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kotykmax [81]
1 year ago
5

A large number of genetic codes are stored as binary values in a list. Which one of the following conditions must be true in ord

er for a researcher to obtain the correct result when using a binary search algorithm to determine if a given genetic code is in the list?
A. The genetic codes must be converted from binary to decimal numbers.
B. The list must be sorted based on the genetic code values.
C. The number of genetic code values in the list must be a power of 2.
D. The number of genetic code values in the list must be even.
Computers and Technology
2 answers:
Bad White [126]1 year ago
3 0

Answer:

D

Explanation:

grandymaker [24]1 year ago
3 0

Answer:

D

Explanation:

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Answer:

Category 6 Cabling

Explanation:

Category 6 cables are High-performance UTP cable that can transmit data 10/100/1000Mbps transfer and up to 10Gbps over shorter distances. They usually make uses of a longitudinal separator, this seperators always tend to separates each of the four pairs of wires from each other.

A twisted-pair cable is a cables that is mostly made of copper wires that are twisted around each other and are surrounded by a plastic jacket, twisted pair cables are majorly used for networking.

There are two types of twisted pair cable,

UTP (Unshielded Twisted Pair) are a type of twisted pair cabling that does not include shielding around its conductors.

STP(shielded twisted pair) are types of twisted pair cabling that does include shield around its conductors.

A category 6 cable can eithet be pure copper UTP or STP.

And they are best for networking.

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MAXImum [283]

Answer:

The program in C++ is as follows:

#include <iostream>

using namespace std;

double *sum_n_avg(double n1, double n2, double n3);

int main () {

   double n1, n2, n3;

   cout<<"Enter 3 inputs: ";

   cin>>n1>>n2>>n3;

   double *p;

   p = sum_n_avg(n1,n2,n3);

   cout<<"Sum: "<<*(p + 0) << endl;

   cout<<"Average: "<<*(p + 1) << endl;

  return 0;

}

double *sum_n_avg(double n1, double n2, double n3) {

  static double arr[2];

  arr[0] = n1 + n2 + n3;

  arr[1] = arr[0]/3;

  return arr;

}

Explanation:

This defines the function prototype

double *sum_n_avg(double n1, double n2, double n3);

The main begins here

int main () {

This declares variables for input

   double n1, n2, n3;

This prompts the user for 3 inputs

   cout<<"Enter 3 inputs: ";

This gets user inputs

   cin>>n1>>n2>>n3;

This declares a pointer to get the returned values from the function

   double *p;

This passes n1, n2 and n3 to the function and gets the sum and average, in return.

   p = sum_n_avg(n1,n2,n3);

Print sum and average

<em>    cout<<"Sum: "<<*(p + 0) << endl;</em>

<em>    cout<<"Average: "<<*(p + 1) << endl;</em>

  return 0;

}

The function begins here

double *sum_n_avg(double n1, double n2, double n3) {

Declare a static array

  static double arr[2];

Calculate sum

  arr[0] = n1 + n2 + n3;

Calculate average

  arr[1] = arr[0]/3;

Return the array to main

  return arr;

}

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Explanation:

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