Achemical company makes two brands of antifreeze. the first brand is 60% pure antifreeze, and the second brand is 85% pure antifreeze. in order to obtain 110 gallons of a mixture that contains 75% pure antifreeze, how many gallons of each brand of antifreeze must be used

Let’s say we have x gallons of brand 1 antifreeze and y gallons of brand 2 antifreeze. We know we need a total of 140 gallons, so one equation to relate the variables would be x+y=140. We also know that our final mixture should have 60% antifreeze, so another equation would be the weighted average of the two brands: .55x+.80y140=.60. (The weighted average is basically multiplying each % (55 and 80) by the number of gallons you have (x and y

) and adding those together, and dividing that by the total number of gallons (140). This equals the % of the final mixture (60).)

Now we have a system of two equations that we can solve.

x+y=140

.55x+.80y140=.60

*rearrange first equation to solve for y

:

y=140−x

*substitute this value in for y

in the second equation:

.55x+.80(140−x)140=.60

*use algebra and solve for x

:

The algebra shouldn’t be too complex, and I’m hoping you’re asking this about the setup rather than the actual algebra, and I’m lazy, and I used a calculator to solve this, and this is probably a long run-on sentence, and I got x=112

.

*plug this in to first equation and solve for y

:

y=140−112=28

x=112,y=28

112 gallons of brand 1 antifreeze, 28 gallons of brand 2 antifreeze. That’s a whole lot of antifreeze.

thanx hey

i can

step-by-step explanation:

the data of the detroit pistons in numerical order from least to greatest is 0.30, 0.37, 0.55, 0.62, 0.69, 0.72, 0.73, 0.75, 0.77, 0.78, 0.80, 0.81, 0.81, 0.83, 0.84, 0.89

the data of the miami heat in numerical order from least to greatest is 0.50, 0.50, 0.56, 0.57, 0.67, 0.72, 0.72, 0.72, 0.75, 0.75, 0.76, 0.81, 0.82, 0.90, 1.00, 1.00

the detroit pistons median : 1.52

the miami heat median: 1.47

the iqr of this data is 0.05