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Setler79 [48]
2 years ago
6

Every javafx main class __________. group of answer choices implements javafx.application.applicatio extends javafx.application.

application overrides start(scene s) method overrides start() method
Computers and Technology
1 answer:
sergiy2304 [10]2 years ago
5 0

Two concepts central to JavaFX are a stage and a scene.

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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
Mekhanik [1.2K]

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.

7 0
3 years ago
Half of the integers stored in the array data are positive, and half are negative. Determine the
jonny [76]

Answer:

Check the explanation

Explanation:

Below is the approx assembly code for above `for loop` :-

1). mov ecx, 0

2). loop_start :

3).    cmp ecx, ARRAY_LENGTH

4).    jge loop_end

5).    mv temp_a, array[ecx]

6).    cmp temp_a, 0

7).    branch on nge

8).        mv array[ecx], temp_a*2

9).   add ecx, 1

10).   jmp loop_start

11). loop_end :

Assumptions :-

*ARRAY_LENGTH is register with value 1000000

*temp_a is a register

Frequency of statements :-

1) will be executed one time

3) will be executed 1000000 times

4) will be executed 1000000 times

5) will be executed 1000000 times

6) will be executed 1000000 times

7) `nge` will be executed 1000000 times, branch will be executed 500000 times

8) will be executed 500000 times

9) will be executed 1000000 times

10) will be executed 1000000 times

Cost of statements :-

1) 10 ns

3) 10ns + 10ns + 10ns [for two register accesses and one cmp]

4) 10ns [for jge ]

5) 10ns + 100ns + 10ns [10ns for register access `ecx`, 100ns for memory access `array[ecx]`, 10ns for mv]

6) 10ns + 10ns [10ns for register_access `temp_a`, 10ns for mv]

7) 10ns for nge, 10ns for branch

8) 30ns + 110ns + 10ns

10ns + 10ns + 10ns for temp_a*2 [10ns for moving 2 into a register, 10ns for multiplication],

110ns for array[ecx],

10ns for mv

9) 10ns for add, 10ns for `ecx` register access

10) 10ns for jmp

Total time taken = sum of (frequency x cost) of all the statements

1) 10*1

3) 30 * 1000000

4) 10 * 1000000

5) 120 * 1000000

6) 20 * 1000000

7) (10 * 500000) + (10 * 1000000)

8) 150 * 500000

9) 20 * 1000000

10) 10 * 1000000

Sum up all the above costs, you will get the answer.

It will equate to 0.175 seconds

7 0
2 years ago
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