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;
}
Im sure this means add because on a number line when you move to the right its adding. And since the amount of units were moving up by is 6 we would add 0+6 which equals 6.
Answer:
2/5
Step-by-step explanation:
She completed 8 out of 20 so we divide 8 by 20
8/20 can be simplified if we divide both the numerator and denominator by 4
which equals 2/5
Picture? I cant answer when there us no picture?
A. (7/13)·(1/3)=7/39. P=7/39
B. (6/13)·(2/3)=12/39. P=12/39
C. (7/13)·(2/3)=14/39. P=14/39
D. (6/13)·(1/3)=6/39. P=6/39
All the numbers from Bag B were written as thirds cause their probabilities could be simplified to make multiplying easier
