You need 1045 mL of a 70% alcohol solution. On hand, you have a 5% alcohol mixture. How much of the 5% alcohol mixture and pure alcohol will you need to obtain the desired solution?

x=330 ml

y=715 ml

Step-by-step explanation:

If we let x = the amount of 5% solution and y = the amount of pure alcohol, we can set up that;

0.05x + y is the amount of alcohol in our resulting mixture, and x+y is the total amount in our resulting mixture (which we know is 1045 mL)

Now, we also know that:

(0.05x + y)/0.7 = 1045

Thus,

If we simplify this we come up with:

0.05x + y = 731.5 - - - - (eq1)

and we know x + y = 1045 - - - (eq2)

Let's subtract eq(1) to eq(2) to obtain;

0.95x = 1045 - 731.5

0.95x = 313.5

x = 313.5/0.95

x = 330 ml

Put x = 330 ml for y in eq 2 to get;

330 + y = 1045

y = 1045 - 330

y = 715ml

So,

x=330 ml

y=715 ml

c. 4 1/4 pounds

two data sets with the same median will not have the same varaibility

step-by-step explanation: