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ra1l [238]
3 years ago
14

An aquarium tank can hold 5400 liters of water. There are two pipes that can be used to fill the tank. The first pipe alone can

fill the tank in 90 minutes. The second pipe can fill the tank in 60 minutes by itself. When both pipes are working together, how long does it take them to fill the tank?
Mathematics
1 answer:
Nady [450]3 years ago
8 0

Answer:

36 minutes when both pipes are working together

Step-by-step explanation:

capacity of tank = 5400 liters

Pipe A flow per mint. = 5400/90 = 60 liters per mint.

Pipe B flow per mint. = 5400/60 = 90 liters per mint.

Flow of A + B per mint. 60 + 90 = 150 liter per mint.

Therefore, 5400 / 150 = 36 minutes to fill the tank

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the graph will be compressed vertically

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Answer:

The first answer is 1/3 the second answer is also 1/3

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If you count the number of names there are and take two out that would round down from 2/6 to 1/3. The answer is the same for both questions.

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The price of gasoline has increased from $2.00 per gallon to $3.00 per gallon. how would this price change be represented on the
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If the price of gasoline has increased from $2.00 per gallon to $3.00 per gallon. how would this price change be represented on the demand curve is: a movement from one point on the line to a higher point on the line.

<h3>What is demand curve?</h3>

Demand curve can be defined as the curve that show price of goods and services produced as well as the quantity demanded for the goods produce at a particular period of time.

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5 0
1 year ago
The average weight of the entire batch of the boxes of cereal filled today was 20.5 ounces. A random sample of four boxes was se
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Answer:

s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And replacing we got:

s= 0.286

And then the estimator for the standard error is given by:

SE= \frac{0.286}{\sqrt{4}}= 0.143

Step-by-step explanation:

For this case we have the following dataset given:

20.05, 20.56, 20.72, and 20.43

We can assume that the distribution for the sample mean is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And the standard error for this case would be:

SE= \frac{\sigma}{\sqrt{n}}

And we can estimate the deviation with the sample deviation:

s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And replacing we got:

s= 0.286

And then the estimator for the standard error is given by:

SE= \frac{0.286}{\sqrt{4}}= 0.143

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3 years ago
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