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3 years ago

7

At a fixed operating setting, the pressure in a line downstream of a reciprocating compressor has a mean value of 12.0 bar with

a standard deviation of 1.0 bar based on a very large data set obtained from continuous monitoring.
Required:
What is the probability that the line pressure will exceed 13.0 bar during any measurement?

1 answer:

Komok [63]3 years ago

5 0

Answer:

P [ Z > 13 ] = 0,1587 or P [ Z > 13 ] = 15,87 %

Step-by-step explanation:

The probability that the line pressure will exceed 13.0 bar is:

P [ Z > 13 ] = 1 - P [ Z ≤ 13]

And P [ Z ≤ 13] will be find in z table for z (score) =

z ( score) = ( Z - μ₀ ) / σ

z (score) = (13 - 12 )/1

Then from z-table we find

P [ Z ≤ 13] = 0,8413

Finally

P [ Z > 13 ] = 1 - P [ Z ≤ 13]

P [ Z > 13 ] = 1 - 0,8413

P [ Z > 13 ] = 0,1587 or P [ Z > 13 ] = 15,87 %

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Find the missing value of the direct variation (2,5) (5,y)

Karolina [17]

Answer:

The missing value y=12.5 and we have (5,12.5)

Step-by-step explanation:

The formula used for direct variation is:

$y\: \alpha \:x\\y=kx$

We need to find missing value (2,5)(5,y)

First we will find k, and then y

We have x=2, y=5

Find k:

$y=kx\\5=k(2)\\k=\frac{5}{2}\\k=2.5$

Now, we cam find missing value in (5,y)

We have x=5, k=2.5 and y=?

$y=kx\\y=2.5(5)\\y=12.5$

So, the missing value y=12.5 and we have (5,12.5)

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2 years ago

How is the series 5+11+17…+251 represented in summation notation?

NISA [10]

Answer:

$\displaystyle \large{\sum_{n=1}^{42}(6n-1)$

Step-by-step explanation:

Given:

Series 5+11+17+...+251

To find:

Summation notation of the given series

Summation Notation:

$\displaystyle \large{\sum_{k=1}^n a_k}$

Where n is the number of terms and $\displaystyle \large{a_k}$ is general term.

First, determine what kind of series it is, there are two main series that everyone should know:

Arithmetic Series

A series that has common difference.

Geometric Series

A series that has common ratio.

If you notice and keep subtracting the next term with previous term:

11-5 = 6
17-11 = 6

Two common difference, we can in fact say that the series is arithmetic one. Since we know the type of series, we have to find the number of terms.

Now that brings us to arithmetic sequence, we know that first term is 5 and last term is 251, we’ll be finding both general term and number of term using arithmetic sequence:

<u>Arithmetic Sequence</u>

$\displaystyle \large{a_n=a_1+(n-1)d}$

Where $\displaystyle \large{a_n}$ is the nth term, $\displaystyle \large{a_1}$ is the first term and $\displaystyle \large{d}$ is the common difference:

So for our general term:

$\displaystyle \large{a_n=5+(n-1)6}\\\displaystyle \large{a_n=5+6n-6}\\\displaystyle \large{a_n=6n-1}$

And for number of terms, substitute $\displaystyle \large{a_n}$ = 251 and solve for n:

$\displaystyle \large{251=6n-1}\\\displaystyle \large{252=6n}\\\displaystyle \large{n=42}$

Now we can convert the series to summation notation as given the formula above, substitute as we get:

$\displaystyle \large{\sum_{n=1}^{42}(6n-1)$

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2 years ago

What is the deviation of 12334489152622262929293238

Gre4nikov [31]

The deviation is unidentifiable.

8 0

3 years ago

Estimate the correlation coefficient for each of the above scatter plots?

nasty-shy [4]

Answer:

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We see that the overall trend is going from up to down. If you were to draw a best line of fit, it would have a negative slope. Therefore, our correlation is negative.

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3 years ago

In order to estimate the difference between the average hourly wages of employees of two branches of a department store, the fol

Julli [10]

Answer:

$0.071,1.928$

Step-by-step explanation:

Downtown Store North Mall Store

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Sample standard deviation s $2 $1

$n_1=25\\n_2=20$

$\bar{x_1}=9\\ \bar{x_2}=8$

$s_1=2\\s_2=1$

$x_1-x_2=9-8=1$

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Standard error of difference of means = $0.458$

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Degree of freedom =36

So, z value at 95% confidence interval and 36 degree of freedom = 2.0280

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Confidence interval = $1-(2.0280)\times 0.458,1+(2.0280)\times 0.458$

Confidence interval = $0.071,1.928$

Hence Option A is true

Confidence interval is $0.071,1.928$

4 0

2 years ago