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.
For the first one is has greater then 3 terms
The middle one is has exactly one term
And the last one is has two terms
I believe I hope this helps
Answer:
The answer is 62.17
Step-by-step explanation:
Because 3 in the thousandths is not equal to or greater than 5
Answer:
32%
Step-by-step explanation:
X=16
Y=-6
im not good at factorization but this is what i got
