Well it depends. what the problem
Your rise over run will be -3 over 1 your going to start at the point 5 on the y-axis when your at the point 5 go down 3 then over to the right 1 keep going till it off the graph. Now do the opposite way and go up 3 from 5 and to the left 1 and keep doing that till it’s off the graph.
Hope this helps
9y = 6x + 54
divide through by 9:-
y = 6/9y + 54/9
y = 2/3 x + 6 <---- answer
Answer:

Step-by-step explanation:
When you find mean, you must add all the numbers together then divide by the amount of numbers.


There are 10 numbers.


To find median, you have to either find the middle number or find the two numbers that makes it equal. Since this range set is even, we use the second option.



Divide by 2


Step-by-step explanation:
now you can solve it oohccbkoo