If the median of the following data is 33 then, find the value of m Class Frequency 0-15 17 15-30 35 30-45 40 45-60 18 60-75 m 75-90 2​

Answer:

Refer to the attachment↑

Hope it helps...

Answer:

(a)

[tex]2 {x}^{2} - 19x + 24 = 0[/tex]

(b)

[tex]x = \frac{3}{2} [/tex]

Step-by-step explanation:

Refer to the attachment↑↑

Answer: dimensions of pool are 4m and 9m.

Step-by-step explanation:

• Given,the width of the path is x m.

• And the outside edges of pool ,easures 7m and 12m.

• So,the dimensions of pool will decrease by 2x i.e., x from each side.

• Dimensions of pool= (7-2x)m and (12-2x)m

• Given,area of pool= 36 sq.m

• Now,according to question

       (7-2x)×(12-2x)=36

    ⇒7(12-2x)-2x(12-2x)=36

    ⇒84-14x-24x+4=36

    ⇒4-38x+84-36=0

    ⇒4-38x+48=0

    ⇒2(2-19x+24)=0

    ⇒(2-19x+24)=0

    ⇒2-16x-3x+24=0

    ⇒2x(x-8)-3(x-8)=0

    ⇒(2x-3)(x-8)=0

    ⇒If 2x-3=0 then x=  and if x-8=0 then x=8(not possible)

     ∴ x=1.5m

• (7-2x)=(7-2*1.5)=(7-3)=4m and (12-2x)=(12-3)=9m

• So, the dimensions of pool are (7-2x)m and (12-2x)m i.e.,4m and 9m.

Answer:

(a)

[tex]2 {x}^{2} + 19x + 24 = 0[/tex]

(b)

[tex]x = \frac{3}{2} [/tex]

Step-by-step explanation:

Refer to the attachment↑↑

Hope it helps...

[tex]\large\underline{\sf{Solution-}}[/tex]

We know,

Mode of the continuous series is given by

[tex]{\boxed{ \boxed{\sf{Mode = l + \bigg(\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \bigg) \times h }}}} \\ \\ [/tex]

where,

l is lower limit of modal class.

[tex] \sf{f_1} [/tex] is frequency of modal class

[tex] \sf{f_0} [/tex] is frequency of class preceding modal class

[tex] \sf{f_2} [/tex] is frequency of class succeeding modal class

h is class height.

Now, from given data we concluded that

[tex]\rm \: f_0 = 6 \\ [/tex]

[tex]\rm \: f_1 = f \\ [/tex]

[tex]\rm \: f_2 = 8 \\ [/tex]

[tex]\rm \: l = 65 \\ [/tex]

[tex]\rm \: h = 15 \\ [/tex]

[tex]\rm \: Mode = 75 \\ [/tex]

So, on substituting the values in above formula, we get

[tex]\rm \: 75 = 65 + \bigg[\dfrac{f - 6}{2 \times f - 6 - 8} \bigg] \times 15\\ [/tex]

[tex]\sf \: 75 - 65 = \dfrac{f - 6}{2f - 14} \times 15 \\ \\ [/tex]

[tex]\sf \: 10 = \dfrac{f - 6}{2f - 14} \times 15 \\ \\ [/tex]

[tex]\sf \: 2 = \dfrac{f - 6}{2f - 14} \times 3 \\ \\ [/tex]

[tex]\sf \: 2(2f - 14) = 3(f - 6) \\ \\ [/tex]

[tex]\sf \: 4f - 28 = 3f -18 \\ \\ [/tex]

[tex]\sf \: 4f - 3f = 28 -18 \\ \\ [/tex]

[tex]\bf\implies \:f = 10 \\ \\ [/tex]

[tex]\rule{190pt}{2pt} \\ [/tex]

Additional Information :-

1. Mean using Direct Method

[tex]\boxed{ \rm{ \:Mean = \dfrac{ \sum f_i x_i}{ \sum f_i} \: }} \\ \\ [/tex]

2. Mean using Short Cut Method

[tex]\boxed{ \rm{ \:Mean =A + \dfrac{ \sum f_i d_i}{ \sum f_i} \: }} \\ \\ [/tex]

3. Mean using Step Deviation Method

[tex]\boxed{ \rm{ \:Mean =A + \dfrac{ \sum f_i u_i}{ \sum f_i} \times h \: }} \\ \\ [/tex]

Answer:

• The mode of a grouped frequency distribution is 75 and the modal class is 65 - 80.
• The frequency of the class preceding the modal class is 6 and the frequency of the class succeeding the modal class is 8.

• What is frequency of the modal class.

[tex]\clubsuit[/tex] Mode Formula :-

[tex]\bigstar \: \: \sf\boxed{\bold{Mode =\: l + \bigg[\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2}\bigg] \times h}}\: \: \: \bigstar\\[/tex]

where,

• l = Lower Limit of the Modal Class
• h = Class Size of the Class Interval
• f₁ = Frequency of the Modal Class
• f₀ = Frequency of the Class Preceding the Modal Class
• f₂ = Frequency of the Class Succeeding the Modal Class

Let,

[tex]\mapsto \bf Frequency_{(Modal\: Class)} =\: f_1\\[/tex]

Given :

• Mode = 75
• Lower limit = 65
• Class Size = 15
• Frequency of the class preceding = 6
• Frequency of the class succeeding = 8

According to the question by using the formula we get,

[tex]\implies \sf\boxed{\bold{Mode =\: l + \bigg[\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2}\bigg] \times h}}\\[/tex]

[tex]\implies \sf 75 =\: 65 + \bigg[\dfrac{f_1 - 6}{2f_1 - 6 - 8}\bigg] \times 15\\[/tex]

[tex]\implies \sf 75 =\: 65 + \bigg[\dfrac{f_1 - 6}{2f_1 - 14}\bigg] \times 15\\[/tex]

[tex]\implies \sf 75 =\: 65 + \dfrac{f_1 - 6}{2f_1 - 14} \times 15\\[/tex]

[tex]\implies \sf 75 =\: 65 + \dfrac{15(f_1 - 6)}{2f_1 - 14}\\[/tex]

[tex]\implies \sf 75 =\: 65 + \dfrac{15f_1 - 90}{2f_1 - 14}\\[/tex]

[tex]\implies \sf 75 - 65 =\: \dfrac{15f_1 - 90}{2f_1 - 14}[/tex]

[tex]\implies \sf 10 =\: \dfrac{15f_1 - 90}{2f_1 - 14}[/tex]

By doing cross multiplication we get,

[tex]\implies \sf 15f_1 - 90 =\: 10(2f_1 - 14)[/tex]

[tex]\implies \sf 15f_1 - 90 =\: 20f_1 - 140[/tex]

[tex]\implies \sf 15f_1 - 20f_1 =\: - 140 + 90[/tex]

[tex]\implies \sf {\cancel{-}} 5f_1 =\: {\cancel{-}} 50[/tex][tex]\implies \sf 5f_1 =\: 50[/tex]

[tex]\implies \sf f_1 =\: \dfrac{\cancel{50}}{\cancel{5}}[/tex]

[tex]\implies \sf f_1 =\: \dfrac{10}{1}[/tex]

[tex]\implies \sf\bold{f_1 =\: 10}[/tex]

Hence, the required frequency of the modal class is :

[tex]\dag[/tex] Frequency Of Modal Class :

[tex]\dashrightarrow \sf Frequency_{(Modal\: Class)} =\: f_1\\[/tex]

[tex]\dashrightarrow \sf\bold{\underline{Frequency_{(Modal\: Class)} =\: 10}}\\[/tex]

[tex]\therefore[/tex] The frequency of the modal class is 10 .