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makvit [3.9K]
2 years ago
14

a big water tank contained 200 litres of water. during the day, 88 l 580 ml of water was used and 200 ml leaked. how much water

was left in the water tank?
Mathematics
1 answer:
algol132 years ago
3 0

The quantity of water left in the tank as described in the task content is; 111 l 220ml.

<h3>What is the amount of water left in the tank?</h3>

The tank described in the task content initially contains 200 liters, in which case, 1 liter corresponds to 1000ml.

Hence, when 88l 580ml was used, the remaining quantity of water is; 200l - (88l 580ml) = 111 l 420ml.

And finally, since 200 ml leaked, the quantity of water left is; 111l 220ml.

Read more on subtraction;

brainly.com/question/144291

#SPJ1

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7) PG &amp; E have 12 linemen working Tuesdays in Placer County. They work in groups of 8. How many
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Part A

Since order matters, we use the nPr permutation formula

We use n = 12 and r = 8

_{n}P_{r} = \frac{n!}{(n-r)!}\\\\_{12}P_{8} = \frac{12!}{(12-8)!}\\\\_{12}P_{8} = \frac{12!}{4!}\\\\_{12}P_{8} = \frac{12*11*10*9*8*7*6*5*4*3*2*1}{4*3*2*1}\\\\_{12}P_{8} = \frac{479,001,600}{24}\\\\_{12}P_{8} = 19,958,400\\\\

There are a little under 20 million different permutations.

<h3>Answer: 19,958,400</h3>

Side note: your teacher may not want you to type in the commas

============================================================

Part B

In this case, order doesn't matter. We could use the nCr combination formula like so.

_{n}C_{r} = \frac{n!}{r!(n-r)!}\\\\_{12}C_{8} = \frac{12!}{8!(12-8)!}\\\\_{12}C_{8} = \frac{12!}{4!}\\\\_{12}C_{8} = \frac{12*11*10*9*8!}{8!*4!}\\\\_{12}C_{8} = \frac{12*11*10*9}{4!} \ \text{ ... pair of 8! terms cancel}\\\\_{12}C_{8} = \frac{12*11*10*9}{4*3*2*1}\\\\_{12}C_{8} = \frac{11880}{24}\\\\_{12}C_{8} = 495\\\\

We have a much smaller number compared to last time because order isn't important. Consider a group of 3 people {A,B,C} and this group is identical to {C,B,A}. This idea applies to groups of any number.

-----------------

Another way we can compute the answer is to use the result from part A.

Recall that:

nCr = (nPr)/(r!)

If we know the permutation value, we simply divide by r! to get the combination value. In this case, we divide by r! = 8! = 8*7*6*5*4*3*2*1 = 40,320

So,

_{n}C_{r} = \frac{_{n}P_{r}}{r!}\\\\_{12}C_{8} = \frac{_{12}P_{8}}{8!}\\\\_{12}C_{8} = \frac{19,958,400}{40,320}\\\\_{12}C_{8} = 495\\\\

Not only is this shortcut fairly handy, but it's also interesting to see how the concepts of combinations and permutations connect to one another.

-----------------

<h3>Answer: 495</h3>
5 0
2 years ago
3x - 3y = -15<br> -x + y = 5
Katyanochek1 [597]

Answer:

I don't understand what you're asking. can you put a question on this?

4 0
3 years ago
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