A consumer is trying to decide between two long-distance callingplans. The first one charges a flat rate of $0.10 per minute, whereas the second charges a flat rate of $0.99 for calls up to 20minutes in duration and then $0.10 for each additional minuteexceeding 20 (assume that calls lasting a noninteger number ofminutes are charged proportionately to a whole-minute'scharge). Suppose the consumer's distribution of call durationis exponential with parameter λ.

Which plan is better if expected call duration is 10 minutes? 15minutes? [Hint: Let h1(x) denote the cost for the firstplan when call duration is x minutes and let h2(x) bethe cost function for the second plan. Give expressions forthese two cost functions, and then determine the expected cost foreach plan.]

For 10 minutes: E[h1(x)] = $1 and E[h2(x)] = $1.125

For 15 minutes: E[h1(x)] = $1.5 and E[h2(x)] = $1.385

Step-by-step explanation:

Given:

- Flat rate 1st charge = $ 0.10 / min

- Flat rate 2nd charge = $ 0.99/min     0< t < 20 mins

- Overhead rate for 2nd charge = $0.10 / min

- They follow an exponential distribution λ.

Find:

For each plan

- Which plan is better if expected call duration is 10 minutes?

- Which plan is better if expected call duration is 15 minutes?

Solution:

- The expected duration for flat plans of 10 and 15 mins are:

            10 mins: E(h_1(x)) = mins * Rate

                          E(h_1(x)) = 10 * 0.10 = $1

            15 mins: E(h_1(x)) = mins * Rate

                           E(h_1(x)) = 15 * 0.10 = $ 1.5  

- For second plan, you need to work out the probability of a call exceeding 20 minutes. Calls under 20 minutes cost 99 cents. Calls over 20 minutes 100 cents, and then have an expected duration of 10 or 15 minutes for (a) and (b) respectively-- So calls longer than 20 minutes have an expected cost of 99+100 for part (a), and 99+150 for part (b).

- So calls longer than 20 minutes have an expected cost of 99+100 for part (a), and 99+150 for part (b).

- The CDF is the probability that the time will be LESS THAN t, and we will call that probability P.

So for part (a):

        P = p(time < 20) = CDF(20) = 1 - e^(-20/10) = 1 - e^(-2) = 0.865

        E(h2(x)) = P * .99 + (1-P)(.99 + 1) = 0.865*0.99 + (1 - .865)*(1.99) = $1.125

for part (b):

        P = p(time < 20) = CDF(20) = 1 - e^(-20/15) = 1 - e^(-4/3) = 0.736

    E(h2(x)) = P * .99 + (1-P)(.99 + 1.5) = 0.736*0.99 + (1 - .736)*(2.49) = $1.385

- Hence, for 10 mins call Plan 1 is better. However, for 15 mins call its plan 2

a.) list the factors of 40.

1/2/4/5/8/10/20/40

b.) list the factors of 40 that represent the number of cans that could be on each shelf.

5

c.) how many different arrangements of cans are possible?

just one because if you put more on one shelf you will go way over 40 and the cabinet will probably break.

d.) what are the possible arrangements of the cans?

put 5 cans on each self to reach 40 cans in total.

e.) which arrangement uses all the shelves?

5 because you can put 5 cans on each shelf, and since there are 8 shelfs you are equally going to get 40 cans.

answer: 486in^2

step-by-step explanation: