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Afina-wow [57]
3 years ago
7

Find the interquartile range (IQR) of the data set. 1,1,3,4,4,5,5,5,6,7,9

Mathematics
2 answers:
iragen [17]3 years ago
8 0
The interquartile range is 3
Vlad [161]3 years ago
8 0

Answer:

The interquartile range is 3.

Step-by-step explanation:

1, 1, 3, 4, 4, 5, 5, 5, 6, 7, 9

Median: 5

(Q1) Lower quartile: 3

(Q2) Upper quartile: 6

(IQR) Interquartile range: 6 - 3 = 3

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Write the inequality shown in the diagram on the left.
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7 0
3 years ago
Evaluate each finite series for the specified number of terms. 1+2+4+...;n=5
zaharov [31]

Answer:

31

Step-by-step explanation:

The series are given as geometric series because these terms have common ratio and not common difference.

Our common ratio is 2 because:

1*2 = 2

2*2 = 4

The summation formula for geometric series (r ≠ 1) is:

\displaystyle \large{S_n=\frac{a_1(r^n-1)}{r-1}} or \displaystyle \large{S_n=\frac{a_1(1-r^n)}{1-r}}

You may use either one of these formulas but I’ll use the first formula.

We are also given that n = 5, meaning we are adding up 5 terms in the series, substitute n = 5 in along with r = 2 and first term = 1.

\displaystyle \large{S_5=\frac{1(2^5-1)}{2-1}}\\\displaystyle \large{S_5=\frac{2^5-1}{1}}\\\displaystyle \large{S_5=2^5-1}\\\displaystyle \large{S_5=32-1}\\\displaystyle \large{S_5=31}

Therefore, the solution is 31.

__________________________________________________________

Summary

If the sequence has common ratio then the sequence or series is classified as geometric sequence/series.

Common Ratio can be found by either multiplying terms with common ratio to get the exact next sequence which can be expressed as \displaystyle \large{a_{n-1} \cdot r = a_n} meaning “previous term times ratio = next term” or you can also get the next term to divide with previous term which can be expressed as:

\displaystyle \large{r=\frac{a_{n+1}}{a_n}}

Once knowing which sequence or series is it, apply an appropriate formula for the series. For geometric series, apply the following three formulas:

\displaystyle \large{S_n=\frac{a_1(r^n-1)}{r-1}}\\\displaystyle \large{S_n=\frac{a_1(1-r^n)}{1-r}}

Above should be applied for series that have common ratio not equal to 1.

\displaystyle \large{S_n=a_1 \cdot n}

Above should be applied for series that have common ratio exactly equal to 1.

__________________________________________________________

Topics

Sequence & Series — Geometric Series

__________________________________________________________

Others

Let me know if you have any doubts about my answer, explanation or this question through comment!

__________________________________________________________

7 0
3 years ago
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