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Answer:
(a) There are 32,760 different ways to appoint the officers.
(b) 3003 different ways can the committee be appointed.
(c) Probability of getting the five youngest of the qualified candidates is 0.00033.
Step-by-step explanation:
We are given that a corporation must appoint a president, chief executive officer (CEO), chief operating officer (COO), and chief financial officer (CFO). It must also appoint a planning committee with five different members.
There are 15 qualified candidates, and officers can also serve on the committee.
(a) The number of officers are 4, i.e; A president, chief executive officer (CEO), chief operating officer (COO), and chief financial officer (CFO).
There are 15 qualified candidates.
To find the number of ways to appoint the officers, we will use permutation because here the order of selecting the officer's matters.
So, the number of ways to appoint the officers =
= {
}
= = 32,760 ways
(b) The number of committee numbers to appoint include five members.
There are 15 qualified candidates.
To find the number of ways in which the committee can be appointed, we will use combination because here the order of selecting the members doesn't matter.
So, the number of ways to appoint the committee =
= {
}
= = 3003 ways
(c) The probability of randomly selecting the committee members and getting the five youngest of the qualified candidates is given by;
Number of ways of selecting the committee = 3003
Selecting the five youngest of the qualified candidates = 1
So, the required probability =
= <u>0.00033</u>
The height at t seconds after launch is
s(t) = - 16t² + V₀t
where V₀ = initial launch velocity.
Part a:
When s = 192 ft, and V₀ = 112 ft/s, then
-16t² + 112t = 192
16t² - 112t + 192 = 0
t² - 7t + 12 = 0
(t - 3)(t - 4) = 0
t = 3 s, or t = 4 s
The projectile reaches a height of 192 ft at 3 s on the way up, and at 4 s on the way down.
Part b:
When the projectile reaches the ground, s = 0.
Therefore
-16t² + 112t = 0
-16t(t - 7) = 0
t = 0 or t = 7 s
When t=0, the projectile is launched.
When t = 7 s, the projectile returns to the ground.
Answer: 7 s
s(t) = - 16t² + V₀t
where V₀ = initial launch velocity.
Part a:
When s = 192 ft, and V₀ = 112 ft/s, then
-16t² + 112t = 192
16t² - 112t + 192 = 0
t² - 7t + 12 = 0
(t - 3)(t - 4) = 0
t = 3 s, or t = 4 s
The projectile reaches a height of 192 ft at 3 s on the way up, and at 4 s on the way down.
Part b:
When the projectile reaches the ground, s = 0.
Therefore
-16t² + 112t = 0
-16t(t - 7) = 0
t = 0 or t = 7 s
When t=0, the projectile is launched.
When t = 7 s, the projectile returns to the ground.
Answer: 7 s