The answer to your question would be $18
![\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ h=-16t^2+\stackrel{\stackrel{v_o}{\downarrow }}{65}t](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Binitial%20velocity%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20~~~~~~%5Ctextit%7Bin%20feet%7D%20%5C%5C%5C%5C%20h%28t%29%20%3D%20-16t%5E2%2Bv_ot%2Bh_o%20%5Cend%7Barray%7D%20%5Cquad%20%5Cbegin%7Bcases%7D%20v_o%3D%5Cstackrel%7B%7D%7B%5Ctextit%7Binitial%20velocity%20of%20the%20object%7D%7D%5C%5C%5C%5C%20h_o%3D%5Cstackrel%7B%7D%7B%5Ctextit%7Binitial%20height%20of%20the%20object%7D%7D%5C%5C%5C%5C%20h%3D%5Cstackrel%7B%7D%7B%5Ctextit%7Bheight%20of%20the%20object%20at%20%22t%22%20seconds%7D%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20h%3D-16t%5E2%2B%5Cstackrel%7B%5Cstackrel%7Bv_o%7D%7B%5Cdownarrow%20%7D%7D%7B65%7Dt)
now, take a look at the picture below, so for 2) and 3) is the vertex of this quadratic equation, 2) is the y-coordinate and 3) the x-coordinate.
Answer:
The conclusion "T" logically follows from the premises given and the argument is valid
Step-by-step explanation:
Let us use notations to represent the steps
P: I take a bus
Q: I take the subway
R: I will be late for my appointment
S: I take a taxi
T: I will be broke
The given statement in symbolic form can be written as,
(P V Q) → R
S → (¬R ∧ T)
(¬Q ∧ ¬P) → S
¬R
___________________
∴ T
PROOF:
1. (¬Q ∧ ¬P) → S Premise
2. S → (¬R ∧ T) Premise
3. (¬Q ∧ ¬P) → (¬R ∧ T) (1), (2), Chain Rule
4. ¬(P ∨ Q) → (¬R ∧ T) (3), DeMorgan's law
5. (P ∨ Q) → R Premise
6. ¬R Premise
7. ¬(P ∨ Q) (5), (6), Modus Tollen's rule
8. ¬R ∧ T (4), (7), Modus Ponen's rule
9. T (8), Rule of Conjunction
Therefore the conclusion "T" logically follows from the given premises and the argument is valid.
Answer:
The ratio between two values A and B is just the quotient between these two values:
ratio = A/B
a) $280 in 7m
Here the ratio is:
$280/7m = $40/m
This also can be read as:
$40 per meter.
b) 105 miles in 2 hours
Here the ratio is:
105mi/2h = 52.5 mi/h
This also can be read as:
52.5 miles per hour
c) $33 for 5lb
The ratio is:
$33/5lb = $6.6/lb
This can be read as:
$6.6 per pound.
d) 50 pages in 2 hours
the ratio is:
(50 pages)/2h = 25 pages/h
this can be read as:
25 pages per hour.
Students with D's / total students : 3 / (7 + 9 + 11 + 3 + 2) =
3 / 32 = 0.0937...to turn to a percent, multiply by 100 = 9.375....rounded =
9.4% <== ur probability
