Considering the straight line, which is the diagonal line from 1000 at 12 am going to 400 at 8 am, we can create an equation describing this line. We let be the point 12 am as zero since at 24 H time, 12 am is considered as 0:00. We have two points. (0,1000) and (8,400). We can solve for the slope of the line. We have
[tex]\begin{gathered} m=\frac{y_2-y_1_{}_{}}{x_2-x_1} \\ m=\frac{400-1000}{8-0}=-75 \end{gathered}[/tex]We use the slope-intercept form of the equation to create an equation of the line describing the decrease of the people online at 12 am - 8 am. We already solve m. In this case, I will use the point (0, 1000). We have
[tex]\begin{gathered} y=mx+b \\ 1000=-75(0)+b \\ b=1000 \end{gathered}[/tex]Hence, the equation of the line is
[tex]y=-75x+1000[/tex]Let's check the number of people at 7:15 AM, which is written in decimal form as 7.25. and also for 7:30 AM, which is written in decimal form as 7.50. We have the following:
[tex]\begin{gathered} y=-75(7.25)+1000=456.25\approx456 \\ y=-75(7.5)+1000=437.5\approx438 \end{gathered}[/tex]There are approximately 456 people online at 7:15 AM and 438 people online at 7:30 AM.
Upon looking at this, we can also check the other straight line from 8 am to 12 pm. We can use the same technique used above to solve the equation of the line and check the number of people at 8:30 AM.
The points are (8,400) and (12,800). The slope of the line is
[tex]m=\frac{800-400}{12-8}=100[/tex]The equation of the line for this part is
[tex]\begin{gathered} y=mx+b \\ 400=100\cdot8+b \\ b=-400 \\ \\ y=100x-400 \end{gathered}[/tex]To determine the number of people at 8:30 AM, we use the equation above to solve the value of y. We can rewrite 8:30 Am in decimal form as 8.50. We have
[tex]\begin{gathered} y=100(8.5)-400 \\ y=450 \end{gathered}[/tex]Hence, there are 450 people online at 8:30 AM.
Answer: D) 8:30 AM
