Answer:
a. The time to strike the ground is 5 seconds
b. The time for the ball to be over than 96 ft is 1 second
Step-by-step explanation:
a. When the ball thrown vertically up ward to reach its maximum height at velocity = 0 then complete its motion vertically down ward to strike the ground its displacement = zero
∴ s(t) = 0 ⇒ 
∴
5 - t = 0⇒ ∴ t = 5 seconds
b. To tind the time that the ball will be over 96 ft over the ground substitute in 




t = 3 sec. and t = 2 sec.
The ball will be at height 96 ft at time 2 sec. and 3 sec.
So the ball will be at height over than 96 = 3 - 2 = 1 sec.
Answer:
See below
Step-by-step explanation:
If you are squaring a number, and then taking the square root of it, you are essentially undoing the original operation:
![\displaystyle \sqrt{\biggr(\frac{4}{7}\biggr)^2}=\biggr[\biggr(\frac{4}{7}\biggr)^2\biggr]^{\frac{1}{2}}=\biggr(\frac{4}{7}\biggr)^{2*\frac{1}{2}}=\frac{4}{7}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csqrt%7B%5Cbiggr%28%5Cfrac%7B4%7D%7B7%7D%5Cbiggr%29%5E2%7D%3D%5Cbiggr%5B%5Cbiggr%28%5Cfrac%7B4%7D%7B7%7D%5Cbiggr%29%5E2%5Cbiggr%5D%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%3D%5Cbiggr%28%5Cfrac%7B4%7D%7B7%7D%5Cbiggr%29%5E%7B2%2A%5Cfrac%7B1%7D%7B2%7D%7D%3D%5Cfrac%7B4%7D%7B7%7D)
Hence, we are back starting with the original number
You should look for outliers. If there is an outlier then you should use median because if you use mean the data would be skewed.
Answer:
5/6
Step-by-step explanation:
1/3 has two more thirds until it's a whole
17/20 has three more twentieths until it's a whole
1/4 has four more fourths until it's a whole
5/6 has <u>one</u> more sixth until it's a whole