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
Draw a histogram for the intervals 16-18, 19-21, 22-24, and 25-27 using the following data: 26, 16, 22, 27, 20, 24, 21, 27, 22,
wolverine [178]
Answer:
Step-by-step explanation:
Class Interval Variate Frequency
16-18 16 1
19-21 0 0
22-24 0 0
25-27 26,27 2
5 times 40 times 49 equals 9800 because you do 5 time 4 and that's 20 and you add a 0 and you multiply by 49 to get 9800.
Answer:
why thank you :)
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