4 ounces
13% of $15,000.00= 13% * 15000= 0.13 * 15000= $1,950.00
200ml of solution x 20% of alcohol = 40ml of alcohol..
210Type your answer here...
Let a be the number of ounces of 25% alcohol required. Then, 25a + (30x9) = 28(9 + a) 25a + 270 = 252 + 28a 3a = 18 a = 6 Then 6 ounces of 25% alcohol + 9 ounces of 30% alcohol produces 15 ounces of 28% alcohol.
Let x = ounces of 50% solution, and y = ounces of 1% solution. So that we have: 0.5x + 0.01y = 8(0.2) which is a linear equation in two variables, meaning there are infinitely many choices of mixing those solutions.
15% of 13 is 1.95
If the concentration of alcohol and water solution is 25 percent alcohol by volume, the volume of alcohol in a 200 solution is 50.
16%
195015% of 13000= 15% * 13000= 0.15 * 13000= 1950
13% of $15,000.00= 13% * 15000= 0.13 * 15000= $1,950.00
200ml of solution x 20% of alcohol = 40ml of alcohol..
210Type your answer here...
There are 141.95 ml of alcohol in a 167 ml of 85.0% alcohol solution.
Let x be the ounces of 15% alcohol solution. The amount of alcohol in the 15% solution is 0.15x, and the amount of alcohol in the 23% solution is 0.23(100 - x). Setting up the equation 0.15x + 0.23(100 - x) = 0.15(100) solves for x, which is approximately 38.5 ounces of the 15% alcohol solution needed.
Let a be the number of ounces of 25% alcohol required. Then, 25a + (30x9) = 28(9 + a) 25a + 270 = 252 + 28a 3a = 18 a = 6 Then 6 ounces of 25% alcohol + 9 ounces of 30% alcohol produces 15 ounces of 28% alcohol.
Let x = ounces of 50% solution, and y = ounces of 1% solution. So that we have: 0.5x + 0.01y = 8(0.2) which is a linear equation in two variables, meaning there are infinitely many choices of mixing those solutions.
mary mixed 2l of an 80% acid solution with 6l of a 20% acid solution. what was the percent of acid in the resulting mixture