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3 years ago

Which equation represents the sentence, "The sum of three times p and twenty-five is q"?

1 answer:

7 0

It is D because we said sum of three times p and twenty five is q it means p is multiplied first with p before adding 25

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A printer prints 24 pages in 3 minutes. Which unit rate is correct?

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Answer: 8 pages per minute

Step-by-step explanation:

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The LaSalle High School senior class raised funds for an end of the year cruise getaway. The city gave them a special package fo

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A survey was conducted that asked 1011 people how many books they had read in the past year. results indicated that x overbarequ

Tcecarenko [31]

Since we do not know the population standard deviation, we will use the t-distribution to construct the 90% confidence interval of the mean number of books people read. The value of $t_{\frac{\alpha }{2}}$ with 1010 degrees of freedom and with alpha equals to 0.10 is 1.64

First, we need to solve for the margin of error, E.
$E=t_{\frac{\alpha }{2}}\cdot \frac{s}{\sqrt{n}}=1.64\cdot \frac{16.6}{\sqrt{1011}}=0.86$

The lower bound of the confidence interval is
$LB=\overline{x}-E=14.8-0.86=13.94$

The upper bound of the confidence interval is
$LB=\overline{x}+E=14.8+0.86=15.66$

Therefore, the 90% confidence interval is (13.94, 15.66).

We are 90% confident that the interval from 13.94 books to 15.66 books does contain the true value of the population mean number of books people read.

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4 years ago

Suppose the coefficient matrix of a linear system of four equations in four variables has a pivot in each column. Explain why th

Veseljchak [2.6K]

If the coefficient matrix has a pivot in each column, it means that it is shaped like this:

$A=\left[\begin{array}{cccc}a_{1,1}&a_{1,2}&a_{1,3}&a_{1,4}\\0&a_{2,2}&a_{2,3}&a_{2,4}\\0&0&a_{3,3}&a_{3,4}\\0&0&0&a_{4,4}\end{array}\right]$

So, the correspondant system

$Ax = b$

will look like this:

$\left[\begin{array}{cccc}a_{1,1}&a_{1,2}&a_{1,3}&a_{1,4}\\0&a_{2,2}&a_{2,3}&a_{2,4}\\0&0&a_{3,3}&a_{3,4}\\0&0&0&a_{4,4}\end{array}\right]\cdot \left[\begin{array}{c}x_1\\x_2\\x_3\\x_4\end{array}\right] = \left[\begin{array}{c}b_1\\b_2\\b_3\\b_4\end{array}\right]$

This turn into the following system of equations:

$\begin{cases}a_{1,1}x_1+a_{1,2}x_2+a_{1,3}x_3+a_{1,4}x_4=b_1\\a_{2,2}x_2+a_{2,3}x_3+a_{2,4}x_4=b_2\\a_{3,3}x_3+a_{3,4}x_4=b_3\\a_{4,4}x_4=b_4\end{cases}$

The last equation is solvable for $x_4$ : we easily have

$x_4=\dfrac{b_4}{a_{4,4}}$

Once the value for $x_4$ is known, we can solve the third equation for $x_3$ :

$x_3 = \dfrac{b_3-a_{3,4}x_4}{a_{3,3}}$

(recall that $x_4$ is now known)

The pattern should be clear: you can use the last equation to solve for $x_4$ . Once it is known, the third equation involves the only variable $x_3$ . Once

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3 years ago