Respuesta :
Answer:
On average, cars enter the highway during the first half hour of rush hour at a rate 97 per minute.
Step-by-step explanation:
Given that, the rate R(t) at which cars enter the highway is given the formula
[tex]R(t)= 100(1-0.0001t^2)[/tex]
The average rate of car enter the highway during first half hour of rush hour is the average value of R(t) from t=0 to t=30.
[tex]\therefore \int_0^{30} 100(1-0.0001t^2)\ dt[/tex]
[tex]=[100(t-0.0001\frac{t^3}{3})]_0^{30}[/tex]
[tex]=100[(30-0.0001\frac{30^3}{3})-(0-0.0001\frac{0^3}{3})][/tex]
=2901
The average rate of car is [tex]=\frac{\textrm{The number of car}}{Time}[/tex]
                      [tex]=\frac{2910}{30}[/tex]
                     =97
On average, cars enter the highway during the first half hour of rush hour at a rate 97 per minute.
Applying the integral, the average rate is of 10 cars per minute.
The average value of a function F(t) in an interval [a,b] is given by:
[tex]\frac{\int_{a}^{b} F(t) dt}{b - a}[/tex]
In this problem, the rate is given by:
[tex]R(t) = 100 - 0.01t^2[/tex]
Over the first half-hour the interval is [0,30], hence [tex]a = 0,b = 30[/tex].
Then:
[tex]\int_{a}^{b} F(t) dt[/tex]
[tex]\int_{0}^{30} (100 - 0.01t^2) dt[/tex]
[tex]100t - 0.01\frac{t^3}{3}_{t = 0}^{t = 3}[/tex]
[tex]100(3) - 0.01\frac{3^3}{3} = 299.91[/tex]
[tex]\frac{299.91}{30} \approx 10[/tex]
The average rate is of 10 cars per minute.
A similar problem is given at https://brainly.com/question/13218582
