The three numbers are 6, 7 and 9
<u>Explanation:</u>
First of all exclude the numbers which aren't correct.
From the 4th data,We can conclude the digit 5, 2 and 3 are wrong.
Now,
- 147 - One digit is right but wrong placed
- 189 - One digit is right and placed correctly
- 964 - Two digits are correct but at wrong place
- 523 - All digits are wrong
- 286 - One digit is right but in the wrong place
We've to find the three digits :
From case 4: 5, 2 and 3 are wrong
So, remove 5, 2 and 3 from all the cases
The numbers are:
- 147 - One digit is right but wrong placed
- 189 - One digit is right and placed correctly
- 964 - Two digits are correct but at wrong place
- _86 - one digit is right but in the wrong place
From 1st and 3rd we can say that: 4 is present in both the case but wrongly placed.
Now the digits become:
- 1_7 - One digit is right but wrong placed
- 189 - One digit is right and placed correctly
- 96_ - Two digits are correct but at wrong place
- _86 - one digit is right but in the wrong place
From 3rd and 4th case we can say that: 9 and 6 are correct but wrongly placed, 8 is not the number and 6 is placed at first
So, the two numbers of a 3 digit number are 6 and 9
From 2nd, 3rd and 4th case, the position of 6 and 9 are:
6 _ 9
From 1st case, 7 is the digit but wrongly placed. So, the 3 digit number becomes:
6 7 9
Answer:
I don't give answers sorry.
Step-by-step explanation:
What you would do is take 1 and subract the two probabilites from it. The equation should be 
. You can find a common denominator by making it both 15.
1.
The ordered pairs (-2,3) and (5,1) make the inequality true.
2.
The simplified form of the expression is 3y-30.
It’s c. 5. i’m sorry if it’s wrong
