This histogram shows the percentage of the population that smokes tobacco for 123 countries. Which one of the following values is the most reasonable estimate for the IQR?

9.5

Step-by-step explanation:

The formula for interquartile range = Third Quartile - 1st Quartile

= Q3 - Q1

Looking at the histogram given, the class width = 6

Step 1

We find the number of values = n

n = 1 + 12 + 26 + 27 + 31 + 19 + 6 + 1

= 123

Frequency Cumulative Frequency

1 1

12 13

26 39

27 66

31 97

19 116

6 122

1 123

Step 2

We find Q1 ( First Quartile)

Q1 =1/4 (n + 1) value

Q1 = 1/4(124) value

= 31st value

The first Quartile = 31st value

31 - 13 = 18

31st value = 12 + (18/26 × class width)

= 12 +( 18/26 × 6)

First Quartile = 16.153846154

Third Quartile =Q3 =3/4 (n + 1) value

Q1 = 3/4(124) value

= 93rd value

The first Quartile = 93rd value

93 - 66 = 27

93rd value = 1 + (27/97 × class width)

=24 +( 27/97 × 6)

Third Quartile= 25.670103093

Interquartile range = Q3 - Q1

= 25.670103093 - 16.153846154

= 9.5162569392

Approximately = 9.5

Inter Quartile range = 9.5

perimeter of the first rectangular board is 9 feet.

step-by-step explanation:

let p = perimeter of first board.

l = length of first board

w = width of first board

length of first board = 2 * width of first board - 3

l = 2w - 3

perimeter of first board = 2l + 2w

since l = 2w-3, then

p = 2*(2w-3) + 2w which equals 4w-6 + 2w which equals 6w-6

length and width of the second board are reciprocals of corresponding length and width of the first board.

length of second board = 1/(2w-3)

width of second board = 1/w

perimeter of second board is 1/5th perimeter of the first board.

perimeter of second board = p/5

perimeter of the second board = (2/(2w-3) + 2/w)

since the perimeter of the first board is 5 times the perimeter of the second board, this means that:

6w-6 = 5*(2/(2w-3) + 2/w)

which is the same as:

6w-6 = (10/(2w-3) + 10/w)

if we multiply both sides of this equation by (2w-3), we get:

(6w-6)*(2w-3) = 10 + 10*(2w-3)/w

if we multiply both sides of this equation by w, we get:

(6w-6)*(2w-3)*w = 10*w + 10*(2w-3)

we can simplify this to become:

(6w-6)*(2w-3)*w = 10w + 20w - 30

which becomes:

(6w-6)*(2w-3)*w = 30w-30

if we divide both sides of this equation by (6w-6), we get:

(2w-3)*w = (30w-30)/(6w-6)

which becomes:

(2w-3)*w = 5

this can be simplified to:

2w^2 - 3w = 5

subtract 5 from both sides of this equation to get:

2w^2 - 3w - 5 = 0

which can be factored into:

(2w-5)*(w+1) = 0

which makes:

w = 5/2

or

w = -1.

w can't be negative so the only possible answer is w = 5/2.

if w = 5/2, then

p = 2w-3 = 5-3 = 2

we have:

p = 2

w = 5/2

p = 2l + 2w gets p = 4 + 5 = 9

the perimeter of the first board is 9 feet.

since the perimeter of the second board is 1/5 the perimeter of the first board, the perimeter of the second board is 9/5.

to see if that's correct, we substitute known values for l and w into the second board.

the length of the second board is 1/l.

this becomes 1/2.

the width of the second board is 1/w

this bgecomes 1/(5/2) = (2/5)

let l2 = length of second board.

let w2 = width of second board.

l2 = (1/2)

w2 = (2/5)

let p2= perimeter of second board.

then p2 = 9/5

since p2 = 2*l2 + 2*w2, we should get p2 = 9/5 using the dimensions of l2 and w2.

2 * l2 = 2 * (1/2) = 1

2 * w2 = 2 * (2/5) = 4/5

1 is the same as 5/5

5/5 + 4/5 = 9/5

p2 checks out good.

the dimensions as given are good.

answer to the problem is:

perimeter of the first rectangular board is 9 feet.

answer: a. 74

step-by-step explanation:

it shows divided evenly, so add the 37+37 = 74