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.
Answer:
Around 0.22
<u>FIRST THOUGH, I FEEL LIKE 90 MIGHT HAVE MEANT TO BEEN 0.9, SO IF SO SUBSTITUTE THE 90 BELOW FOR 0.9!</u>
<u></u>
Step-by-step explanation:
Okay, so first.
Company A will be 50 a day plus 0.4 a mile
Company B will be 30 a day plus 90 per mile
We can write an equation like this:
50 + 0.4m = 30 + 90m
m = miles they are the same
Then we solve for m.
50 + 0.4m = 30 + 90m
- 0.4m - 0.4m
50 = 30 + 89.6m
- 30 - 30
20 = 89.6m
Divide both sides by 89.6
Around 0.22!
Answer:
<h2>x > 3+y</h2>
Step-by-step explanation:
The answer is b) :) hope i helped!
