Answer:
Do you have a picture to show ?
Answer:
Please find attached pdf
Step-by-step explanation:
Answer:
26
Step-by-step explanation:
Let's find first the lower quartile in both A and B
steps:
- look for the smallest sum of two consecutive numbers
- divide by 2 to get the lower quartile
73 and 75 are the smallest consecutive numbers
The lower quartile in A is 74
47 and 49 are the smallest consecutive numbers
The lower quartile in B is 48
Sum of n terms of a geometric series is
Sn = a1 (r^n-1)/(r-1)
a1 = 1st term = 4
r = common ratio = 12/4 = 3
so, the sum is
Sn = 4 (3^n-1)/(3-1) = 2(3^n-1)
