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expeople1 [14]
3 years ago
15

There are twice as many liters of water in one container as in another. If 8 liters of water are removed from each of the two co

ntainers, there will be three times as many liters of water in one container as in the other. How many liters of water did the smaller container have to begin with?
Mathematics
1 answer:
melomori [17]3 years ago
3 0
<span> Suppose the # of litres in the small container is 'x'.
This would make '2x' litres of water in the larger container.
After taking 8 litres from both containers, the following equation can be written:

(2x-8) = 3 (x-8)

Open the brackets:

2x-8 = 3x-24

Collect like terms:

x = 16

There were 16 litres in the smaller container</span>
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Find the y-intercept for the graph of the quadratic function. y + 4 = (x + 2)2
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y-intercept is equal the value of function for x = 0.

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The output of a chemical process is continually monitored to ensure that the concentration remains within acceptable limits. Whe
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a)  0.31 = 31%

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Step-by-step explanation:

If F_X(x) is the cumulative distribution function of the random variable X, then by definition the probability P of the random variable is given by

P(X \leq x) = F_X(x)

If additionally the random variable is discrete (only has non-negative integers as outcomes as is the case in this problem) then

P(X=a)=F_X(a)-\lim_{x \to a^-}F_X(x)

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b)

In this case we want P(X>3)

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c)

Now, we are interested in P(X=1)

P(X =1) =F_X(1)-\lim_{x \to 1^-}F_X(x)=0.53-0.17=0.36

d)

The expected number of times that the process is recalibrated during the week is the expected value of the probability distribution:

P(X=1)+2P(X=2)+...+nP(X=n)+...

But it is easy to see that P(X=n) = 0 if n is an integer >4

So, the expected value is

P(X=1)+2P(X=2)+3P(X=3)+4P(X=4)

We already have P(X=1) and P(X=2). Let's compute the rest

P(X =3) =F_X(3)-\lim_{x \to 3^-}F_X(x)=0.97-0.84=0.13

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and the expected value is

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4 years ago
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