A traffic engineer monitors the rate at which cars enter the main highway during the afternoon rush hour. From her data she estimates that between 4:30 P. M. and 5:30 P. M. the rate R(t) at which cars enter the highway is given by the formula R(t) = 100(1 − 0.0001t2) cars per minute, where t is the time (in minutes) since 4:30 P. M. Find the average rate, in cars per minute, at which cars enter the highway during the first half hour of rush hour

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:

In this problem, the rate is given by:

Over the first half-hour the interval is [0,30], hence .

Then:

The average rate is of 10 cars per minute.

A similar problem is given at link

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

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.

=2901

The average rate of car is

                                          =97

On average, cars enter the highway during the first half hour of rush hour at a rate 97 per minute.

option c is correct.

step-by-step explanation:

the rule regarding the sides of triangle is:

(length of longest side)^2 = square of sum of other two sides.

the sides given are:

6, 10 and 8.

applying the rule we will get:

(10)^2 = (6)^2 + (8)^2

100 = 36+64

100 = 100

so, her answer is incorrect.

no, she should have added 6^2 and 8^2 and compared that to 10^2.

so, option c is correct.