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.
The number is 'n'.
Five times the number is 5n .
Nine more than that is 5n + 9 .
Answer:
Arc DB = 146
x = 5
Step-by-step explanation:
Remark
The relationship between the large arc, the small arc and the intersecting angle is
1/2 (large arc - small arc) = intersecting angle.
Givens
Large arc = 25x + 21
Small arc = 96
5x
Solution
1/2*(25x + 21 - 96) = 5x*2 Multiply both sides by 2
25x + 21 - 96 = 10 x Combine
25x - 75 = 10x Subtract 25x from both sides
-75 = 10x - 25x
-15x = -75
x = 5
DB = 25x + 21
DB = 25*5 + 21
DB = 125 + 21
DB = 146
Tell me if this is not in the choices.
I think 30 because I did all the equation and I got 30
