Question

1. a company advertises on a website. a worker tracked the number of visits to the website and the number of clicks on the advertisement. the table shows the data for several days. a linear function can be used to model the data. website advertisement based on the table, what is the best prediction of the number of clicks on the advertisement if 1,500 people visit the website?

Explanation:

Step1: Find the slope \(m\)

Use the formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Let's take two points \((x_1,y_1)=(153,14)\) and \((x_2,y_2)=(629,38)\)
\(m=\frac{38 - 14}{629 - 153}=\frac{24}{476}=\frac{6}{119}\approx0.0504\)

Step2: Find the \(y -\)intercept \(b\)

Use the point - slope form \(y - y_1=m(x - x_1)\). Using the point \((153,14)\)
\(y-14=\frac{6}{119}(x - 153)\)
\(y=\frac{6}{119}x-\frac{6\times153}{119}+14\)
\(y=\frac{6}{119}x-\frac{918}{119}+14\)
\(y=\frac{6}{119}x-\frac{918}{119}+\frac{14\times119}{119}\)
\(y=\frac{6}{119}x-\frac{918 - 1666}{119}\)
\(y=\frac{6}{119}x-\frac{- 748}{119}\)
\(y=\frac{6}{119}x + 6.2857\)

Step3: Predict \(y\) when \(x = 1500\)

Substitute \(x = 1500\) into \(y=\frac{6}{119}x+6.2857\)
\(y=\frac{6\times1500}{119}+6.2857\)
\(y=\frac{9000}{119}+6.2857\)
\(y\approx75.63+6.2857\)
\(y\approx81.92\approx83\)

Answer:

83