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EleoNora [17]
3 years ago
11

A chemical company makes two brands of antifreeze. The first brand is 60% pure antifreeze, and the second brand is 85% pure anti

freeze. 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
Mathematics
1 answer:
rjkz [21]3 years ago
3 0

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 heyaaaaaa

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(b) The probability of getting the correct combination in the first try is \frac{1}{216000}.

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The lock has three-cylinder combinations with 60 numbers on each cylinder.

The procedure of the opening lock is to turn to a number on the first​ cylinder, then to a second number on the second​ cylinder, and then to a third number on the third cylinder.

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There are three cylinder combinations on the lock, each with 60 numbers.

It is provided that repetitions are allowed.

Then each cylinder can take any of the 60 numbers.

So there are 60 options for the first cylinder.

There are 60 options for the second cylinder.

And there are 60 options for the third cylinder.

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Thus, the number of different lock combinations is 216,000.

(b)

The events of getting a correct combinations of the three​-cylinder combination lock implies that all the three cylinder are set at the correct numbers.

Each cylinder has 60 numbers.

This implies that there are 60 possible ways to get a correct number for the first cylinder.

The probability of getting the correct number for the first cylinder is:

P (Number on 1st cylinder is correct) = \frac{1}{60}.

Similarly for the second cylinder the probability of getting the correct number is:

P (Number on 2nd cylinder is correct) = \frac{1}{60}.

And similarly for the third cylinder the probability of getting the correct number is:

P (Number on 3rd cylinder is correct) = \frac{1}{60}.

So the probability of getting the correct combination in the first try is:

P (Correct combination in the 1st try) = \frac{1}{60}\times \frac{1}{60}\times \frac{1}{60}=\frac{1}{216000}.

Thus, the probability of getting the correct combination in the first try is \frac{1}{216000}.

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Answer:

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