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Questions for Quiz 2 - Finite Mathematics | MATH 1630, Quizzes of Mathematics

Material Type: Quiz; Class: Finite Mathematics; Subject: Mathematics; University: Walters State Community College; Term: Spring 2009;

Typology: Quizzes

Pre 2010

Uploaded on 08/16/2009

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MATH 1630 Internet
Spring 2009 Quiz 2 Name: 62 possible points
You must show all work to receive full credit.
1. (5 points) Solve the following system of equations using the elimination method.
3x + 4y = -2
2x + 3y = -1
Weโ€™ll eliminate the xโ€™s. To do this, weโ€™ll multiply the top equation by 2 and the bottom equation
by -3
2(3x + 4y = -2)
-3(2x + 3y = -1)
6x + 8y = -4
-6x โ€“ 9y = 3
Add the equations.
-1y = -1
Divide both sides by -1
-1y/-1 = -1/-1
y = 1
To solve for x, weโ€™ll substitute 1 for y in the top equation.
3x + 4(1) = -2
3x + 4 = -2
Subtract 4 from both sides, and then divide both sides by 3.
3x โ€“ 4 โ€“ 4 = -2 โ€“ 4
3x = -6
3x/3 = -6/3
x = -2
The final solution is x = -2 and y = 1.
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Download Questions for Quiz 2 - Finite Mathematics | MATH 1630 and more Quizzes Mathematics in PDF only on Docsity!

MATH 1630 Internet Spring 2009 Quiz 2 Name: 62 possible points

You must show all work to receive full credit.

  1. (5 points) Solve the following system of equations using the elimination method.

3x + 4y = - 2x + 3y = -

Weโ€™ll eliminate the xโ€™s. To do this, weโ€™ll multiply the top equation by 2 and the bottom equation by -

2(3x + 4y = -2) -3(2x + 3y = -1)

6x + 8y = - -6x โ€“ 9y = 3

Add the equations.

-1y = -

Divide both sides by -

-1y/

-1 =^

y = 1

To solve for x, weโ€™ll substitute 1 for y in the top equation.

3x + 4(1) = - 3x + 4 = -

Subtract 4 from both sides, and then divide both sides by 3.

3x โ€“ 4 โ€“ 4 = -2 โ€“ 4

3x = -

3x /

3 =^

3

x = -

The final solution is x = -2 and y = 1.

  1. (5 points) Solve the following system of equations using the substitution method.

2x โ€“ 3y = 11 3x + y = 11

The easiest path is to solve the second equation for y.

3x + y = 11

Subtract 3x from both sides.

3x โ€“ 3x + y = -3x + 11

y = -3x + 11

Substitute -3x + 11 for y in the top equation.

2x โ€“ 3 (-3x + 11) = 11

2x + 9x โ€“ 33 = 11 11x โ€“ 33 = 11

Add 33 to both sides, then divide both sides by 11.

11x โ€“ 33 + 33 = 11 + 33 11x = 44

11x /

11 =^

11

x = 4

To solve for y, substitute 4 for x in the bottom equation.

3(4) + y = 11 12 + y = 11

Subtract 12 from both sides.

12 โ€“ 12 + y = 11 โ€“ 12 y = -

So the solution is x = 4 and y = -

  1. Answer the following questions regarding the following supply and demand equations. Demand: 2 p + q = 56 Supply: 3 p โˆ’ q = 34

a. (5 points) Determine the price and quantity for the market equilibrium point.

The equilibrium point is the point of intersection for the supply and demand functions. To determine this point, solve the above system of equations.

Add the equations 5 90 5 90 5 5 $ 2 18 56 36 56 36 36 56 36 20

p q p q

p p p q q q q

Market equilibrium is achieved if the price is set at $18. Twenty units will be made and sold.

b. (5 points) If the price is set at $14, determine whether there will be a market surplus or market shortage. Calculate the number of units of surplus or shortage.

Demand at $14:

q q q q

Supply at $14:

q q q q q q โˆ’ = โˆ’ = โˆ’ โˆ’ = โˆ’ โˆ’ = โˆ’ โˆ’ (^) =โˆ’ โˆ’ โˆ’ = The customers will demand 28 units, the company will only want to make 8 units. Since demand is greater than supply there will be a shortage of goods. The amount of shortage is 28 โ€“ 8 = 20.

c. (2 points) Based on your answers to part b. above and the principles of supply and demand, explain whether the price will more likely remain at $14, fall below $14, or climb above $14.

Since there is a shortage of goods, the price will like climb above $14. This will encourage the company to manufacture more goods will dampening demand.

  1. (4 points) Textbook Pg 132, Problem 23d

3 5;^2

f x x g x x f f x f f x f x x x x

D

  1. (4 points) Textbook Pg 132, Problem 34

Equation of the line passing through (-2, 7) and (6, -4) y = mx + b

2 1 2 1

Plug in a point to solve for b. I'll use (-2,7). 7 11 2 8 7 11 4 7 11 11 11 4 4 4 28 11 17 4 4 4 11 17 8 4

m y^ y x x y x b

b

b

b

b

y x

= โˆ’^ = โˆ’ โˆ’ = = โˆ’

  1. (6 points) Solve the following equation. Round any decimal answer to 4 places.

2

x x

2

2

2 2 2

2 2

4 3 3 4 5^45

x x

x x

x x

x x

x x

a b c

b b ac

x

a

x

x

x

โŽœ โˆ’^ +^ โŽŸ=

โˆ’ ยฑ โˆ’ โˆ’ โˆ’^ ยฑ^ โˆ’^ โˆ’^ โˆ’

= โˆ’^ = โˆ’ = โˆ’ โˆ’ 714962686 โ†’ โˆ’2.

Two solutions: x = -2.7150 and x = 3.

  1. (4 points) Textbook pg 149, Problem 46a.

P = โˆ’0.01 x^2 + 50 x โˆ’ 300 Set P = $250 and solve for x.

2

2 2

2 2

4 50 50 4 0.01^550

x x

x x x x a b c

x b^ b^ ac a

x

x

โˆ’ ยฑ โˆ’ โˆ’^ ยฑ^ โˆ’^ โˆ’^ โˆ’

= โˆ’^ ยฑ^ โˆ’^ = โˆ’^ ยฑ^ =โˆ’^ ยฑ

= โˆ’^ ยฑ = โ†’

x = โˆ’^ ยฑ = โ†’ โˆ’

Two levels of production yield a profit of $250, 11 units and 4989 units.

Due to rounding, we donโ€™t get a profit of exactly $250, but weโ€™re pretty close.