Vanessa earns a base salary of $400.00 every week with an additional %5, percent commission on everything she sells. Vanessa sol
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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 <em>x</em> on the table given will correspond to the value of <em>y</em> 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.
Since the value of <em>y</em> 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 <em>y</em>
Since the value of <em>y</em> is not equal to 2.25 and the deviation is too large. this equation is not a good model,
c) We put <em>x</em> = 3 on the third equation and solve for <em>y,</em>
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 <em>x</em>. If x = 5, we get
which has a slight deviation on the given value of <em>y</em> on the table for <em>x</em> = 5. let's try for <em>x</em> = 7. We have
and the answer has a small deviation compared to the actual value given. The other values of <em>x</em> can again be put on the equation and check their corresponding value of <em>y</em>, and the resulting values are as follows
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 <em>x</em> = 3 on the third equation and solve for <em>y,</em>
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 <em>x</em>. If x = 5, we get
where the answer's deviation is too large compared to the value of <em>y</em> 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.