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Lena [83]
3 years ago
7

Which is the least technically experienced technical support group?

Computers and Technology
1 answer:
Nezavi [6.7K]3 years ago
5 0

Answer:

Tier 1 support

Explanation:

Of the given options, tier 1 support technical group is the least experienced group. This group of technicians is also referred to as level 1 technical group.

The tier 1 or level support are made up of junior technician, and they have few technical understandings.

Their roles include email response and basic troubleshooting, attending to phone calls, among others.

When a problem cannot be solved by tier 1 support  technicians, they pass the problem to tier 2 support technicians,  

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Match each option to the description of the task it accomplishes.
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write a program, using python, to determine the mean, median, and standard deviation of a list of numbers. In addition, count th
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Answer:

The program to this question can be given as:

Program:

import math   #import packages

import sys

number =  [3, 6, 7, 2, 8, 9, 2, 3, 7, 4, 5, 9, 2, 1, 6, 9, 6]  #define list  

n = len(number)   # taking length of list

print("Sample Data: ",number)  #print list

#find Mean

print("Mean is: ")  #message  

add = sum(number)  # addition of numbers.

mean = add / n  #add divide by total list number.

print(str(mean))  #print mean

number.sort()   #short list by using sort function

#find Median

print("Median is: ")

if n % 2 == 0:        #conditional statement.

   median1 = number[n//2]  

   median2 = number[n//2 - 1]  

   median = (median1 + median2)/2

else:  

   median = number[n//2]  

print(str(median)) #print median

#find Standard deviation.

print("Standard Deviation :")

def SD(number):  #define function SD

   if n <= 1:      #check condition

       return 0.0   #return value.

   mean,SD = average(number), 0.0  

   # find Standard deviation(SD).

   for j in number:

       SD =SD+(float(j)-mean)**2

       #calculate Standard deviation holing in variable SD  

       SD =math.sqrt((SD)/float(n-1))#using math sqrt function.

   return SD  #return value.

def average(av):         #define function average

   n,mean =len(av), 0.0  

   if n <= 1:

       return av[0]   #return value.

   # calculate average

   for I in av:

       mean +=float(i)

       mean /= float(n)

   return mean             #return value.

print(SD(number)) #print value

Output:

Sample Data:  [3, 6, 7, 2, 8, 9, 2, 3, 7, 4, 5, 9, 2, 1, 6, 9, 6]

Mean is:  

5.235294117647059

Median is:  

6

Standard Deviation :

2.140859045465338

Explanation:

In the above python program first, we import packages to perform maths and other functions. Then we declare the list that is number in the list the elements are given in the question. Then we declare variable n in this variable we take the number of elements present in the list and print the list first. In the list, we perform three operations that can be given as:

Mean:  

In the mean operation firstly we define the variable add in this variable we add all elements of the list by using the math sum function then we divide the number by number of elements present in the list. and sore the value in the mean variable and print it.

Median:

In the median section firstly we sort the list by using the sort function. Then we use the conditional statements for calculating median. In the if block first we modules the number by 2 if it gives 0. it calculates first and second median the add-in median .else it will print that number.

Standard Deviation:

In the standard deviation part first, we define two functions that is SD() function and average().In the SD() function we pass the list as the parameter. In this function we use the conditional statement in the if block we call the average() function and use for loop for return all values. In the average() function we calculate the average of the all number and return value. At last, we print all values.

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