A scientist has two acid solutions: Artesia, which is a $70\%$ acid solution, and Borromeo, which is a $20\%$ acid solution. (a) Find the amount of Artesia (in mL) that must be added to $500$ mL of Borromeo in order to produce a $60\%$ acid solution. (b) Find the amount of Artesia and Borromeo (in mL) that can be combined in order to form a $100$ mL solution that is $55\%$ acid. (c) Does there exist a combination of Artesia and Borromeo that is $80\%$ acid?

(a) 2000 mL

(b) 70 mL, 30 mL

(c) No

Step-by-step explanation:

Given that Artesia is a 70% acid solution, and Borromeo is a 20% acid solution.

(a) Let x mL of Artesia is added to 500  mL of Borromeo to produce a 60% acid solution.

So, 70% of x + 20% of 500 = 60% of (x+500)

mL.

(b) Let x mL of Artesia and y mL be combined in order to form a 100 mL solution having 55% acid solution.

So, x+y=100\Rightarrow y= 100-x

And, 70% of x + 20% of y = 55% of (100)\\\\

mL.

and y= 100-x= 100-70=30 mL.

(c)  Let x mL of Artesia and y mL of Borromeo be combined in order to make 80% acid solution.

And, 70% of x + 20% of y = 80% of (x+y)

As x and y are amount of Artesia and Borromeo, so x>0 and y>0

, so the negative value of x/y is not possible.

Ad the negative value of x/y is not so x/y=-6 is not possible, hence,  any combination of Artesia and Borromeo that is 80% acid solution is not possible.

c

step-by-step explanation:

it isn't d, because the interquartile range is from 27-57. and the interquartile range is basically half the data, it's pretty obvious.

it isn't b, because like i said earlier, the interquartile range is from 27-57. 57-27 is 30. the real range has to be longer

not a either. although it looks like it, 3/4 of the data is not greater than 27. but if c is wrong than this should be correct

the correct answer is c, because 13 is the point in which the plot starts

a complex number is a number that can be expressed in the form a + bi, where a and b are real numbers, and i is a solution of the equation x2 = −1. because no real number satisfies this equation, i is called an imaginary number. has no real solution, since the square of a real number cannot be negative.