Identify the class width, class midpoints, and class boundaries for the given frequency distribution. daily low tempurature frequencey daily low tempurature full data set32-24 1 44-46 735-37 3 47-49 738-40 5 50-52 141-43 11

I . class width is 3

ii. Class midpoints are 33,36,39,42,45,48,51.

iii. Class boundaries are: 31.5-34.5, 34.5-37.5, 37.5-40.5, 40.5-43.5, 43.5-46.5, 46.5-49.5, 49.5-52.5

Step-by-step explanation:

I. The class width is the difference between upper class boundary and lower class boundary.

ii. Class midpoints is the summation of upper and lower class limits, divided by 2.

iii. Class boundaries is the subtraction of 0.5 from the lower classe limits and the addition of 0.5 to upper class limits.

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2