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sammy [17]
2 years ago
15

Ian recorded the time, in seconds, it took each contestant to solve the riddle. He recorded the results. 68 seconds 89 seconds 1

22 seconds 72 seconds 139 seconds 49seconds 115 seconds 83 seconds 97 seconds 56 seconds 142 seconds What is the third quartile of the data?
Mathematics
1 answer:
jeka57 [31]2 years ago
4 0
Your answer: 122

First we have to put the data in ascending order.

49, 56, 68, 72, 83, 89, 97, 115, 122, 139, 142.

I added spaces around the 89 because that’s the median/middle number, and it splits the data into upper and lower halves. The first quartile is the middle number of the bottom half (68), and the third quartile is the middle number of the top half (122)

Does that all make sense? Happy to answer questions.

If they ask for inner quartile range/IQR, that is the 3rd quartile minus the 1st quartile. 122 - 68 = 54
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His first payment is $100, thus a₁ = 100.

the next "term", month will be 1.1 times more than the one before, namely r = 1.1, the common ratio.

\bf \qquad \qquad \textit{sum of a finite geometric sequence}
\\\\
S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
r=\textit{common ratio}\\
----------\\
a_1=100\\
r=1.1\\
n=20
\end{cases}

\bf \sum\limits_{i=1}^{20}~100(1.1)^{i-1}\qquad \qquad\qquad  \qquad S_{20}=100\left( \cfrac{1-1.1^{20}}{1-1.1} \right)
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S_{20}=100\left( \cfrac{1-\stackrel{\approx}{6.727499949}}{-0.1} \right)\implies S_{20}\approx 100(57.27499949)
\\\\\\
S_{20}\approx 5727.4999493256

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