Every morning I take either bus number 5 or bus number 8 to work. Every morning the waiting time for the number 5 is exponential
with mean 10 minutes, while the waiting time for the number 8 is exponential with mean 20 minutes. Assume all waiting times are independent of each other. Let S be the total amount of bus-waiting (in minutes) that I have done duringn mornings, and let T, be the number of times I have taken the number 5 bus during n mornings. a) Find the limit i, PS, s 7nl (b) Find the limit lim P(T 0.6n). 1-00 Hint. Recall Examples 6.33 and 6.34.
1 answer:
You might be interested in
Answer with explanation:
Price of Product on the Website = $ 412.50
It is given that, product is offered at a discount of 25% on the website.
Let actual price of product on the store = $ x
Writing the above statement in terms of equation
→Price at store - Discount= Price at Website
Price at store of that Product = $ 550
Answer:
1) You can rely solely on your brakes because when doing so the car will just travel 250ft from the point you hit your brakes till the point the car stopped completely, leaving you 50ft away from the cow.
2) See attached picture.
j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet
j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.
j"(t) represents the acceleration of the car t seconds after the brakes have been hit in
3) yes, any time after t=5.28 will not accurately model the path of the car since at that exact time the car will reach a velocity of 0ft/s and unless another force is applied to the car, then the car will not move after that time.
4)
(see attached picture for graph)
Step-by-step explanation:
1) In this part of the problem we need to find the time when the speed of the car is 0. Gets to a complete stop. For this we will need to take the derivative of the position function so we get:
and we set the first derivative equal to zero so we get:
95-18t=0
and solve for t
-18t=-95
t=5.28s
so now we calculate the position of the car after 5.28 seconds, so we get:
so we have that the car will stop 250.69ft after he hit the brakes, so there will be about 50ft between the car and the cow when the car stops completely, so he can rely just on the breaks.
2) For answer 2 I take the second derivative of the function so I get:
j"(t)=-18
and then we graph them. (See attached picture)
j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet
j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.
j"(t) represents the acceleration of the car t seconds after the brakes have been hit in
3) yes, any time after t=5.28 will not accurately model the path of the car since at that exact time the car will reach a velocity of 0ft/s and unless another force is applied to the car, then the car will not move after that time.
4)
(see attached picture for graph)
