Answer:
since i dont have data this is the best i can do
Step-by-step explanation:
non-filled seats diveded by total number of seats, everything multilped by 100 gives you percentage of non-filled seats.
Filled seats diveded by total number of seats, everything multilped by 100 gives you percentage of filled seats.
total should add up to 100
Answer:
The lengths of Gajge’s runs have greater variability because there is a greater difference between his longest and shortest runs is the answer.
Step-by-step explanation:
given that Ty and Gajge are football players.
Carries is 15 for both and average is the same 4 for both.
But on scrutiny we find that maximum and minimum and 6 and 2 for Ty.
Hence range for Ty = 6-2 =4 (2 runs on eithre side of mean)
But for Gajge, highest is 19 and lowest is 2.
i.e. range = 19-2 =17 very much higher than that of Ty
The lengths of Gajge’s runs have greater variability because there is a greater difference between his longest and shortest runs.
First, we have to figure out the expression for the total price of potatoes. We find that by multiplying the price of potatoes per pound by how many pounds were bought. Robert bought x pounds of potatoes at $1.99 per pound. This can be written as:
$1.99x
Next, we have to find the equation for Robert's change. Change is defined as amount paid minus cost. In this scenario, Robert paid $10, and we already know the cost is $1.99x. So using this information, we can now write the equation:
$10 - $1.99x
So the answer is a.
This must be on the moon as the acceleration due to gravity in this equation must be around 1/8 that on earth. :) Anyway...
h=-2t^2+9t+11
A)
dh/dt=-4t+9, when velocity, dh/dt=0, it is the maximum height reached
dh/dt=0 only when 4t=9, t=2.25 seconds
h(2.25)=21.125 ft (21 1/8 ft)
B)
As seen in A), the time of maximum height was 2.25 seconds after the squirrel jumped.
C)
The squirrel reaches the ground when h=0...
0=-2t^2+9t+11
-2t^2-2t+11t+11=0
-2t(t+1)+11(t+1)=0
(-2t+11)(t+1)=0, since t>0 for this problem...
-2t+11=0
-2t=-11
t=5.5 seconds.
