Fire load (mj/m²) is the heat energy that could be released per square meter of floor area by combustion of contents and the structure itself.
the following cumulative percentages are for fire loads in a sample of 388 rooms:

value: 0 150 300 450 600 750 900
cumulative %: 0 19.1 37.6 62.4 77.2 86.9 93.9
value: 1050 1200 1350 1500 1650 1800 1950
cumulative %: 95.4 97.9 98.4 99.5 99.6 99.8 100.0

a. construct a relative frequency histogram and comment on interesting features.
b. what proportion of reloads are less than 600? at least 1200?
c. what proportion of the loads are between 600 and 1200?

Step-by-step explanation:

Hello!

a. See attachment.

b. "The proportion of reloads that are less than 600."

Symbolically:

P(X<600)

This means that is all the accumulated proportions without including the relative frequency of the value 600.

P(<600)= 0.624

"At least 1200"

If you say "at least" you could also say "1200 or more", symbolically:

P(X≤1200)= 0.997

It represents the accumulated relative frequencies until and including 1200

c. To calculate the proportion between 600 and 1200, you have to calculate the difference of the accumulated frequency until the biggest value and the accumulated frequency until the smallest value.

Symbolically:

P(600 ≤ X ≤ 1200)= P(X ≤ 1200) - P(X ≤ 600) = 0.997 - 0.772= 0.225

I hope it helps!

answer:   x = 3 + √11     and     x = 3 - √11

step-by-step explanation:

x² - 6x + 9 = 11

(x - 3)² = 11

x - 3 = ±√11

    x = 3 ± √11

x = 3 + √11     and     x = 3 - √11

see explanation

step-by-step explanation:

a

(6r + 7) + (13 + 7r) ( remove brackets and collect like terms )

= 6r + 7 + 13 + 7r

= 13r + 20

b

remove brackets and collect like terms

13 - r - 1 + r

= 12 - r

c

remove brackets and collect like terms

= - 8 - r + 2r - 4

= - 12 + r

d

remove brackets and collect like terms

= 7r - - - 6r

[-   - = - - = - ]

= r -