Question

Plz solve Q 8,9 fast

Answer 1

8. here, highest frequency is 50.

so, modal class : 35 - 40,

lower limit of modal class, l = 35

limit size , h = 40 - 35 = 5

frequency of modal class, $f_1$ = 50

frequency of preceding modal class, $f_0$ = 34

frequency of proceeding modal class, $f_2$ = 42

now, mode = $l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h$

= 35 + (50 - 34)/(2 × 50 - 34 - 42) × 5

= 35 + 16/(100 - 76) × 5

= 35 + 10/3

= 38.33

9. Let small angle = x

larger angle = 18° + x

sum of two supplementary angles = 180°

or, 18° + x + x = 180°

or, 2x = 180° - 18° = 162°

or, x = 81°

hence, angles are ; 81°, 99°

Answer 2

Given: two supplementary angles, one is greate than other by 18°

To find: both the angles.

Solution:

• So, we have given that there are two angles, so let first angle be 'x' and second angle be 'y'.

• Since the sum of angles is supplementary, so

             x + y = 180°     ....................(i)

• Now, we have provided that one angle is greater than other by 18°, so

             x =  y + 18°     .....................(ii)

• Now, put the value of 'x' in the equation (i), we get

            y + 18° + y = 180°

• Solving further we get,

            2y = 180° - 18°

            =  162°

            y = 162/2

            = 81°

• So y = 81°

• Lets put y in equation (ii), we get,

            x = 81 + 18

             x = 99°

Answer:

             So the first angle is 81° and second angle is 99°.