1. wetlands offer a diversity of benefits. they provide habitat for wildlife, spawning grounds for u. s. commercial fish, and
renewable timber resources. in the last 200 years, the united states has lost more than half its wetlands. environmental
almanac gives the percentage of wetlands lost in each state in the last 200 years. for the lower 48 states, the percentage loss
of wetlands per state is as follows:
46 37 36 42 81 20 73 59 35 50 87 52
24 27 38 56 39 74 56 31 27 91 46 9
54 52 30 33 28 35 35 23 90 72 85 42
59 50 49 48 38 60 46 87 50 89 49 67
make a stem-and-leaf display of these data. be sure to indicate the scale.
how are the percentages distributed? is the distribution skewed? are there any gaps?

Step-by-step explanation:

Percentage of wetlands lost in each state is shown in stem and leaf plot as 0-10, 10-20 etc.

We have gaps for stem 1.

Stem and leaf plot would be as follows:

StemLeaf

09

1  

20 3 4 7 7 8

30 1 3 5 5 5 6 7 8 8 9

42 2 6 6 6 8 9 9

50 0 0 2 2 4 6 6 9 9

60 7

72 3 4

81 5 7 7 9

90 1

Let us calculate the summary for the data as

Minimum:9

Maximum:91

Count:48

Sum:2407

Mean:50.15

Median:48.5

Mode:46, 35, 50

Standard Deviation:20.72

Variance:429.1

Skewness = 0.466538

Yes this is skewed.

18

step-by-step explanation:

ik this stuph and i am smart donkey

tell you what. i'll find the length of the medians and you can fill in the blanks where they belong.

br

b = (2a,0)

r = (a/2,b/2) remember these things are medians. they go to the 1/2 way point of the line opposite the vertex they face.

br^2 =   (2a - a/2)^2 + (0 - b/2)^2

br^2 = (3/2 a) ^2 + b^2 / 4

br^2 = 9/4 a^2 + b^2 / 4   we need to find some relationship between a and b.

let's try ab = bc  

ab = 2a

bc = (2a - a)^2 + (b - 0)^2

bc = sqrt(a^2 + b^2)

ab = bc

2a = sqrt(a^2 + b^2) square both sides.

4a^2 = a^2 + b^2 subtract a^2 from both sides.

sqrt(3a^2) = sqrt(b^2)

sqrt(3)a = b

let's put b into br^2

br^2 = 9/4 a^2 + 3b^2 / 4

br^2 = 12 a^2 / 4

br^2 = 3a^2 

br = sqrt(3) * a

cp

c = (a,b) ; p = (a,0) this is another application of the distance formula, and it is a good one.

cp^2 = (a - a)^2 + (b - 0)^2

cp^2 = 0 + b^2

cp^2 = b^2 which you can see from the diagram. 

cp = sqrt(b^2)

cp = b but b = sqrt(3) * a 

cp = sqrt(3)*a

aq

a = (0,0)

q = (3/2) a, b/2

aq^2 = (3a/2 -0)^2 + (b/2 - 0)^2

aq^2 = 9a^2/4 + b^2/4

but b = sqrt(3) * a 

aq^2 = 9a^2 / 4 + (sqrt(3)*a)^2 /4

aq^2 = 9 a^2/ 4 + 3 a^2 / 4

aq^2 = 12 a^2/4

aq^2 = 3 a^2

aq = (sqrt  3) * a which agrees with the other two.