The correct Answer is B Morning Newspaper
Answer
(a) 
(b) 
Step-by-step explanation:
(a) 
δ(t)
where δ(t) = unit impulse function
The Laplace transform of function f(t) is given as:
where a = ∞
=> 
where d(t) = δ(t)
=> 
Integrating, we have:
=> 
Inputting the boundary conditions t = a = ∞, t = 0:
(b) 
The Laplace transform of function f(t) is given as:
Integrating, we have:
![F(s) = [\frac{-e^{-(s + 1)t}} {s + 1} - \frac{4e^{-(s + 4)}}{s + 4} - \frac{(3(s + 1)t + 1)e^{-3(s + 1)t})}{9(s + 1)^2}] \left \{ {{a} \atop {0}} \right.](https://tex.z-dn.net/?f=F%28s%29%20%3D%20%5B%5Cfrac%7B-e%5E%7B-%28s%20%2B%201%29t%7D%7D%20%7Bs%20%2B%201%7D%20-%20%5Cfrac%7B4e%5E%7B-%28s%20%2B%204%29%7D%7D%7Bs%20%2B%204%7D%20-%20%5Cfrac%7B%283%28s%20%2B%201%29t%20%2B%201%29e%5E%7B-3%28s%20%2B%201%29t%7D%29%7D%7B9%28s%20%2B%201%29%5E2%7D%5D%20%5Cleft%20%5C%7B%20%7B%7Ba%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
Inputting the boundary condition, t = a = ∞, t = 0:
Answer:
Equal to 440
Step-by-step explanation:
Answer:
The number of cars passing or the people arriving during lunch hour will be around 220
Step-by-step explanation:
First we find the for ten minutes period from given data:
Average Cars in 10 minutes = (18+38+87+16+56+5)/6
Average Cars minutes = 36.67
Now, for average cars in 1 hr, we must multiply this value by 6. As, 10 minutes multiplied by 6 equals 1 hr.
Average cars in an hour = 36.67*6
<u>Average Cars in an hour = 220</u>
Hence, the number of cars or people arriving during lunch hour will also be equal to average number of cars arriving during an hour.
