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tamaranim1 [39]
3 years ago
15

The first quartile of the data set represented by the box plot is (8, 9, 12,14) . The median of the data set is (8, 12,16,19)

Mathematics
2 answers:
Andrei [34K]3 years ago
7 0
The first quartile is 12. The median is 16!
blondinia [14]3 years ago
5 0

Answer: 16

Step-by-step explanation:

  • The median is the middle number of a set of data  (if there are an odd number of data values) or the average of the two middle numbers (if there are an even number of data values).

  • A  box-whisker plot gives us information of the shape, variability, and median of a data set.

From the given box-whisker plot a vertical line goes through the box at 16 i.e. the median.

Hence, the median of the data set = 16.

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Binomial Expansion/Pascal's triangle. Please help with all of number 5.
Mandarinka [93]
\begin{matrix}1\\1&1\\1&2&1\\1&3&3&1\\1&4&6&4&1\end{bmatrix}

The rows add up to 1,2,4,8,16, respectively. (Notice they're all powers of 2)

The sum of the numbers in row n is 2^{n-1}.

The last problem can be solved with the binomial theorem, but I'll assume you don't take that for granted. You can prove this claim by induction. When n=1,

(1+x)^1=1+x=\dbinom10+\dbinom11x

so the base case holds. Assume the claim holds for n=k, so that

(1+x)^k=\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k

Use this to show that it holds for n=k+1.

(1+x)^{k+1}=(1+x)(1+x)^k
(1+x)^{k+1}=(1+x)\left(\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k\right)
(1+x)^{k+1}=1+\left(\dbinom k0+\dbinom k1\right)x+\left(\dbinom k1+\dbinom k2\right)x^2+\cdots+\left(\dbinom k{k-2}+\dbinom k{k-1}\right)x^{k-1}+\left(\dbinom k{k-1}+\dbinom kk\right)x^k+x^{k+1}

Notice that

\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!}{\ell!(k-\ell)!}+\dfrac{k!}{(\ell+1)!(k-\ell-1)!}
\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)}{(\ell+1)!(k-\ell)!}+\dfrac{k!(k-\ell)}{(\ell+1)!(k-\ell)!}
\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)+k!(k-\ell)}{(\ell+1)!(k-\ell)!}
\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(k+1)}{(\ell+1)!(k-\ell)!}
\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{(k+1)!}{(\ell+1)!((k+1)-(\ell+1))!}
\dbinom k\ell+\dbinom k{\ell+1}=\dbinom{k+1}{\ell+1}

So you can write the expansion for n=k+1 as

(1+x)^{k+1}=1+\dbinom{k+1}1x+\dbinom{k+1}2x^2+\cdots+\dbinom{k+1}{k-1}x^{k-1}+\dbinom{k+1}kx^k+x^{k+1}

and since \dbinom{k+1}0=\dbinom{k+1}{k+1}=1, you have

(1+x)^{k+1}=\dbinom{k+1}0+\dbinom{k+1}1x+\cdots+\dbinom{k+1}kx^k+\dbinom{k+1}{k+1}x^{k+1}

and so the claim holds for n=k+1, thus proving the claim overall that

(1+x)^n=\dbinom n0+\dbinom n1x+\cdots+\dbinom n{n-1}x^{n-1}+\dbinom nnx^n

Setting x=1 gives

(1+1)^n=\dbinom n0+\dbinom n1+\cdots+\dbinom n{n-1}+\dbinom nn=2^n

which agrees with the result obtained for part (c).
4 0
3 years ago
You are going shopping. You are given 40 dollars to buy clothing. You buy
Oduvanchick [21]

Answer:

6 dollars

Step-by-step explanation:

40-10=30

30/5= 6

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Musya8 [376]

Answer:

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3 years ago
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Find all the zeros of the equation.<br><br> -3<img src="https://tex.z-dn.net/?f=%20%20x%5E%7B4%7D%20%7D%20" id="TexFormula1" tit
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make the equation equal to 0
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3 years ago
Rene made a 90 % on her math test. If she answered 20 problems correctly , how many problems were on the test ?
MakcuM [25]

Answer:

Rene made = 90 on her math test

If she answer =20problems correctly

we know,

how many problem were there = ?

there where = 90-20

= 70

There where 70 problems

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