It’s a little complicated but here’s how it works:
Imagine a table with the intervals
0:4 , 4:6 , 6:7 , 7:10 , 10:13 (10 year intervals)
Then we have different rows
Class width: 4 , 2 , 1 , 3 , 3
Freq density: 0.2 , 0.5 , 1.2 , 0.7 , 0.3
So now calculate frequency where freq = class width * density
Freq: 0.8 , 1 , 3.6 , 2.1 , 0.9
So to find median find cumulative frequency
(Add all freq)
Cfreq = 8.4 now divide by 2 = 4.2
So find the interval where 4.2 lies.
0.8 + 1 = 1.8 + 3.6 = 5.6
So 4.2 (median) will lie in that interval 60-70 years.
11/20 = 55/100 = 0.55
1/2 = 50/100 = 0.50
0.51
1/2. 11/20, 0.51 (least to greatest)
C is your answer
hope this helps
One of the zeroes would be : x = 1
f (1) = 1^3 + 6(1) ^2 + 3 (1) - 10
f(1) = 1 + 6 + 3 - 10
f (1) = 10 - 10
f (1) = 0
Hope this helps
i am not sure but it think its 6 root 2
