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fgiga [73]
2 years ago
10

Juana is interested in estimating how long it takes students at her school to complete a maze she designed. She takes a random s

ample of 4 students and records how long it takes each of them to complete the maze. They work separately from each other, so Juana is willing to assume independence. Here are the data with summary
statistics:
Mathematics
1 answer:
NemiM [27]2 years ago
3 0

The 95 percent confidence interval for the mean time (in seconds) it takes students to complete this maze is: 39±1.96(8.83/√4).

<h3>Confidence interval</h3>

Using this formula

Confidence interval=Mean ± z × Standard deviation√Sample size

Where:

  • xMean=39
  • Z value for 95% CI=1.96
  • Standard deviation=8.83
  • Sample size=4

Let plug in the formula

Confidence interval=39±1.96×8.83/√4

Confidence interval=39±8.65

Inconclusion the 95 percent confidence interval for the mean time (in seconds) it takes students to complete this maze is: 39±1.96(8.83/√4).

Learn more about confidence interval here:brainly.com/question/16974109

brainly.com/question/15712887

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A=\left[\begin{array}{cc}3&-5\\-1&2\end{array}\right] ,\\\\ X=\left[\begin{array}{c}x&y\end{array}\right],\\\\B=\left[\begin{array}{c}2&0\end{array}\right]

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Adj.A=\left[\begin{array}{cc}2&1\\5&3\end{array}\right] ,\\\\ |A|=6-5\\\\|A|=1\\\\\left[\begin{array}{c}x&y\end{array}\right]=\left[\begin{array}{cc}2&5\\1&3\end{array}\right] \times \left[\begin{array}{c}2&0\end{array}\right]\\\\x=4, y=2

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The given equations are

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   A=\left[\begin{array}{ccc}1&1&-1\\1&0&1\\2&1&1\end{array}\right]\\\\ X=\left[\begin{array}{ccc}x\\y\\z\end{array}\right]\\\\B= \left[\begin{array}{ccc}2\\7\\13\end{array}\right]\\\\\rightarrow X=A^{-1}B\\\\\rightarrow X=\frac{Adj.A}{|A|}\times B\\\\a_{11}=-1,a_{12}=1,a_{13}=1,a_{21}=-2,a_{22}=3,a_{23}=1,a_{31}=1,a_{32}=-2,a_{33}=-1\\\\|A|=1\times(0-1)-1\times(1-2)-1\times(1-0)\\\\=-1+1-1\\\\|A|=-1\\\\Adj.A=\left[\begin{array}{ccc}-1&-2&1\\1&3&-2\\1&1&-1\end{array}\right]

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