Answer:
a. N(500, 100)
Step-by-step explanation:
The normal probability distribution, with mean M and standard deviation S, can be represented in the following notation.
N(M,S).
In this problem, we have that:
Mean = 500
Standard deviation = 100
Which of the following options would be the correct way to represent the information?
a. N(500, 100)
Answer: It's easy and simple!
Step-by-step explanation: Split it into rectangles and multiply the height and length. Than add it together and Then you have your answer.
There are 75 people and you need to give each of them $120 for their graduation day.
Answer:
M=v^2r / 2G
Step-by-step explanation:
Open the file, it is solved for you, your welcome.
Answer:
Algorithm
Start
Int n // To represent the number of array
Input n
Int countsearch = 0
float search
Float [] numbers // To represent an array of non decreasing number
// Input array elements but first Initialise a counter element
Int count = 0, digit
Do
// Check if element to be inserted is the first element
If(count == 0) Then
Input numbers[count]
Else
lbl: Input digit
If(digit > numbers[count-1]) then
numbers[count] = digit
Else
Output "Number must be greater than the previous number"
Goto lbl
Endif
Endif
count = count + 1
While(count<n)
count = 0
// Input element to count
input search
// Begin searching and counting
Do
if(numbers [count] == search)
countsearch = countsearch+1;
End if
While (count < n)
Output count
Program to illustrate the above
// Written in C++
// Comments are used for explanatory purpose
#include<iostream>
using namespace std;
int main()
{
// Variable declaration
float [] numbers;
int n, count;
float num, searchdigit;
cout<<"Number of array elements: ";
cin>> n;
// Enter array element
for(int I = 0; I<n;I++)
{
if(I == 0)
{
cin>>numbers [0]
}
else
{
lbl: cin>>num;
if(num >= numbers [I])
{
numbers [I] = num;
}
else
{
goto lbl;
}
}
// Search for a particular number
int search;
cin>>searchdigit;
for(int I = 0; I<n; I++)
{
if(numbers[I] == searchdigit
search++
}
}
// Print result
cout<<search;
return 0;
}