Question

please answer to my question​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Class Interval

• 0 - 10
• 10 - 20
• 20 - 30
• 30 - 40

Frequency

• 0
• 12 -- $f_0$ )
• 20 --( $f_1$)
• 9 --- ( $f_2$ )
• 4

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Now we have to take a higest frequency, Highest Frequency is 20 we can write it like this $20f$ and the Procedure frequency consider as 12 $f_0$ , And Successor frequency is Can be Consider as 9 $f_2$.

○ Formula of Mode

• $\boxed{ \displaystyle \: l + ( \frac{2f_1 - f_0}{2f_1 - f_0 - f_2} )}$

We Know higest frequency is 20 so, Corresponding Class interval is 20 - 30 So take it's lower limit Which is 20

• $l = 20$

The Size of Class interval is 10

• $h = 10$

Now We have,

• $l = 20$
• $f_1 = 20$
• $f_0 = 12$
• $f_2 = 9$
• $h = 10$

Put these given Values in formula

$\rightarrow \bf \: 20 + ( \frac{20 - 12}{2 \times 20 - 2 - 9} ) \times 10 \\ \\ \rightarrow \bf20 + ( \frac{8}{40 - 21} ) \times 10 \\ \\ \rightarrow \bf20 + \frac{8}{19} \times 10 \\ \\ \rightarrow \bf20 + \frac{ \cancel{80}}{ \cancel{19}} \\ \\ \rightarrow \bf20 + 4.21 \\ \\ : \implies \boxed{ \green{ \sf24.21}}$

24.21 is the mode of the following distribution table .

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