P(not 5) =5/6 and P(5) = 1/6
If the first 5 is rolled on the 5th roll then the first four rolls were not 5
P(5 on fifth roll) = 5/6 x 5/6 x 5/6 x 5/6 x 1/6 = 5^4/6^5 = 625/7776 = 0.080375...
The two numbers are 14 and 11
The question is defective, or at least is trying to lead you down the primrose path.
The function is linear, so the rate of change is the same no matter what interval (section) of it you're looking at.
The "rate of change" is just the slope of the function in the section. That's
(change in f(x) ) / (change in 'x') between the ends of the section.
In Section A:Length of the section = (1 - 0) = 1f(1) = 5f(0) = 0change in the value of the function = (5 - 0) = 5Rate of change = (change in the value of the function) / (size of the section) = 5/1 = 5
In Section B:Length of the section = (3 - 2) = 1 f(3) = 15f(2) = 10change in the value of the function = (15 - 10) = 5Rate of change = (change in the value of the function) / (size of the section) = 5/1 = 5
Part A:The average rate of change of each section is 5.
Part B:The average rate of change of Section B is equal to the average rate of change of Section A.
Explanation:The average rates of change in every section are equalbecause the function is linear, its graph is a straight line,and the rate of change is just the slope of the graph.
