How Is a Salesperson's Wage Calculated Based on Sales?

Thread starter nycmathguy
Start date Jun 19, 2021
Tags

Linear Linear equation

AI Thread Summary

A pharmaceutical salesperson's monthly wage is calculated as a base salary of $5000 plus a commission of 7% on sales. The linear equation representing this wage structure is W(S) = 5000 + 0.07S, where W is the monthly wage and S is the monthly sales. This formula effectively combines the fixed salary with the variable commission based on sales performance. The equation suggests that as sales increase, the total wage will also rise proportionally. Understanding this calculation is essential for both salespeople and employers in the pharmaceutical industry.

Jun 19, 2021

nycmathguy

Homework Statement: Write a linear equation for the monthly wages in terms of monthly sales.

Relevant Equations: y = mx + b

A pharmaceutical salesperson receives a monthly salary of $5000 plus a commission of
7% of sales. Write a linear equation for the salesperson’s monthly wage W in terms of monthly sales S.

I will let W(S) = monthly wage W in terms of monthly sales S.

$5000 plus a commission of 7% of sales =
5000 + 0.07S.

We now put it all together:

W(S) = 5000 + 0.07S

You say?

Physics news on Phys.org

Jun 19, 2021

Doc Al

Mentor

Looks good to me.

Thread 'Finding the number of ways to arrange identical balls in a circle (3 different colors)'

I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks

View full post »

Thread 'Making the denominator of a certain fraction real'

Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...

View full post »