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sergij07 [2.7K]
3 years ago
13

What are the 1st and 3rd quartiles? what is the interquartile range?

Mathematics
2 answers:
PolarNik [594]3 years ago
8 0

Answer:

The first quartile is 5

The third is 8

The interquartile is 3

Step-by-step explanation:

boblepop8 is correct

FrozenT [24]3 years ago
7 0

Answer:

The first quartile is 5

The third is 8

The interquartile is the difference between the first and third quartile.

The interquartile is 3

Step-by-step explanation:

First, make a list of all the numbers. To find the first quartile, you have to first find the median. The median is the first 7 in the list.

Everything before 7 is the lower percentile of the range.

The firsr quartile is the median of the lower percentile.

The lower list is:

2,3,4,4,5,6,6,6

The median of this list is 5.

The process for finding the third quartile is the same except the list is the higher percentile which is the list of numbers above the median.

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A figure is translated &lt;3, -3&gt;. Which translation will move the image back to the
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<h3>Answer:   < -3, 3 ></h3>

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8 0
3 years ago
If S_1=1,S_2=8 and S_n=S_n-1+2S_n-2 whenever n≥2. Show that S_n=3⋅2n−1+2(−1)n for all n≥1.
Snezhnost [94]

You can try to show this by induction:

• According to the given closed form, we have S_1=3\times2^{1-1}+2(-1)^1=3-2=1, which agrees with the initial value <em>S</em>₁ = 1.

• Assume the closed form is correct for all <em>n</em> up to <em>n</em> = <em>k</em>. In particular, we assume

S_{k-1}=3\times2^{(k-1)-1}+2(-1)^{k-1}=3\times2^{k-2}+2(-1)^{k-1}

and

S_k=3\times2^{k-1}+2(-1)^k

We want to then use this assumption to show the closed form is correct for <em>n</em> = <em>k</em> + 1, or

S_{k+1}=3\times2^{(k+1)-1}+2(-1)^{k+1}=3\times2^k+2(-1)^{k+1}

From the given recurrence, we know

S_{k+1}=S_k+2S_{k-1}

so that

S_{k+1}=3\times2^{k-1}+2(-1)^k + 2\left(3\times2^{k-2}+2(-1)^{k-1}\right)

S_{k+1}=3\times2^{k-1}+2(-1)^k + 3\times2^{k-1}+4(-1)^{k-1}

S_{k+1}=2\times3\times2^{k-1}+(-1)^k\left(2+4(-1)^{-1}\right)

S_{k+1}=3\times2^k-2(-1)^k

S_{k+1}=3\times2^k+2(-1)(-1)^k

\boxed{S_{k+1}=3\times2^k+2(-1)^{k+1}}

which is what we needed. QED

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Read 2 more answers
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