When there are a number of overlapping shaded areas on the graph, I find it convenient to use the reverse of the inequalities. That makes the <em>unshaded</em> area the solution space. Here, the vertices of the triangular solution space are ...

(2, 9), (2, 13), (6, 9)

Any of the grid points within (or on) this triangle is a possible solution. One of them is (2, 9) corresponding to 2 nickels and 9 dimes.

Three solutions are shown:

(x, y) = (2, 9), (3, 10), (4, 11)

4 0

3 years ago

Problem 5. Buses arrive at 116th and Broadway at the times of a Poisson arrival process with intensity λ arrivals per hour. Thes

Shtirlitz [24]

Answer:

Step-by-step explanation:

The problem states that there are only two types of busses - M104 and M6 with probable occurence of 0.6 and 0.4 respectively.

If the average number of busses arriving per hour is λ, the average number of M6 busses per hour is 0.4λ

Now consider a set of 3 M6 busses as an event. The average number of such events per hour will be

μ = 0.4λ / 3

The expected number of hours for the event "THIRD M6 arrives", let's say X is

E[X] = 1 / μ ( exponential distribution) = 3 / 0.4λ

= 7.5 / λ

The variance of event X is =

$Var[x] = \frac{1}{U^2} = \frac{56.25}{\lambda ^2}$

6 0

3 years ago

If you know this please help me out!!

-Dominant- [34]

Answer:

e b

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers