Question

School is making digital backups of old reels of film in its library archives the table shown approximate run Times of the films for a given diameter of film in the reel. Which of the following equations is a good model for the run time, y, as a function of the diameter, X?

Answer 1

One technique that you can apply when solving such a problem is trial and error. We try to use each equation to prove that a given value of x on the table given will correspond to the value of y on the table.

a) Let's try to put x = 3 for the first equation and we must get an answer equal to 2.25.

`y=7.72(3)-29.02=-5.86_{}`

Since the value of y is not equal to 2.25 and the deviation is too large. this equation is not a good model,

b) We put x = 3 on the second equation and solve for y

`y=-7.52(3)^2+0.19(3)+3.26=-63.85`

Since the value of y is not equal to 2.25 and the deviation is too large. this equation is not a good model,

c) We put x = 3 on the third equation and solve for y,

`y=0.4(3)^2+0.79(3)-4.93=1.04`

Again, the value that we get is not equal to 2.25, hence, this equation is not a good model. But since its value is close to 2.25, we try to other values of x. If x = 5, we get

`y=0.4(5)^2+0.79(5)-4.93=9.02`

which has a slight deviation on the given value of y on the table for x = 5. let's try for x = 7. We have

`y=0.4(7)^2+0.79(7)-4.93=20.2`

and the answer has a small deviation compared to the actual value given. The other values of x can again be put on the equation and check their corresponding value of y, and the resulting values are as follows

`\begin{gathered} y=0.4(8)^2+0.79(8)-4.93=26.99 \\ y=0.4(12)^2+0.79(12)-4.93=62.15 \\ y=0.4(14)^2+0.79(14)-4.93=84.53 \end{gathered}`

And as you can see, the deviation of values from the table to calculated becomes smaller. Hence, this is the best model.

d) We put x = 3 on the third equation and solve for y,

`y=4.19(1.02)^3=4.45_{}_{}`

Again, the value that we get is not equal to 2.25, hence, this equation is not a good model. But since its value is close to 2.25, we try to other values of x. If x = 5, we get

`y=4.19(1.02)^5=4.63`

where the answer's deviation is too large compared to the value of y if x = 5 on the table given.

Based on the calculations used above, the best equation that can be a good model is equation 3.