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UNO [17]
2 years ago
5

Molly wrote a computer program that generates two random numbers between one and 10. When she runs it, what is the probability t

hat both values will be less than 7? (Type your answer as a rati You can use a number more than once. SUBMIT​
Mathematics
1 answer:
Igoryamba2 years ago
3 0

Answer:

Step-by-step explanation:

The accepted values are  2 3 4 5 6   One is not counted. It says between.            

The total number of ways each dice can be thrown is  2 3 4 5 6 7 8 9 Ten is not counted. Same reason.

5/8 * 5/8 = 25/64 = 0.39

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kupik [55]

Answer:

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Step-by-step explanation:

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3 years ago
In ΔUVW, the measure of ∠W=90°, the measure of ∠V=71°, and VW = 55 feet. Find the length of UV to the nearest tenth of a foot.
Katarina [22]

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Step-by-step explanation:

7 0
3 years ago
1. Let L be a list of numbers in non-decreasing order, and x be a given number. Describe an algorithm that counts the number of
e-lub [12.9K]

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;

}

8 0
3 years ago
Can someone help explain this?
bogdanovich [222]
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2 years ago
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