+0  
 
0
26
2
avatar+362 

 

The number of students enrolled in a new course as a function of time can be represented by the function.

 
AngelRay  Nov 15, 2017

Best Answer 

 #2
avatar+5254 
+1

f(2)  =  42 - 1   =   16 - 1   =   15           At  2  hours, there are  15  students enrolled.

 

f(4)  =  44 - 1   =   256 - 1   =   255       At  4  hours, there are  255  students enrolled.

 

 

The number of students goes from  15  to  255 .

255 - 15  =  240      So the number of students increased by  240 .

 

The number of hours goes from  2  to  4 .

4 - 2  =  2      So the number of hours increased by  2 .

 

In  2  hours time, the number of students increased by  240 .

 

On average, that is an increase of  240/2  ,  or  120 ,  students per hour.

 

 

The average increase in the number of students per hour is  120  .

 
hectictar  Nov 15, 2017
Sort: 

2+0 Answers

 #1
avatar+634 
+1

For this, plug values 2 and 4 in for x.

\(f(2)=(4)^2-1, f(4)=(4)^4-1\)

\(f(2)=16-1, f(4)=256-1\)

\(f(2)=15, f(4)=255\)

For the average, add the two values and divide by 2.

\(\frac{15+255}{2}=\frac{270}{2}=135\)

So the average increase is 135 students.

 
AdamTaurus  Nov 15, 2017
 #2
avatar+5254 
+1
Best Answer

f(2)  =  42 - 1   =   16 - 1   =   15           At  2  hours, there are  15  students enrolled.

 

f(4)  =  44 - 1   =   256 - 1   =   255       At  4  hours, there are  255  students enrolled.

 

 

The number of students goes from  15  to  255 .

255 - 15  =  240      So the number of students increased by  240 .

 

The number of hours goes from  2  to  4 .

4 - 2  =  2      So the number of hours increased by  2 .

 

In  2  hours time, the number of students increased by  240 .

 

On average, that is an increase of  240/2  ,  or  120 ,  students per hour.

 

 

The average increase in the number of students per hour is  120  .

 
hectictar  Nov 15, 2017

14 Online Users

avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details