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1 year ago

A 4/7 inch pipe is to be shortened to 3/8 inch. How much must be removed?

Mathematics

1 answer:

11Alexandr11 [23.1K]1 year ago

6 0

11/56 inches or 0.196 inches must be removed.

The pipe measures 4/7 inches but needs to be reduced to 3/8 inches.

In order to find out the inches to be removed, you must subtract the length that the pipe should be from the length that it currently is.

Length to be removed = 4/7 - 3/8

You need a common denominator so find the lowest common factor of both denominators:

= 56

In the shared fraction, multiply the numerator by the number you get when you divide 56 by the denominator.

= 56/7 = 8 8 x 4 = 32

= 56 / 8 = 7 7 x 3 = 21

= (32 - 21) / 56

= 11 / 56 inches

= 0.196 inches

In conclusion, 11/56 inches must be removed to get the pipe to 3/8 inches

Find out more at brainly.com/question/4681199.

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We can't find what PT exactly is because we don't know TQ or x. However, we can find PT in terms of TQ!

Multiply the second equation by seven sixths to get:

$\frac{7TQ}{6}=7x- \frac{7}{3}$

Subtract $\frac{65}{3}$ from both sides to get:

$\frac{7TQ}{6}- \frac{65}{3}=7x-24$

Substituting PT for 7x-24, we see that:

$PT= \frac{7TQ}{6} - \frac{65}{3}$ , which is our answer.

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Answer:

a) $\bar X=10.65$

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c) $IQR = Q_3 -Q_1 = 11.05-10.55=0.5$

And we can find the usual limits with:

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And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

d) The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.

And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

Step-by-step explanation:

We have the following data:

11.8 10.3 10.7 10.6 11.5 8.3 10.5 10.9 10.7 11.2

Part a

We can calculate the sample mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And if we replace we got: $\bar X=10.65$

For the median we need to sort the values on increasing order and we have:

8.3 10.3 10.5 10.6 10.7 10.7 10.9 11.2 11.5 11.8

Since n =10 we can calculate the median as the average between the 5th and 6th position of the dataset ordered and we got:

$Median =\frac{10.7+10.7}{2}=10.7$

The mode would be the most repeated value on this case:

$Mode= 10.7$

Part b

The range is defined as $Range =Max-Min$ and if we replace we got:

$Range = 11.8-8.3=3.5$

We can calculate the standard deviation with the following formula:

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

And if we replace we got:

$s= 0.948$

Part c

For this case we can use the IQR method in order to determine if 8.3 is an outlier or not.

We can calculate the first quartile with these values: 8.3 10.3 10.5 10.6 10.7 10.7 and $Q_1= \frac{10.6+10.7}{2}=10.55$

And for the Q3 we can use: 10.7 10.7 10.9 11.2 11.5 11.8 and we got $Q_3 = \frac{10.9+11.2}{2}=11.05$

Then we can find the IQR like this:

$IQR = Q_3 -Q_1 = 11.05-10.55=0.5$

And we can find the usual limits with:

$Lower = Q_1 -1.5 IQR = 10.55 -1.5*0.5=9.8$

$Upperer = Q_3 +1.5 IQR = 10.55 +1.5*0.5=11.3$

And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

Part d

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And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

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Step-by-step explanation:

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