Evaluate an appropriate measure of dispersion for the following data: class interval. no of people
<50
50- 70
70-90
90- 100
110 - 130
130 - 150
>150​

b

step-by-step explanation:

  answer:

  write 548 as 500 + 40 + 8. multiply each addend by 5. then add the three products to get 2,740.

answer:

given the linear inequality :

add 2y to both sides we get;

simplify:

add 12 to both sides of an equation;

simplify:

divide both sides by 2 we get;

or

the related equation for the inequality equation [1] is;

y-intercepts defined as the graph crosses the y-axis.

substitute x =0 in [2] to solve for y;

y = 6

x-intercepts defined as the graph crosses the x-axis.

substitute value y = 0 to solve for x;

subtract 6 from both sides we get

multiply both sides by 2 we get;

or

x = -12

now, plot these point (0, 6) and (-12, 0) on graph as shown below and shade the graph below the solid line.