<h2>
Maximum area is 25 m²</h2>
Explanation:
Let L be the length and W be the width.
Aidan has 20 ft of fence with which to build a rectangular dog run.
Fencing = 2L + 2W = 20 ft
L + W = 10
W = 10 - L
We need to find what is the largest area that can be enclosed.
Area = Length x Width
A = LW
A = L x (10-L) = 10 L - L²
For maximum area differential is zero
So we have
dA = 0
10 - 2 L = 0
L = 5 m
W = 10 - 5 = 5 m
Area = 5 x 5 = 25 m²
Maximum area is 25 m²
Answer:
Step-by-step explanation:
The missing image is attached below.
From the given information
The number of players is 44.
From the box plot;
The minimum weight = 154 pounds
The first quartile Q₁ = 159 which is 25% of the data below 159 pounds.
The second quartile Q₂ = 213 which is 50% of the data below 213 pounds
The third quartile Q₃ = 253 which is 75% of the data below 253 pounds
The maximum weight = 268 pounds.
Here, the value of 213 is the middle value and which signifies the median.
The 50% of the data are located both on the left side and the right side of the median.
Thus, the percentage of players weighting greater than or equal to 213 pounds is 50%.
Answer:
117 inches squared
Step-by-step explanation:
perimeter= 44 inches
L=W+4
44=w+w+(w+4)+(w+4)
44=4w+8
36=4w
9=width
9+4=13
width=9
Length=13
area=L*W
13*9=117
The coach can select the captain in 20 ways (i.e. from all the available 20 players), and he can select the co-captain in 19 ways (i.e. after selecting the captain, there will be 19 people left).
Therefore, the coach can select the captain and the co-captain in 20 x 19 = 380 ways.