Darren wanted to see how contagious yawning can be. To better understand this, he conducted a social experiment in which he yawned in front of a large, random crowd and observed how many people yawned as a result. The relationship between the elapsed time ttt, in minutes, since Darren yawned, and the number of people in the crowd, P(t)P(t)P, left parenthesis, t, right parenthesis, who yawned as a result is modeled by the following function: P(t)=5⋅4t10.5

10.5 minutes

Step-by-step explanation:

Thinking about the problem

The modeling function is of the form P(t)=A⋅Bf(t), where B=4B=4B, equals, 4 and f(t)=\dfrac{t}{10.5}f(t)=

10.5

t

f, left parenthesis, t, right parenthesis, equals, start fraction, t, divided by, 10, point, 5, end fraction.

Note that each time f(t)f(t)f, left parenthesis, t, right parenthesis increases by 111, the quantity is multiplied by B=4B=4B, equals, 4.

Therefore, we need to find the ttt-interval over which f(t)f(t)f, left parenthesis, t, right parenthesis increases by 111.

Hint #22 / 3

Finding the appropriate unit interval

fff is a linear function whose slope is \dfrac{1}{10.5}

10.5

1

start fraction, 1, divided by, 10, point, 5, end fraction.

This means that whenever ttt increases by \Delta tΔtdelta, t, f(t)f(t)f, left parenthesis, t, right parenthesis increases by \dfrac{\Delta t}{10.5}

10.5

Δt

start fraction, delta, t, divided by, 10, point, 5, end fraction.

Therefore, for f(t)f(t)f, left parenthesis, t, right parenthesis to increase by 111, we need \Delta t=10.5Δt=10.5delta, t, equals, 10, point, 5. In other words, the ttt-interval we are looking for is 10.510.510, point, 5 minutes.

Hint #33 / 3

Summary

The number of people who yawned quadruples every 10.510.510, point, 5 minutes.

36/60

36/2=18

60/2=30

18/30

18/3=6

30/3=10

6/10

6/2=3

10/2=5

your final answer would be 3/5 of an

hope this !