Pls Answer Quickly!! At a college basketball game, one lucky ticket holder will win an all-expenses-paid trip to the conference championship game. There are 4,000 ticket holders at the game. The organizers want to place chips, each labeled with the seat number of a ticket holder, in a bin and draw the winner from the bin. Unfortunately, the bin will hold only 400 chips.

Which method assures both that the organizers can use the bin for the drawing and that each ticket holder will have a fair chance of winning?

A.
Randomly assign the 4,000 ticket holders to 40 equal-sized groups. Then randomly select 4 of the groups and place their seat numbers in the bin. Randomly select the winner of the trip from the bin.
B.
Place 400 chips labeled with the seat numbers of the first 400 ticket holders into the bin, and randomly select the winner of the trip from the bin.
C.
Place chips labeled with the seat numbers of each ticket holder into 10 groups based on age. Then randomly select one of the groups and place it in the bin. Randomly select the winner of the trip from the bin.
D.
Place 400 chips labeled with the seat numbers of ticket holders from a certain section of the auditorium into the bin, and randomly select the winner of the trip from the bin.

A sounds correct to me.

Step-by-step explanation:

Have a great day! :)

answer:  

step-by-step explanation:

here x represents the height of the pole,

and, the height of the shadow of the pole is 12 feet.

also, the angle of elevation from the top of the pole is 40°,

therefore,

by the low of sine,

which is the required equation for finding the value of x,

step-by-step explanation:

i found the life tables for the year 2012. 

life table of males at age 30. probability of dying between ages x and x+1

total population:  

29–30 . . . . . . . . . . . . 0.001030

30–31 . . . . . . . . . . . . 0.001061

males:  

29–30 . . . . . . . . . . . . 0.001420

30–31 . . . . . . . . . . . . 0.001448

white population

29–30 . . . . . . . . . . . . 0.001005

30–31 . . . . . . . . . . . . 0.001037

white males

29–30 . . . . . . . . . . . . 0.001372

30–31 . . . . . . . . . . . . 0.001399

non hispanic white population

29–30 . . . . . . . . . . . . 0.001083

30–31 . . . . . . . . . . . . 0.001116

non hispanic white males

29–30 . . . . . . . . . . . . 0.001477

30–31 . . . . . . . . . . . . 0.001503

jacob, non hispanic white male. 30 years old.

probability that jacob lives to be 30 years old. 

since the given probabilities are probabilities of dying, i think we simply subtract it from 1 to get the probability of living. we will use the probability of non hispanic white male.

                                          probability of living to be 30 y.o.

29 - 30 : 1 - 0.001477 = 0.998523

30 - 31 : 1 - 0.001503 = 0.998497