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adelina 88 [10]
3 years ago
11

A plumber uses 16 inches of tubing to connect each washing machine in the laundry source he wants to install 18 washing machine

how many yards of tubing will he need
Mathematics
2 answers:
34kurt3 years ago
8 0

Answer:

Step-by-step explanation:

Since each washing machine needs 16 inches of tubing we are going to multiply 16 by 18, the amount of laundry sources he wants to install. The answer to the multiplication problem is 288. Next we have to divide 288 by 36 since 36 inches are in a yard. The final answer is 8 yards. I hope this helped you out!

Ymorist [56]3 years ago
5 0
Since there are 36 inches in a yard, multiply 16 by 18 and divide that answer by 36

18*16=288
288/36=8
He needs 8 yards of tubing
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On a​ map, 1 inch equals 10 miles. Two cities are 6 inches apart on the map. What is the actual distance between the​ cities?
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Answer:

The answer is 60 miles.

Step-by-step explanation:

Each inch is 10 miles and there are 6 inches. This means that the cities and 60 miles apart.

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The alternating current (AC) breakdown voltage of an insulating liquid indicates its dielectric strength. The article "Testing P
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Answer:

Step-by-step explanation:

Hello!

Given the variable

X: breakdown voltage of an insulating liquid under certain conditions.

a)

Graphic in attachment.

Box: The median is closer to the third quartile than the first quartile showing skewness to the left.

The black dot in the middle of the box represents the sample mean (X[bar]) as you can see it is also moved to the right side of the distribution, affected by the presence of an outlier.

Whiskers:

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Altogether you can say that the distribution of the data set is skewed to the left.

b)

To estimate the population mean you need the population to be at least normal. The box plot shows that the distribution is not symmetrical but, considering that the sample size is n= 48 you can apply the Central Limit Theorem and approximate the sampling distribution to normal.

X[bar]≈N(μ;σ²)

As the distribution is approximate, using the sample standard deviation (since we don't know the population value) for the distribution standard deviation is also valid ⇒ Z= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } }

X[bar] ± Z_{1-\alpha /2} * \frac{S}{\sqrt{n} }

∑X= 2626 ∑X²= 144950 n= 48

X[bar]= ∑X/n= 2626/48= 54.71

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[53.22; 56.18]kV

c)

95% CI

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The semi amplitude or margin of error for the CI can be calculated as:

d= Z_{1-\alpha /2} * \frac{S}{\sqrt{n} }

From this formula you can clear the sample size

\frac{d}{Z_{1-\alpha /2}} = \frac{S}{\sqrt{n} }

(\frac{d}{Z_{1-\alpha /2}} )*\sqrt{n} = S

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