The function for this problem is:
h(t) = -16(t)^2 + vt + s
h= the height
t= time
v= velocity
s= starting height
With the information given, we know that the starting height is 0, since it was from the ground, and the velocity of the ball is 35 feet per second. Inserting the these information into the equation, we get:
h(t) = -16(t)^2 + 35t
Now the question asks to find the maximum height. It can be done by using a grapher to graph the maximum of the parabola. It could also be done by finding the vertex, which would be the maximum, of the graph by using x= -b/(2a), where b is equal to 35 and a is equal to -16. We get x=35/32, the x-value of where the vertex lies. You can use this value as the t-value in the previous equation to find the h-value of the vertex. When you do, you get h= 19.1 feet, or answer D.
Answer:
The correct option is (b).
Step-by-step explanation:
If X N (µ, σ²), then , is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, Z N (0, 1).
The distribution of these z-variate is known as the standard normal distribution.
The mean and standard deviation of the active minutes of students is:
<em>μ</em> = 60 minutes
<em>σ </em> = 12 minutes
Compute the <em>z</em>-score for the student being active 48 minutes as follows:
Thus, the <em>z</em>-score for the student being active 48 minutes is -1.0.
The correct option is (b).
Answer: c
Step-by-step explanation:
5.00 can be divided by 2 and this the out come 2.5(y)
same goes with 12.50 but by 5 and equal 2.5 (x)
if you have trouble theirs a calculator called Desmos.
You don't now you are a student or what