Answer:
a.) 1.38 seconds
b.) 17.59ft
Step-by-step explanation:
h(t) = -16t^2 + 22.08t + 6
if we were to graph this, the vertex of the function would be the point, which if we substituted into the function would give us the maximum height.
to find the vertex, since we are dealing with something called "quadratic form" ax^2+bx+c, we can use a formula to find the vertex
-b/2a
b=22.08
a=-16
-22.08/-16, we get 1.38 when the minuses cancel out. since our x is time, it will be 1.38 seconds
now since the vertex is 1.38, we can substitute 1.38 into the function to find the maximum height.
h(1.38)= -16(1.38)^2 + 22.08t + 6 -----> is maximum height.
approximately = 17.59ft -------> calculator used, and rounded to 2 significant figures.
for c the time can be equal to (69+sqrt(8511))/100, as the negative version would be incompatible since we are talking about time. or if you wanted a rounded decimal, approx 1.62 seconds.
Answer:
![P(A) = \frac{3}{10}](https://tex.z-dn.net/?f=P%28A%29%20%3D%20%5Cfrac%7B3%7D%7B10%7D)
Step-by-step explanation:
Given
![S = \{11,12,13,14,15,16,17,18,19,20\}](https://tex.z-dn.net/?f=S%20%3D%20%5C%7B11%2C12%2C13%2C14%2C15%2C16%2C17%2C18%2C19%2C20%5C%7D)
Required
The probability of having a multiple of 3
Let the event of having a multiple of 3 be represented as: A
So:
![A = \{12,15,18\}](https://tex.z-dn.net/?f=A%20%3D%20%5C%7B12%2C15%2C18%5C%7D)
![n(A) = 3](https://tex.z-dn.net/?f=n%28A%29%20%3D%203)
So, the probability is:
![P(A) = \frac{n(A)}{n(S)}](https://tex.z-dn.net/?f=P%28A%29%20%3D%20%5Cfrac%7Bn%28A%29%7D%7Bn%28S%29%7D)
Where
i.e. the sample size
So:
![P(A) = \frac{n(A)}{n(S)}](https://tex.z-dn.net/?f=P%28A%29%20%3D%20%5Cfrac%7Bn%28A%29%7D%7Bn%28S%29%7D)
![P(A) = \frac{3}{10}](https://tex.z-dn.net/?f=P%28A%29%20%3D%20%5Cfrac%7B3%7D%7B10%7D)
Well to isolate the variable you should subtract 6 from both sides resulting in -4y > 16
Now just divide by -4, however, keep in mind that when you divide or multiply by a negative number when there is a sign, always switch the sign, so just divide by -4 and get y < 4
To the nearest hundredth is 3 digits to the left of the decimal point.
We look at the digit to the right of that number.
If that number is 5 or higher, we round up. Else, round down.
5 is our third digit to the left of the decimal point. 5 is right beside that, therefore we round up.
555.555 rounded to the nearest hundredth is 600.
Refer to the figure shown below.
The feasible region that satisfies all the constraints is the shaded region.
The bounding vertices are
A (0, 3)
B (0, 0)
C (5, 0)
D (1.5, 3.5)
All the functions that define the constraints are either linear or constant.
The maximum value is at vertex D, and equal to 3.5.