Question about ArcMedia JavaScript Source Code 3000 Pro (gar448) for PC

For int i=0; i< array.count; i++

{

sum+= array[i];

}

average = sum/5

Posted on Dec 30, 2009

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Posted on Jan 02, 2017

Mean absolute deviation.

Let's break it down into its components.

Mean - Average

Absolute - absolute value - sign doesn't matter - if negative, make positive

Deviation - difference from the mean

Let's do an example. Population 3 7 4 2. Total 16. Mean (Average) 16/4 = 4.

Deviations

3 - 4 = -1 absolute value 1

7 - 4 = 3

4 - 4 = 0

2 - 4 = -2 absolute value 2

Sum of these deviations 1 + 3 + 0 + 2 = 6

Mean (average) of these deviations is 6 / 4 = 1.5.

Good luck,

Paul

Let's break it down into its components.

Mean - Average

Absolute - absolute value - sign doesn't matter - if negative, make positive

Deviation - difference from the mean

Let's do an example. Population 3 7 4 2. Total 16. Mean (Average) 16/4 = 4.

Deviations

3 - 4 = -1 absolute value 1

7 - 4 = 3

4 - 4 = 0

2 - 4 = -2 absolute value 2

Sum of these deviations 1 + 3 + 0 + 2 = 6

Mean (average) of these deviations is 6 / 4 = 1.5.

Good luck,

Paul

Apr 05, 2016 | Texas Instruments TI-30Xa Scientific...

Mean= average =(sum of data values)/(number of data values)

Mode is the most frequent data value.

Median: Sort the data in ascending order. The element in the middle is the median.

If the number of data is odd, there will be one MIDDLE element. It is the median.

If the number of data is even, there will be two elements in the middle. Add the values of these middle elements and divide by 2. The result will be the median. It is not part of the data.

Mode is the most frequent data value.

Median: Sort the data in ascending order. The element in the middle is the median.

If the number of data is odd, there will be one MIDDLE element. It is the median.

If the number of data is even, there will be two elements in the middle. Add the values of these middle elements and divide by 2. The result will be the median. It is not part of the data.

Apr 22, 2014 | Texas Instruments TI-34II Explorer Plus...

Sum of data values =6+6+11+12=35

Number of data =4

Average =Mean =35/4=8.75

Mar 13, 2014 | Office Equipment & Supplies

Hello,

The following code will dynamically calculate the some of the first two fields:

Enter first number:

Enter second number:

Their sum is:

Best Regards, Ben

The following code will dynamically calculate the some of the first two fields:

Enter first number:

Enter second number:

Their sum is:

Best Regards, Ben

Jun 29, 2011 | Computers & Internet

You could do this by creating two arrays as such:

int firstArray[5]; int secondArray[5]

and then the sum variable:

int sum = 0;

then you could do a for loop to find the sum of the second array:

int firstArray[5]; int secondArray[5]

and then the sum variable:

int sum = 0;

then you could do a for loop to find the sum of the second array:

Feb 22, 2011 | Microsoft Visual C++ Professional Edition...

2nd [CLR TVM] (clear TVM registers)

2 4 0 0 +/- PV ($2400 initial investment, negative because you're paying it out)

6 I/Y (6% annual interest)

1 N (one year)

CPT FV (compute future value, see 2544.00, the value after one year)

5 N (five years)

CPT FV (see 3211.74, the value after five years)

1 0 N (ten years)

CPT FV (see 4298.03, the value after ten years)

2 4 0 0 +/- PV ($2400 initial investment, negative because you're paying it out)

6 I/Y (6% annual interest)

1 N (one year)

CPT FV (compute future value, see 2544.00, the value after one year)

5 N (five years)

CPT FV (see 3211.74, the value after five years)

1 0 N (ten years)

CPT FV (see 4298.03, the value after ten years)

Feb 19, 2011 | Texas Instruments BA II PLUS Financial...

The TI-30XA does not have to be switched to a statistics mode, the operations are available all the time. The choice of operations are somewhat limited; the calculator offers statistics functions for one dimensional data sets only (no linear interpolation of (x,y) value pairs, for example):

Example: calculate the average and standard deviation of the set { 3, 4, 5, 5, 5, 5, 6 }.

- [2nd] [CSR] - erase statistics data
- [] - add a number to the sample set
- [2nd] [] - remove a number from the sample set
- [2nd] [FRQ] - multiple entry of the same sample value
- [2nd] [n] - sample set size
- [2nd] [] - sum of all sample elements
- [2nd] [] - sum of the squares of all sample elements
- [2nd] [] - average of the sample set
- [2nd] [] - standard deviation (sample set is a complete set)
- [2nd] [] - standard deviation (sample set is a subset of a larger set)

Example: calculate the average and standard deviation of the set { 3, 4, 5, 5, 5, 5, 6 }.

- [2nd] [CSR] - erase stat memory
- 3 [] 4 [] 5 [2nd] [FRQ] 4 [] 6 [] - enter the data, note the use of [2nd] [FRQ] to enter the four occurrences of "5". Entering 5 [] four times would have done the same.
- [2nd] [] - display the average of the set, (4.714...)
- [2nd] [] - display the standard deviation of the set (0.8806...)

Jan 25, 2011 | Texas Instruments TI-30XA Calculator

Hi.

I suggest creating an Array of Cards (cardOne,cardTwo,etc. to five)

If you can enter them in one at a time, after declaring the array, assign the values to the array.

cin < cardOne; (Psuedo code by the way, hope you don't mind)

Afterwards, you'll have an array with 5 cards.

Declare what full houses, flushes, etc would require.

Use an if statement (If you don't mind lots of code) to compare the entered Array against possible hands.

I didn't necessarily give you any specific C++ code. However, if you Google C++ user input, you should find plenty of commands that may help you.

Good Luck

(And not sure about Linear search to be honest, you can use a Linear search, but if you did it the way I explained, you just have to compare hand to a series of conditionals.)

I suggest creating an Array of Cards (cardOne,cardTwo,etc. to five)

If you can enter them in one at a time, after declaring the array, assign the values to the array.

cin < cardOne; (Psuedo code by the way, hope you don't mind)

Afterwards, you'll have an array with 5 cards.

Declare what full houses, flushes, etc would require.

Use an if statement (If you don't mind lots of code) to compare the entered Array against possible hands.

I didn't necessarily give you any specific C++ code. However, if you Google C++ user input, you should find plenty of commands that may help you.

Good Luck

(And not sure about Linear search to be honest, you can use a Linear search, but if you did it the way I explained, you just have to compare hand to a series of conditionals.)

Sep 23, 2009 | Computers & Internet

the best thing to do here is to separate the steps for calculating the variance into different functions. To compute the variance the first thing to do is to compute the mean or the average of the numbers in the array. so your first function will loop through each of the numbers and divide it by the length of your array. the output is the returned. The next thing to do is create another function for the variance. loop through each of the numbers again and subtract to each number the mean. square the value then add to the previous. so that means you have another variable that starts from zero and just adds the answer. after that divide it by the length of the array again and what you are going to get is the value for the variance.

May 13, 2009 | Computers & Internet

Hi,

I am writing in C code here

#include<stdio.h>

#include<conio.h>

void main()

{

int a[5],i,sum=0;

float avg=0;

printf("enter the five values");

for(i=0;i<5;i++)

scanf("%d",&a[i]);

for(i=0;i<5;i++)

sum+=a[i];

printf("Sum=%d",sum);

avg=sum/5;

printf("Average=%f",avg);

getch();

}

I am writing in C code here

#include<stdio.h>

#include<conio.h>

void main()

{

int a[5],i,sum=0;

float avg=0;

printf("enter the five values");

for(i=0;i<5;i++)

scanf("%d",&a[i]);

for(i=0;i<5;i++)

sum+=a[i];

printf("Sum=%d",sum);

avg=sum/5;

printf("Average=%f",avg);

getch();

}

Mar 18, 2009 | ArcMedia JavaScript Source Code 3000 Pro...

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