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3 years ago

ANSWER ONLY IF YOU KNOW ( no links or ill report you)

Mathematics

1 answer:

Elis [28]3 years ago

4 0

Step-by-step explanation:

$C=2\pi r$

$C=2*\pi *28$

$C=56\pi$

Yes, it is because the circumference is always $2\pi$ times the amount of the radius.

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Helppp me plsssssssss<br><br>

Oliga [24]

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,

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

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

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

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

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

7 0

3 years ago

biked 4 ½ miles today, and biked 6 ¾ miles. How many times further did bike compared to ’ ride? SHOW YOUR WORK.

Tresset [83]

Answer: 2 1/4

Step-by-step explanation: first off change the 1/2 to 2/4. then subtract 6 3/4 to 4 2/4. which gives you 2 1/4

5 0

2 years ago

Read 2 more answers

There are 40 rows of seats in the Varsity Theater, with 20 seats in each row. During

Natali5045456 [20]

Answer: $12x - $275 = p

Step-by-step explanation:

$12x < how much it cost per seat/ticket

$275< How much they paid to rent the place

p< The profit

7 0

3 years ago

Andrew is evaluating his current financial position by totaling his monthly cash inflows and outflows. His cash inflows each mon

jeka57 [31]

Answer:

Andrew has $15 left over each month

Step-by-step explanation:

Inflows=$240+$60=$300/mo.

Outflows=$170+$70+45=$285/mo.

Inflows-Outflows=left over

$300-$285=$15

8 0

3 years ago

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

ss7ja [257]

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

6 0

3 years ago