1 litre of 10% solution contains 100mL of iodine
1 litre of 90% solution contains 900mL of iodine
Mixing them will produce 2 litres of solution containing 1000mL of iodine which is 50% iodine.
To produce 8 litres of 50% iodine they can mix 4 litres of 10% and 4 litres of 90%
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:
15.1
Step-by-step explanation:
29.34
-14.24
---------
15.10
Step-by-step explanation:
x + y = 9. => 2x + 2y = 18.
2x + 2y = 18
- (2x - 3y = -12)
=> 5y = 30, y = 6.
Therefore x + (6) = 9, x = 3.
The solution is x = 3 and y = 6.
