Answer:
Skewed to the right
Step-by-step explanation:
As we can see in the table, the first three intervals have smaller frequency and last 4 intervals have higher frequency values. When the class intervals and frequency will be plotted on graph while taking class intervals on x-axis and frequency at y-axis, the graph will be skewed to the right because of the larger frequency values in the last intervals..
The answer to (r -3s2t)4 is 4r- 24st
Step :
5 x (-4) - 1 x 13 -20 - 13 -33
--------------------- = ---------- = -----
2 x 5 + 1 11 11
Ans ; -33/11
Answer:
The length and width of the plot that will maximize the area of the rectangular plot are 54 ft and 27 ft respectively.
Step-by-step explanation:
Given that,
The length of fencing of the rectangular plot is = 108 ft.
Let the longer side of the rectangular plot be x which is also the side along the river side and the width of the rectangular plot be y.
Since the fence along the river does not need.
So the total perimeter of the rectangle is =2(x+y) -x
=2x+2y-y
=x+2y
So,
x+2y =108
⇒x=108 -2y
Then the area of the rectangle plot is A = xy
A=xy
⇒A= (108-2y)y
⇒ A = 108y-2y²
A = 108y-2y²
Differentiating with respect to x
A'= 108 -4y
Again differentiating with respect to x
A''= -4
For maximum or minimum, A'=0
108 -4y=0
⇒4y=108

⇒y=27.

Since at y= 27, A''<0
So, at y=27 ft , the area of the rectangular plot maximum.
Then x= (108-2.27)
=54 ft.
The length and width of the plot that will maximize the area of the rectangular plot are 54 ft and 27 ft respectively.