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
175/5= 35
The answer is 35 minutes.
Hope this helps!
Q+12-2q+44=0
-q=-44-12
q=56
Answer:
a) X[bar]₁= 1839.20 cal
b) X[bar]₂= 1779.07 cal
c) S₁= 386.35 cal
Step-by-step explanation:
Hello!
You have two independent samples,
Sample 1: n₁= 15 children that did not eat fast food.
Sample 2: n₂= 15 children that ate fast food.
The study variables are:
X₁: Calorie consumption of a kid that does not eat fast food in one day.
X₂: Calorie consumprion of a kid that eats fast food in one day.
a)
The point estimate of the population mean is the sample mean
X[bar]₁= (∑X₁/n₁) = (27588/15)= 1839.20 cal
b)
X[bar]₂= (∑X₂/n₂)= (26686/15)= 1779.07 cal
c)
To calculate the sample standard deiation, you have to calculate the sample variance first:
S₁²=
[∑X₁² - (( ∑X₁)²/n₁)]=
= 149263.4571 cal²
S₁= 386.35 cal
I hope it helps!
Answer:
11.8 in.
Step-by-step explanation: