H (t) = - 16t ^ 2 + 16t + 480
For this case, the first thing to do is to match the polynomial to zero to find the roots.
We have then:
0 = -16t ^ 2 + 16t + 480
From here, we obtain the following roots of the polynomial:
t1 = -5
t2 = 6
We ignore the negative root because the time is always greater than zero.
Answer:
it takes Rose to hit the water about
t = 6 seconds
Answer is C
14.2+6.5(p+3q)-4.05q
14.2+6.5p+19.5q-4.05q
14.2+6.5p+15.45q→6.5p+15.45q+14.2
so the answer is C
Answer:
answer is

Step-by-step explanation:
After working this way for 6 months he takes a simple random sample of 15 days. He records how long he walked that day (in hours) as recorded by his fitness watch as well as his billable hours for that day as recorded by a work app on his computer.
Slope is -0.245
Sample size n = 15
Standard error is 0.205
Confidence level 95
Sognificance level is (100 - 95)% = 0.05
Degree of freedom is n -2 = 15 -2 = 13
Critical Value =2.16 = [using excel = TINV (0.05, 13)]
Marginal Error = Critical Value * standard error
= 2.16 * 0.205
= 0.4428

In order to solve this, you have to set up a systems of linear equations.
Let's say that children = c and adults = a
30a + 12c = 19,080
a + c = 960
I'm going to show you how to solve this system of linear equations by substitution, the easiest way to solve in my opinion.
a + c = 960
- c - c
---------------------- ⇒ Step 1: Solve for either a or c in either equation.
a = 960 - c
20(960 - c)+ 12c = 19,080
19,200 - 20c + 12c = 19,080
19,200 - 8c = 19,080
- 19,200 - 19,200
---------------------------------- ⇒ Step 2: Substitute in the value you got for a or c
8c = -120 into the opposite equation.
------ ---------
8 8
c = -15
30a + 12(-15) = 19,080
30a - 180 = 19,080
+ 180 + 180
-------------------------------
30a = 19,260
------- -----------
30 30
a = 642
__________________________________________________________
I just realized that there can't be a negative amount of children, so I'm sorry if these results are all wrong.
Answer:
C
Step-by-step explanation:
Statement C of the following statements is true
C). Spin-locks can be used to prevent busy waiting in the implementation of semaphore.