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Sonbull [250]
3 years ago
9

What is the median value of the data set shown on the line plot? Enter your answer in the box.

Mathematics
2 answers:
KengaRu [80]3 years ago
8 0

Answer:

Median is 110

Step-by-step explanation:

Let us write the data shown on the line plot

They are in ascending order as follows:

70,80,90,90,100,100,100,110,110,110,110,120,1200,120,120,130,130,140

There are 18 entries

The middle entries are 9 and 10

Hence median = average of 9th and 10th

= 110

kotegsom [21]3 years ago
5 0
The median value of this data set is 110 (:
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Write the explicit formula that represents the geometric sequence -2,8,-32,128.
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Answer:

a_{n} = - 2(-4)^{n-1}

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The n th term ( explicit formula ) for a geometric sequence is

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where a is the first term and r the common ratio

Here a = - 2 and r = 8 ÷ - 2 = - 4, thus

a_{n} = - 2(-4)^{n-1}

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Prove that the diagonals of a parallelogram bisect each other​
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Answer:

[ See the attached picture ]

The diagonals of a parallelogram bisect each other.

✧ Given : ABCD is a parallelogram. Diagonals AC and BD intersect at O.

✺ To prove : AC and BD bisect each other at O , i.e AO = OC and BO = OD.

Proof :\begin{array}{ |c| c |  c |  } \hline \tt{SN}& \tt{STATEMENTS} & \tt{REASONS}\\ \hline 1& \sf{In  \: \triangle ^{s}  \:AOB \: and \: COD  } \\  \sf{(i)}&  \sf{ \angle \: OAB =  \angle \: OCD\: (A)}& \sf{AB \parallel \: DC \: and \: alternate \: angles} \\  \sf{(ii)} &\sf{AB = DC(S)}& \sf{Opposite \: sides \: of \: a \: parallelogram} \\  \sf{(iii)} &\sf{ \angle \: OBA=  \angle \: ODC(A)} &\sf{AB \parallel \:DC \: and \: alternate \: angles} \\  \sf{(iv)}& \sf{ \triangle \:AOB\cong \triangle \: COD}& \sf{A.S.A \: axiom}\\ \hline 2.& \sf{AO = OC \: and \: BO = OD}& \sf{Corresponding \: sides \: of \: congruent \: triangle}\\ \hline 3.& \sf{AC \: and \: BD \: bisect \: each \: other \: at \: O}& \sf{From \: statement \: (2)}\\ \\ \hline\end{array}.          Proved ✔

♕ And we're done! Hurrayyy! ;)

# STUDY HARD! So, Tomorrow you can answer people like this , " Dude , I just bought this expensive mobile phone but it is not that expensive for me" [ - Unknown ] :P

☄ Hope I helped! ♡

☃ Let me know if you have any questions! ♪

\underbrace{ \overbrace  {\mathfrak{Carry \: On \: Learning}}} ☂

▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁

5 0
3 years ago
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