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Leno4ka [110]
3 years ago
9

PLEASE HELP AND SHOW WORK!!

Mathematics
1 answer:
Nastasia [14]3 years ago
7 0

Answer:

Pipe 1 alone fills a tank in 10 min

Pipe 2 alone fills the tank in 20 min

If both are turned on together, how long will it take to fill the tank???

Pipe 1 fills at the RATE of 1/10th tank per min

Pipe 2 fills at the RATE of 1/20th tank per min

Pipe 1 + Pipe 2 fills at the rate of 1/10+1/20 or 3/20th tank per min

Let x= time to fill the tank

(3/20)(x)=1 Where 1 denotes a full tank (Multiply both sides by 20)

3x=20

x=6 2/3 min----------both working together

NOW YOUR PROBLEM

Let x= amount of time for smaller pipe to fill the tank

Then (x-5)=amount of time for larger pipe to fill the tank

NOW HERE'S WHERE THE RECIPROCAL COMES IN:

The smaller pipe fills the tank at the RATE of 1/x cu units per min

The larger tank fills the tank at the RATE of 1/(x-5) cu units per min

Together, they fill the tank at the rate of 1/x+(1/(x-5)) cu units per min

But now we are told that together the fill the tank in 11 1/9 minutes. so our equation to solve is:

(1/x+(1/(x-5))(11 1/9)=1 where 1 denotes a full tank. In fact, in each of the above reciprocals, the 1 denotes a full tank.

So, dividing both sides by 11 1/9, we get:

1/x+(1/(x-5)=1/(11 1/9)-------------- which is your equation

I assume that you have no problems solviing this eq.

Hope this helps----ptaylor

Step-by-step explanation:

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Answer:

a) For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

b) For this case the data is skewed to the left and we can't assume that we have the normality assumption.

c) This last case the histogram is not symmetrical and the data seems to be skewed.

Step-by-step explanation:

For this case we have the following data:

(a)data = c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

We can use the following R code to get the histogram

> x1<-c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

> hist(x1,main="Histogram a)")

The result is on the first figure attached.

For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

(b)data = c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> x2<- c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> hist(x2,main="Histogram b)")

The result is on the first figure attached.

For this case the data is skewed to the left and we can't assume that we have the normality assumption.

(c)data = c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> x3<-c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> hist(x3,main="Histogram c)")

The result is on the first figure attached.

This last case the histogram is not symmetrical and the data seems to be skewed.

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Step-by-step explanation:

<u>Step </u>:-

The equation of straight line having slope 'm' and passing through the point

(x_{1} ,y_{1} ) is y-y_{1} = m (x-x_{1} )

Slope of horizontal line(that is x-axis) is m = 0

slope of vertical line (that is y-axis) is m is not defined

The equation of straight line having slope is not defined and passing through the point (1,7).

y-7 = \frac{1}{0} (x-1)

cross multiplication we will get equation

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