RTT + (0.50 MB / Bandwidth)

Compression Time = 1 sec

Situation B:

Total Time = Compression Time + Transmission Time

Transmission Time = RTT + (1 / Bandwidth) xTransferSize

Transmission Time = RTT + (0.40 MB / Bandwidth)

Transfer Size = 0.40 MB

Total Time = Compression Time + RTT + (0.40 MB /Bandwidth)

Total Time = 2 s + RTT + (0.40 MB / Bandwidth)

Compression Time = 2 sec

Setting the total times equal:

1 s + RTT + (0.50 MB / Bandwidth) = 2 s + RTT + (0.40 MB /Bandwidth)

As the equation is simplified, the RTT term drops out(which will be discussed later):

1 s + (0.50 MB / Bandwidth) = 2 s + (0.40 MB /Bandwidth)

Like terms are collected:

(0.50 MB / Bandwidth) - (0.40 MB / Bandwidth) = 2 s - 1s

0.10 MB / Bandwidth = 1 s

Algebra is applied:

0.10 MB / 1 s = Bandwidth

Simplify:

0.10 MB/s = Bandwidth

The bandwidth, at which the two total times are equivalent, is 0.10 MB/s, or 800 kbps.

(2) . Assume the RTT for the network connection is 200 ms.

For situtation 1:

Total Time = Compression Time + RTT + (1/Bandwidth) xTransferSize

Total Time = 1 sec + 0.200 sec + (1 / 0.10 MB/s) x 0.50 MB

Total Time = 1.2 sec + 5 sec

Total Time = 6.2 sec

For situation 2:

Total Time = Compression Time + RTT + (1/Bandwidth) xTransferSize

Total Time = 2 sec + 0.200 sec + (1 / 0.10 MB/s) x 0.40 MB

Total Time = 2.2 sec + 4 sec

Total Time = 6.2 sec

Thus, latency is not a factor.

5 0

2 years ago