Answer:
4 litres of 12% solution
2 litre of 24% solution
Step-by-step explanation:
Let the 12% solution = l
Let the 24% solution = h
Dimitri needs a total of 6 litres of both :
l + h = 6 - - - - - - - - (1)
He needs a mixture of both volumes of both to give a 16% volume
12%l + 24%h = 16% * 6
0.12l + 0.24h = 0.96 ----(2)
From (1)
l + h = 6;
l = 6 - h
Substitute l =6-h into (2)
0.12(6 - h) + 0.24h = 0.96
0.72 - 0.12h + 0.24h = 0.96
0.72 + 0.12h = 0.96
0.12h = 0.96 - 0.72
0.12h = 0.24
h = 0.24 / 0.12
h = 2
From ;
l = 6 - h
l = 6 - 2
l = 4
Naturally, any integer

larger than 127 will return

, and of course

, so we restrict the possible solutions to

.
Now,

is the same as saying there exists some integer

such that

We have

which means that any

that satisfies the modular equivalence must be a divisor of 120, of which there are 16:

.
In the cases where the modulus is smaller than the remainder 7, we can see that the equivalence still holds. For instance,

(If we're allowing

, then I see no reason we shouldn't also allow 2, 3, 4, 5, 6.)
The probability of a hand of 13 cards contains all 4 of one of the denominations, say Aces, is given by:

The same probability applies to getting all 4 of each of the other denominations.
Therefore the required probability is:

The answer is 0.0343.
She made her first mistake in step 2
-2x+5 =-11
add 5 to each side
-2x-5+5 = -11+5
-2x = -6 (not -16)
divide by -2
-2x/-2 = -6/-2
x = 3
She added 5 to 11 and got 16 and made it negative instead of subtracting 5 from 11 and making it negative