Answer: The decimal number 278910, expressed in binary, is as follows:
1000100000101111110
Explanation:
Any decimal number can be expressed as a linear combination of powers of 2, as follows:
N = aₙ* 2ⁿ +.....+ a₀*2⁰, where the coefficients aₓ can be 0 or 1.
This means that any number, can be decomposed in powers of 2, so a useful and at the same time simple way to find the binary equivalent of a decimal number, is simply to substract from the number the maximum power of 2 that gives a positive outcome, and put a "1" in the most left position, filling with zeros to the right till finding the following power of two (obtained repeating the process with the result from the first substraction).
For the first substraction, we try different choices, until we get a positive result substracting 2¹⁸ from 278910, as follows:
278,910-262,144= 16,766.
Intuitively, we know that as being 16 a power of 2, it's possible that a number close to the one we have as a result, be a power of 2 indeed.
Trying with 2¹⁴, we find that we are right, because the result is a small number:
16,766 - 16,384 = 382
Now. it's very easy, as the greatest power of 2 smaller than 382, is 2⁸=256.
382-256= 126.
126 can be written as 64+32+16+8+4+2, so all that we need now is, going from left to right, put "1", as the coefficient of the powers of 2 18, 14, 8, 6, 5,4,3,2 and 1, filling with zeros the remaining ones.
The final number can be written as follows:
1000100000101111110
Answer:
T3.
Explanation:
A T3 is an acronym for Transmission system 3 and it is also known as Digital Signal Level 3 (DS3). T3 is a point-to-point physical circuit connection which is capable of transmitting up to 44.736Mbps.
This simply means that, when using a Transmission system 3 (T3), it is very much possible or easier to transmit data such as video and audio at the rate of 44.736Mbps.
Please note, Mbps represents megabit per seconds and it is a unit for the measurement of data transmission rate. The Transmission system 3 is an internet connection that has up to 672 circuit channels having 64Kbs.
Also, worthy of note is the fact that a T3 line is made up of twenty-eight (28) Transmission system 1 lines (T1) each having a data transfer rate of 1.544Mbps.
Additionally, Transmission 3 lines are symmetrical and duplex and thus, has equal upload and download speeds. Therefore, transmissions can be done on T3 lines simultaneously without the data lines being clogged or jammed.
If you need speeds of 16 Mbps between two corporate sites in the United States, you would need a T3 leased line.
Generally, the T3 lines are mostly used for multichannel applications and uninterrupted high bandwidth consumptions such as Telemedicine, internet telephony, video conferencing, e-commerce etc.
Mobile phone
Workstation
Server
b) source address
c) type of application
The most significant protocol at layer 3, often known as the network layer, is the Internet Protocol, or IP.The IP protocol, the industry standard for packet routing among interconnected networks, is the source of the Internet's name. Thus, option C is correct.
Application-layer firewalls operate at the TCP/IP stack's application level (all browser traffic, or all telnet or ftp traffic, for example), and thus have the ability to intercept any packets going to or from an application. They stop different packets (usually dropping them without acknowledgment to the sender).
Firewalls are frequently positioned at a network's edge. An external interface is the one that is located outside the network, while an internal interface is the one that is located inside the firewall.
Therefore, The terms “unprotected” and “protected,” respectively, are sometimes used to describe these two interfaces.
Learn more about TCP/IP here:
#SPJ2
The answer is c) type of application
Answer:
vowels = ("a", "e", "i", "o", "u")
word = input("Enter a word: ")
is_all = True
for c in vowels:
if c not in word:
is_all = False
if is_all == True:
print(word + " is a vowel word.")
else:
print(word + " is not a vowel word.")
Explanation:
Initialize a tuple, vowels, containing every vowel
Ask the user to enter a word
Initially, set the is_all as True. This will be used to check, if every vowel is in word or not.
Create a for loop that iterates through the vowels. Inside the loop, check if each vowel is in the word or not. If one of the vowel is not in the vowels, set the is_all as False.
When the loop is done, check the is_all. If it is True, the word is a vowel word. Otherwise, it is not a vowel word.
Answer:
Here the code is given as,
Explanation:
Code:
#include <math.h>
#include <cmath>
#include <iostream>
using namespace std;
int main() {
int v_stop = 0,count = 0 ;
int x;
double y;
int t_count [100];
double p_item [100];
double Total_rev = 0.0;
double cost_trx[100];
double Largest_element , Smallest_element;
double unit_sold = 0.0;
for( int a = 1; a < 100 && v_stop != -99 ; a = a + 1 )
{
cout << "Transaction # " << a << " : " ;
cin >> x >> y;
t_count[a] = x;
p_item [a] = y;
cost_trx[a] = x*y;
v_stop = x;
count = count + 1;
}
for( int a = 1; a < count; a = a + 1 )
{
Total_rev = Total_rev + cost_trx[a];
unit_sold = unit_sold + t_count[a];
}
Largest_element = cost_trx[1];
for(int i = 2;i < count - 1; ++i)
{
// Change < to > if you want to find the smallest element
if(Largest_element < cost_trx[i])
Largest_element = cost_trx[i];
}
Smallest_element = cost_trx[1];
for(int i = 2;i < count - 1; ++i)
{
// Change < to > if you want to find the smallest element
if(Smallest_element > cost_trx[i])
Smallest_element = cost_trx[i];
}
cout << "TRANSACTION PROCESSING REPORT " << endl;
cout << "Transaction Processed : " << count-1 << endl;
cout << "Uints Sold: " << unit_sold << endl;
cout << "Average Units per order: " << unit_sold/(count - 1) << endl;
cout << "Largest Transaction: " << Largest_element << endl;
cout << "Smallest Transaction: " << Smallest_element << endl;
cout << "Total Revenue: $ " << Total_rev << endl;
cout << "Average Revenue : $ " << Total_rev/(count - 1) << endl;
return 0;
}
Output:
Answer:
Allison missed 58.21% of the times.
Explanation:
The first step is to divide 28 by 67 to get the answer in decimal form:
28 / 67 = 0.4179
Then, we multiplied the answer from the first step by one hundred to get the answer as a percentage:
0.4179 * 100 = 41.79%
Then 100(%) - 41.79(%) = 58.21%