Vehicle B needs fewer repairs on average but the number of repairs per vehicle has greater variation
Answer:
P_1st year = P0 x (1 + r/n) = P0 x (1 + r/n)^1
P_2nd year = P0 x (1 + r/n) (1 + r/n) = P0 x (1 + r/n)^2
...
P_n year = P0 x (1 + r/n)^n
Hope this helps!
:)
The range of values is =5m^4/p that is your answer
X +(x+1) = 291
2x=290
x=145
145+1= 146
145+146 =291
