Answer:
For maximum area of the rectangular exercise run dimensions will be 50ft by 25ft.
Step-by-step explanation:
Let the length of the rectangular exercise run = l ft
and width of the run = w ft
Sinoman has to cover a rectangular exercise run from three sides with the fencing material,
So length of the material = (l + 2w) ft
l + 2w = 100
l = 100 - 2w --------(1)
Area of the rectangular area covered = Length × width
A = lw
A = w(100 - 2w) [(l = 100 - 2w)from equation (1)
For maximum area we find the derivative of area and equate it to zero.
![\frac{dA}{dw}=\frac{d}{dw}[w(100-2w)]](https://tex.z-dn.net/?f=%5Cfrac%7BdA%7D%7Bdw%7D%3D%5Cfrac%7Bd%7D%7Bdw%7D%5Bw%28100-2w%29%5D)

A' = 100 - 4w
For A' = 0
100 - 4w = 0
4w = 100
w = 25 ft
From equation (1)
l = 100 - 2w
l = 100 - 2×(25)
l = 50 ft
Therefore, for maximum area of the rectangular exercise run dimensions will be 50ft by 25ft.
Answer:
Volume = 3/4 x 1/2 x 4 = 3/2 inches³
Answer:
248.96
Step-by-step explanation:
From this regression output we have the MS Residual or mean squared error to be equal to 61983.1
the question requires us to find the standard error of the estimate. The standard error of the estimate can be gotten by finding the square root of the MSE.

= 248.96
the standard error of the estimate = 248.96
thank you!