Mr. Acosta works in the lab at a pharmaceutical company he needs to make 50 L of 31% acid solution to test a new product key supplier only ships the 27% and 32% solution Mr. Costa decides to make the 31% solution by mixing the 21% solution with the 32% solution how much of the 27% solution will Mr. Costa need to use?

10 litres of the 27% acid solution is needed

Step-by-step explanation:

In this question, we are tasked with calculating the amount in litres of the 27% solution that Mr. Costa will need

we proceed as follows;

let us suppose Acosta needs X lts of 27%solution and Y lts of 32% solution to make 50L of 31% acid solution.

So total sum of x and y is 48

x + y = 50 (i)

amount of acid in 27% solution is 0.27Y

amount of acid in 32% solution is 0.32Y

Total amount of acid in 50 liters of 31% solution is 50 × 0.31 = 15.5

So 0.27x + 0.32y = 15.5 (ii)

From i, we can say y = 50 -x , we then substitute this into 2 to yield the following;

0.27x + 0.32(50 - x) = 15.5

0.27x + 16 - 0.32x = 15.5

⇒ 0.32x - 0.27x = 16 - 15.5

0.05x = 0.5

x = 0.5/0.05

x =10

Hence, 10 litres of the 27% acid solution is needed

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !