Salvatore Boone

2021-12-21

A community college offers courses in Applied Algebra, Business Calculus, and Statistics. Each section of Applied Algebra has 50 students and earns the college $20,000 in revenue. Each section of Business Calculus has 20 students and earns the college $50,000 in revenue. Each section of Statistics has 30 students and earns the college $60.000 in revenue. Assuming the college wishes to offer a total of 7 sections, accommodate 230 students, and bring in $320,000 in revenue, how many sections of each course should they offer?

? sections of Applied Algebra

? sections of Business Calculus

? sections of Statistics

? sections of Applied Algebra

? sections of Business Calculus

? sections of Statistics

rodclassique4r

Beginner2021-12-22Added 37 answers

Step 1

let x ,y and z be the number of sections of Applied Algebra, Business Calculus and Statistics.

Total number of sections is 7, so$x+y+z=7$ ... (1)

Total number of students is 230, so

$50x+20y+30z=230$

$5x+2y+3z=23$ ...(2)

Total revenue is 320,000, so

$20,000x+50,000y+60,000z=320,000$

$2x+5y+6z=32$ ...(3)

Step 2

Subtract equation (2) from 5 times equation (1)

$3y+2z=12$ ...(4)

Subtract twice the equation (1) from equation (3)

$3y+4z=18$ ...(5)

Subtract equation (4) from equation (5)

$2z=6$

$z=3$

From equation (4)

$3y+2z=12$

$3y+2\left(3\right)=12$

$3y+6=12$

$3y=6$

$y=2$

From equation (1)

$x+y+z=7$

$x+2+3=7$

$x=2$

Step 3

Ans: The number of sections of Applied Algebra, Business Calculus and Statistics are 2,2 and 3 respectively.

let x ,y and z be the number of sections of Applied Algebra, Business Calculus and Statistics.

Total number of sections is 7, so

Total number of students is 230, so

Total revenue is 320,000, so

Step 2

Subtract equation (2) from 5 times equation (1)

Subtract twice the equation (1) from equation (3)

Subtract equation (4) from equation (5)

From equation (4)

From equation (1)

Step 3

Ans: The number of sections of Applied Algebra, Business Calculus and Statistics are 2,2 and 3 respectively.

yotaniwc

Beginner2021-12-23Added 34 answers

Step 1

Assume number of sections in applied algebra, Business calculus and Statistics be a, b and s respectively.

Step 2

Total number of sections$=7$

$a+b+s=7$ ...(1)

Total number of students$=230$

$50a+20b+30s=230$

$5a+2b+3s=23$ ...(2)

Total revenue$=320,000$

$20,000a+50,000b+60,000s=320,000$

$2a+5b+6s=32$ ...(3)

Step 3

Substitute$a=7-b-s$ into equation (2).

$5(7-b-s)+2b+2s=23$

$35-5b-5s+2b+3s=23$

$-3b-2s=-12$

$3b+2s=12$ ...(4)

Substitute$a=7-b-s$ into equation (3).

$2(7-b-s)+5b+6s=32$

$14-2b-2s+5b+6s=32$

$3b+4s=18$ ...(5)

Step 4

Subtract equation (4) from equation (5).

$(3b+4s)-(3b+2s)=18-12$

$2s=6$

$s=3$

Substitute$s=3$ into the equation (4).

$3b+2\left(3\right)=12$

$3b+6=12$

$3b=6$

$b=2$

Substitute$s=3\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}b=2$ into the equation (1).

$a+2+3=7$

$a=2$

Step 5

Answer:

2 sections of applied algebra

2 sections of business calculus

3 sections of statistics

Assume number of sections in applied algebra, Business calculus and Statistics be a, b and s respectively.

Step 2

Total number of sections

Total number of students

Total revenue

Step 3

Substitute

Substitute

Step 4

Subtract equation (4) from equation (5).

Substitute

Substitute

Step 5

Answer:

2 sections of applied algebra

2 sections of business calculus

3 sections of statistics

nick1337

Expert2021-12-28Added 699 answers

Step 1

Given information

Applied Algebra has 40 student and revenue is $70,000

Business Calculus has 30 student and revenue is $40,000

Statistics has 20 student and revenue is $50,000

And total number of section is 11 and total revenue is $550,000

We have to find the sections of each courses.

Let the number of section of Applied Algebra, Business Calculus and Statistics is x,y and z respectively.

According to given condition

Since, total number of sections is 11

Hence,

And total number of student is 300

NSK

Hence,

And total revenue is $550,000

Hence,

Step 2

NSK

Subtract equation from equation (2), we get

Now, subtract equation

Solve equation (5)

Step 3

Now, substitute

y=4

Hence, y=4

Substitute

=2

Hence, x=

Now, substitute x=2 and y=4 in equation (1)

Hence, z=5

Step 4

Therefore,

2 section of Applied Algebra

4 section of Business Calculus

5 section of Statistic

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