Acar dealership sells 20 cars when there is a $2,000 markup over the factory price of each vehicle. the sales department determines that for each $400 decrease in markup an additional 6 cars are sold. write a function that models the total monthly profit p(n). let n represent the number of markdowns. a) p(n) = 2400n2 + 4000n + 40,000 b) p(n) = 2400n2 − 4000n + 40,000 c) p(n) = −2400n2 + 4000n + 40,000 d) p(n) = −2400n2 − 4000n + 40,000

C)  

Step-by-step explanation:

Since, profit = number of car sold × total markup

Original markup = $2,000

Also, the original number of sold cars = 20,

Let the markup is decreases by n times of $400,

New markup = ( 2000 - 400n ) dollars,

According to the question,

For each $400 decrease in markup an additional 6 cars are sold.

So, new number of sold cars = 20 + 6x,

Hence, the total monthly profit P(n) would be,

P(n) = ( 2000 - 400n )( 20 + 6n )

Option 'C' is correct.

x = 23.78

step-by-step explanation:

1. x> -5

2. x< 8

step-by-step explanation:

1. 2x> -10

divide by 2 from both sides of equation.

2x/2> -10/2

simplify, to find the answer.

-10/2=-5

x> -5 is the correct answer.

2. 3x< 24

divide by 3 from both sides of equation.

3x/3< 24/3

simplify, to find the answer.

24/3=8

x< 8 is the correct answer.

i hope this you, and have a wonderful day!