Question

Urban Community College is planning to offer courses in Finite Math, Applied Calculus, and Computer Methods. Each

section of Finite Math has 40 students and earns the college $40,000 in revenue. Each section of Applied Calculus has 40
students and earns the college $60,000, while each section of Computer Methods has 10 students and earns the college
$29,000. Assuming the college wishes to offer a total of seven sections, accommodate 220 students, and bring in $298,000
in revenues, how many sections of each course should it offer?
Finite Math
Applied Calculus
Computer Methods
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Answer 1

The Urban Community College should offer 2 sections of Finite Math, 3 sections of Applied Calculus, and 2 sections of Computer Methods.

Step-by-step explanation:

To determine how many sections of each course the Urban Community College should offer, we need to set up a system of equations to represent the given information. Let's assume x is the number of sections for Finite Math, y is the number of sections for Applied Calculus, and z is the number of sections for Computer Methods.

We have the following conditions:

x + y + z = 7 (Total number of sections)

40x + 40y + 10z = 220 (Total number of students)

40000x + 60000y + 29000z = 298000 (Total revenue)

Solving these equations, we find that x = 2, y = 3, and z = 2. So, Urban Community College should offer 2 sections of Finite Math, 3 sections of Applied Calculus, and 2 sections of Computer Methods.

Answer 2

Step-by-step explanation:

Let's use the following variables:

Let F be the number of sections of Finite Math

Let A be the number of sections of Applied Calculus

Let C be the number of sections of Computer Methods

We can create a system of equations based on the given information:

F + A + C = 7 (the college wishes to offer a total of seven sections)

40F + 60A + 290C = 298000 (the college wishes to bring in $298,000 in revenues)

40F + 40A + 10C = 220 (the college wishes to accommodate 220 students)

We can solve this system of equations using any method we prefer. Here's one possible way:

Multiply the third equation by 10 to get:

400F + 400A + 100C = 2200

Subtract the third equation from the second equation to eliminate C:

20A + 280C = 76000

Multiply the first equation by 40 and subtract it from the previous equation to eliminate A:

240C = 76000 - 1600

Solve for C:

C = 290

Substitute C = 290 into the third equation to solve for F:

40F + 40A + 10(290) = 220

40F + 40A = -2680

F + A = -67

Substitute C = 290 and F + A = -67 into the first equation to solve for A:

A = 10

Substitute A = 10 and C = 290 into the first equation to solve for F:

F = -3

We have a problem here: we obtained a negative number of sections for Finite Math, which is impossible. This means that the original system of equations is inconsistent, and there is no solution that satisfies all the requirements.

Therefore, the college cannot offer the desired number of sections while accommodating the desired number of students and revenue. Some adjustments need to be made, such as changing the number of students per section or the tuition fees.