Suppose that the base salary of a salesperson is $30,000 per year. The salesperson gets an additional $50 for every sale. Let x represents the number of sales. How can we model the scenario with a linear function? How many items does the salesperson need to sell to earn $50,000

Respuesta :

Answer:

See below

Step-by-step explanation:

Salary (y)       =    30 000   +  50 x        

y = 30 000 + 50x     where x is the number of sales

How many sales for 50 000 ?

50 000 = 30 000 + 50 x

20 000 = 50 x

x = 400 sales

I. Model Problems

A linear model is a linear equation that represents a real-world

scenario. You can write the equation for a linear model in the same way

you would write the slope-intercept equation of a line. The y-intercept

of a linear model is the quantity that does not depend on x. The slope is

the quantity that changes at a constant rate as x changes. The change

must be at a constant rate in order for the equation to be a linear model.

Example 1 A machine salesperson earns a base salary of $40,000 plus

a commission of $300 for every machine he sells. Write an equation

that shows the total amount of income the salesperson earns, if he sells

x machines in a year.

The y-intercept is $40,000; the salesperson earns a $40,000 salary in a

year and that amount does not depend on x.

The slope is $300 because the salesperson’s income increases by $300

for each machine he sells.

Answer: The linear model representing the salesperson’s total income

is y = $300x + $40,000.

Linear models can be used to solve problems.

Example 2 The linear model that shows the total income for the

salesperson in example 1 is y = 300x + 40,000. (a) What would be the

salesperson’s income if he sold 150 machines? (b) How many

machines would the salesperson need to sell to earn a $100,000

income?

(a) If the salesperson were to sell 150 machines, let x = 150 in the linear

model; 300(150) + 40,000 = 85,000.

Answer: His income would be $85,000.

(b) To find the number of machines he needs to sell to earn a $100,000

income, let y = 100,000 and solve for x: