Question

The number of views on an interesting video after it's uploaded is represented by the following table:

Answer 1

Answer:

The model for V(t) ,the number of views, t ,months after it's uploaded, best fits the data is $,V(t)=50(6.25)^t$

Step-by-step explanation:

Time (months)         Views

0                                  50

2                                  313

4                                  1950

6                                  12,210

8                                  76,300

10                                 476,800

$(x_1,y_1)=(0,50)\\(x_2,y_2)=(2,313)\\(x_3,y_3)=(4,1950)$

$Slope = \frac{y_2-y_1}{x_2-x_1}$

$Slope = \frac{313-50}{2-0}$

Slope =131.5

$Slope = \frac{y_3-y_2}{x_3-x_2}$

$Slope = \frac{1950-313}{4-2}$

Slope =818.5

Since the slope is not same . So, it is not a linear relationship

$\frac{y_2}{y_1}=\frac{313}{50}=6.25\\\frac{y_3}{y_2}=\frac{1950}{313}=6.25$

Since the ratio between the consecutive outputs is same so, it is exponential relationship.

Formula : $y=ab^t$

Where a is the initial value = 50

b = rate of change = 6.25

The model for V(t) ,the number of views, t ,months after it's uploaded, best fits the data

So$,V(t)=50(6.25)^t$