Gottfried wanted to see how contagious yawning can be. To better understand this, he conducted a social experiment by yawning in front of a random large crowd and observing how many people yawned as a result.
The relationship between the elapsed time ttt, in minutes, since Gottfried yawned, and the number of people in the crowd, P_{\text{minute}}(t)P
minute

(t)P, start subscript, start text, m, i, n, u, t, e, end text, end subscript, left parenthesis, t, right parenthesis, who yawned as a result is modeled by the following function:
Pminute(t)=5⋅(1.03)t
Complete the following sentence about the hourly rate of change in the number of people who yawn in Gottfried's experiment.
Round your answer to two decimal places.
Every hour, the number of people who yawn in Gottfried's experiment grows by a factor of

Every hour, the number of people who yawn in Gottfried's experiment grows by a factor of 5.89

Step-by-step explanation:

Understanding the problem

The expression for P_{\text{minute}}(t)P  

minute

(t)P, start subscript, start text, m, i, n, u, t, e, end text, end subscript, left parenthesis, t, right parenthesis, the number of people who yawn in Gottfried's experiment after ttt minutes, is 5⋅(1.03)t. This means that every minute, the number of people who yawn in Gottfried's experiment grows by a factor of 1.031.031, point, 03.

Let's change this expression so that the base tells us by which factor the number of people who yawn in Gottfried's experiment grows every hour.

Hint #22 / 3

Changing the unit

We want to find the expression for P_{\text{hour}}(x)P  

hour

(x)P, start subscript, start text, h, o, u, r, end text, end subscript, left parenthesis, x, right parenthesis, which models the number of people who yawn in Gottfried's experiment after xxx hours.

NOTE: The function P_{\text{hour}}P  

hour

P, start subscript, start text, h, o, u, r, end text, end subscript is not equivalent to P_{\text{minute}}P  

minute

P, start subscript, start text, m, i, n, u, t, e, end text, end subscript, although they model the same situation. [Tell me more.]

P_{\text{hour}}  

P, start subscript, start text, h, o, u, r, end text, end subscriptP_{\text{minute}}  

P, start subscript, start text, m, i, n, u, t, e, end text, end subscript

P_{\text{hour}}(1)  

P, start subscript, start text, h, o, u, r, end text, end subscript, left parenthesis, 1, right parenthesis11P_{\text{minute}}(1)  

P, start subscript, start text, m, i, n, u, t, e, end text, end subscript, left parenthesis, 1, right parenthesis11

Since there are \blueD{60}60start color #11accd, 60, end color #11accd minutes in an hour, t=\blueD{60}xt=60xt, equals, start color #11accd, 60, end color #11accd, x, and the expression for P_{\text{hour}}(x)P  

hour

(x)P, start subscript, start text, h, o, u, r, end text, end subscript, left parenthesis, x, right parenthesis is the same as the following expression.

Pminute(60x)=5⋅(1.03)60x=5⋅((1.03)60)x

Evaluating \maroonC{(1.03)^{60}}(1.03)  

60

start color #ed5fa6, left parenthesis, 1, point, 03, right parenthesis, start superscript, 60, end superscript, end color #ed5fa6 and rounding to two decimal places, we find that Phour(x)=5⋅(5.89)x.

Answers:

This forms the equation 1x-5y = -15, which is the same as x-5y = -15

Explanation:

Let's find the slope. I'll pick the first two rows to form the two points needed

(x1,y1) = (0,3)

(x2,y2) = (10,5)

These are then plugged into the slope formula.

m = slope

m = (y2-y1)/(x2-x1)

m = (5-3)/(10-0)

m = 2/10

m = 1/5

m = 0.2

The y intercept is b = 3 since this is the y value when x = 0.

With m = 0.2 and b = 3, we go from y = mx+b to y = 0.2x+3

Let's multiply both sides by 5 to turn that decimal value into a whole number. I'm picking 5 since 0.2 = 1/5

So we get

y = 0.2x+3

5y = 5*(0.2x+3)

5y = 5*0.2x+5*3

5y = x+15

Now let's get x and y together, but move that 15 to its own side

5y = x+15

x+15 = 5y

x = 5y-15

x-5y = -15

This is one way to write the equation in standard form.

Standard form is Ax+By = C, where A,B,C are integers. Some math textbooks insist that A > 0. In this case, A = 1, B = -5, and C = -15.

As a way to check, we can plug in (x,y) = (15,6) from the third row of the table to see that..

x-5y = -15

15-5(6) = -15

15 - 30 = -15

-15 = -15

This verifies the third row. I'll let you check the first two rows of the table to fully confirm this equation works.