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ololo11 [35]
3 years ago
5

collect data (in decibels) of 10 different things in term of loudness. Arrange the data in ascending order & find mean, medi

an & mode for the same.
Mathematics
1 answer:
ss7ja [257]3 years ago
6 0
Whisper-20dB
Quiet Residence-30dB
Soft stereo in Residence-40dB
Average Speech-60dB
Cafeteria-80dB
Pneumatic Jackhammer-90dB
Loud crowd noise-100dB
Accelerating Bike-100dB
Rock concert-120dB
Jet Engine (75 ft away)-140dB

(1)Mode=100dB (since 100 is occurring the maximum no. of times, i.e. it has the highest frequency of 2)
(2)Mean=sum of observations/Total number of observations
=(20+30+40+60+80+90+100+100+120+140)/10
=770/10=10dB
(3)Median= 1/2[{n/2}th observation + {(n/2)+1}th observation], where, n=total no. of observations
So, 1/2[{10/2}th observation + {(10/2)+1}th observation]
=1/2[5th observation+6th observation]
=1/2[80+90]                    [Because:5th observation=80dB and 6th=90dB]
=1/2(170)
=85
Therefore, median=85dB

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Answer:

2 batches of orange water = 6 tablespoon of yellow water and 2 tablespoon of red water

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Step-by-step explanation:

1 batch

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Yellow water : Red water = 3 : 1

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4 batches of orange water = 12 tablespoon of yellow water and 4 tablespoon of red water

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3 years ago
Using a scale of 4cm to 1 unit on both axis, what does that mean?​
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The length of a rectangular carpet is 4 feet greater than twice its width. If the area is 48 square feet, find the carpet’s leng
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Helppp me plsssssssss<br><br>​
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Answer:

The class 35 - 40 has maximum frequency. So, it is the modal class.

From the given data,

  • \sf \:\:\:\:\:\:\:\:\:\:x_{k}=35
  • \sf \:\:\:\:\:\:\:\:\:\:f_{k}=50
  • \sf \:\:\:\:\:\:\:\:\:\:f_{k-1}=34
  • \sf \:\:\:\:\:\:\:\:\:\:f_{k+1}=42
  • \sf \:\:\:\:\:\:\:\:\:\:h=5

{\bf \:\: {By\:using\:the\: formula}} \\ \\

\:\dag\:{\small{\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}}} \\ \\

\sf \:\:\:\:\:\:\:\:\:= 35+ {\bigg(5 \times \dfrac{(50 - 34)}{ ( 2 \times 50 - 34 - 42)}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= 35 +{\bigg(5 \times \dfrac{16}{24}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= {\bigg(35+\dfrac{10}{3}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(35 + 3.33) =.38.33 \\ \\

\:\:\sf {Hence,}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ \large{\underline{\mathcal{\gray{ mode\:=\:38.33}}}} \\ \\

{\large{\frak{\pmb{\underline{Additional\: information }}}}}

MODE

  • Most precisely, mode is that value of the variable at which the concentration of the data is maximum.

MODAL CLASS

  • In a frequency distribution the class having maximum frequency is called the modal class.

{\bf{\underline{Formula\:for\: calculating\:mode:}}} \\

{\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}} \\ \\

Where,

\sf \small\pink{ \bigstar} \: x_{k}= lower\:limit\:of\:the\:modal\:class\:interval.

\small \blue{ \bigstar}\sf \: f_{k}=frequency\:of\:the\:modal\:class

\sf \small\orange{ \bigstar}\: f_{k-1}=frequency\:of\:the\:class\: preceding\:the\;modal\:class

\sf \small\green{ \bigstar}\: f_{k+1}=frequency\:of\:the\:class\: succeeding\:the\;modal\:class

\small \purple{ \bigstar}\sf \: h= width \:of\:the\:class\:interval

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