Answer:
your answer has been attached
Answer:
(a) 65318400
(b) 1080
(c) 311040
Step-by-step explanation:
I think you forgot to put the questions
(a) if the programmes can be perfomed in any order ?
(b) if the programmes of the same kind were perfomed at a stretch?
(c) if the programmes of the same kind were perfomed at a strech and considering the order of performance of the programmes of the same kind?
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Each team needs two people for a 3-legged race.
Therefore, the number of orders (combinations) of 2 people at a time from 8 is
₈C₂ = 8!/(2!6!)
= (8*7*6!)/(2*6!)
= (8*7)/2
= 56/2
= 28
Answer: 28
10.…….……......................
T = 5, so after 5 years
p(t) = t^3 - 14t^2 + 20t + 120
Take derivative to find minimum:
p’(t) = 3t^2 - 28t + 10
Factor to solve for t:
p’(t) = (3t - 2)(t - 5)
0 = (3t - 2)(t - 5)
0 = 3t - 2
2 = 3t
2/3 = t
Plug 2/3 into original equation, this is a maximum. We want the minimum:
0 = t - 5
5 = t
Plug back into original:
5^3 - 14(5)^2 + 20(5) + 120
125 - 14(25) + 100 + 120
125 - 350 + 220
- 225 + 220
p(5) = -5
