find dy/dx when y =
[tex] \sqrt{(1 + cosx)/( 1 - cosx)} [/tex]
​

The dosage needed for this in an  order for morphine sulfate 1/4 gr IM stat is 1.5 ml.

What is an equation?

An equation is an expression that shows the relationship between two or more numbers and variables.

An independent variable is a variable that does not depend on any other variable for its value while a dependent variable is a variable that depends on other variable.

One grain =  60mg, hence:

1/4 gr = 1/4 gr * 60 mg per gr = 15 mg

Hence, for 10 mg per ml:

Dosage = 15 mg /  10 mg per ml = 1.5 ml

The dosage needed for this in an  order for morphine sulfate 1/4 gr IM stat is 1.5 ml.

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1. Solving linear equations:

(a) x = 4.

(b) r = 2/3.

2. Proportion:

(a) When x = 40, y = 100.

(b) When x = 40, y = 4.

3. System of equations:

(a) r = 8, s = 3.

(b) p = 5, q = -2.

4. Graphing lines:

(a) The slope of the line through (3, 4) and (-1, 3) is 1/4.

(b) The slope of the line 3x - 4y = 7 is 3/4.

(c) The slope-intercept form of the line through (5, -2) and (-1, 6) is y = (-4/3)x + 14/3.

5. Introductory quadratics:

(a) The solutions to the equation 4x² = 81 are -9/2, and 9/2.

(b) The solutions to the equation x² + 8x + 12 are -6, and -2.

(c) The solutions to the equation x² - 3x - 88 are -8, and 11.

6. The weight of an orange is 1 unit, the weight of an apple is 5 units, and the weight of a banana is 2 units.

7. x = 3.

1. Solving linear equations:

(a) 3x - 7 = 9 - x,

or, 3x + x = 9 + 7,

or, 4x = 16,

or, x = 16/4 = 4.

Thus, x = 4.

(b) (7 - 2r)/3 = 4r,

or, 7 - 2r = 3*4r = 12r,

or, -2r - 12r = -7,

or, -14r = -7,

or, r = -7/-14 = 1/2.

Thus, r = 1/2.

2. Proportion:

(a) x and y directly proportion, means x/y = constant.

When x = 8, y = 20.

Thus, x/y = constant, or, 8/20 = constant, or, constant = 0.4.

When x = 40,

x/y = 0.4,

or, y = x/0.4 = 40/0.4 = 100.

Thus, when x = 40, y = 100.

(b) x and y indirectly proportion, means xy = constant.

When x = 8, y = 20.

Thus, xy = constant, or, 8*20 = constant, or, constant = 160.

When x = 40,

xy = 160,

or, 40y = 160,

or, y = 160/40 = 4.

Thus, when x = 40, y = 4.

3. System of equations:

(a) r - s = 5 ...(i)

3r - 5s = 9 ... (ii)

3*(i) - (ii) gives:

3r - 3s = 15

3r - 5s = 9

(-) (+)    (-)

_________

2s = 6,

or, s = 3.

Substituting in (i), we get

r - s = 5,

or, r - 3 = 5,

or, r = 8.

Thus, r = 8, s = 3.

(b) 3p + 7q = 1 ...(i).

5p = 14q + 53,

or, 5p - 14q = 53 ...(ii).

2*(i) + (ii) gives:

6p + 14q = 2

5p - 14q = 53

____________

11p = 55,

or, p = 5.

Substituting in (i), we get:

3p + 7q = 1,

or, 3*5 + 7q = 1,

or, 7q = 1 - 15 = -14,

or, q = -14/7 = -2.

Thus, p = 5, q = -2.

4. Graphing lines:

(a) Slope of the line through (3, 4) and (-1, 3) is,

m = (4 - 3)/(3 - (-1)),

or, m = 1/4.

Thus, the slope of the line through (3, 4) and (-1, 3) is 1/4.

(b) The graph given: 3x - 4y = 7.

Representing in the slope-intercept form, y = mx + b, gives:

3x - 4y = 7,

or, 4y = 3x - 7,

or, y = (3/4)x + (-7/4).

Thus, the slope of the line 3x - 4y = 7 is 3/4.

(c) Slope of the line through (5, -2) and (-1, 6) is,

m = (6 - (-2))/(-1 - 5),

or, m = 8/(-6) = -4/3.

Substituting m = -4/3 in the slope-intercept form, y = mx + b, gives:

y = (-4/3)x + b.

Substituting y = 6, and x = -1 gives:

6 = (-4/3)(-1) + b,

or, b = 6 - 4/3 = 14/3.

Thus, the slope-intercept form of the line through (5, -2) and (-1, 6) is y = (-4/3)x + 14/3.

5. Introductory quadratics:

(a) 4x² = 81,

or, 4x² - 81 = 0,

or (2x)² - 9² = 0,

or, (2x + 9)(2x - 9) = 0.

Either, 2x + 9 = 0 ⇒ x = -9/2,

or, 2x - 9 = 0 ⇒ x = 9/2.

Thus, the solutions to the equation 4x² = 81 are -9/2, and 9/2.

(b) x² + 8x + 12 = 0,

or, x² + 2x + 6x + 12 = 0,

or, x(x + 2) + 6(x + 2) = 0,

or, (x + 6)(x + 2) = 0.

Either, x + 6 = 0 ⇒ x = -6,

or, x + 2 = 0 ⇒ x = -2.

Thus, the solutions to the equation x² + 8x + 12 are -6, and -2.

(c) x² - 3x - 88 = 0,

or, x² - 11x + 8x - 88 = 0,

or, x(x - 11) + 8(x - 11) = 0,

or, (x + 8)(x - 11) = 0.

Either, x + 8 = 0 ⇒ x = -8,

or, x - 11 = 0 ⇒ x = 11.

Thus, the solutions to the equation x² - 3x - 88 are -8, and 11.

6. We assume the weight of one orange, one apple, and one banana to be x, y, and z units respectively.

Thus, we have:

3x + 2y + z = 15 ... (i)

5x + 7y + 2z = 44 ... (ii)

x + 3y + 5z = 26 ... (iii)

2(i) - (ii) gives:

6x + 4y + 2z = 30

5x + 7y + 2z = 44

(-)  (-)  (-)        (-)

______________

x - 3y = -14 ... (iv)

5(i) - (iii) gives:

15x + 10y + 5z = 75

x + 3y + 5z = 26

(-) (-) (-)          (-)

________________

14x + 7y = 49 ... (v)

14(iv) - (v) gives:

14x - 42y = -196

14x + 7y = 49

(-)    (-)     (-)

_____________

-49y = -245,

or, y = 5.

Substituting in (v), we get:

14x + 7y = 49,

or, 14x + 35 = 49,

or, x = 14/14 = 1.

Substituting x = 1 and y = 5 in (i), we get:

3x + 2y + z = 15,

or, 3 + 10 + z = 15,

or, z = 2.

Thus, the weight of an orange is 1 unit, the weight of an apple is 5 units, and the weight of a banana is 2 units.

7. 3/(1 - (2/x)) = 3x,

or, 3/((x - 2)/x) = 3x,

or, 3x/(x - 2) = 3x,

or, 3x = 3x(x - 2),

or, 3x = 3x² - 6x,

or, 3x² - 9x = 0,

or, 3x(x - 3) = 0.

Either, 3x = 0 ⇒ x = 0,

or, x - 3 = 0 ⇒ x = 3.

Since we had a term 2/x, x cannot be 0.

Thus, x = 3.

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Answer:last option -5± √17

2

Step-by-step explanation:

Quadratic equations represent any equation that can be rearranged in standard forma (ax² + bx + c( =0(a, b & c) are known.

Quadratic equations can always be solved using the quadratic formula, but sometimes factoring or isolating the variable is also posible.

In the square equation ax² + bx + c = 0

[tex]\boldsymbol{\sf{x=\dfrac{-b\pm\sqrt{\Delta} }{2a} \ ,\Delta=b^{2}-4ac } }[/tex]

Let's calculate the discriminant of the quadratic equation:

Since the discriminant is greater than zero, then the quadratic equation has two real roots.

[tex]\boldsymbol{\sf{x_{1}=\dfrac{-b-\sqrt{\Delta} }{ 2\cdot a}=\dfrac{-5-\sqrt{17} }{2\cdot1} }}[/tex]

[tex]\boldsymbol{\sf{x_{2}=\dfrac{-b+\sqrt{\Delta} }{ 2\cdot a}=\dfrac{-5+\sqrt{17} }{2\cdot1} }}[/tex]

Solution:

[tex]\boldsymbol{\sf{x=\dfrac{-5\pm\sqrt{17} }{2} }}[/tex]

ヘ( ^o^)ノ＼(^_^ )If you want to learn more about mathematics, I share this link to complement your learning:

A. The pair of linear equations are: x + y = 55 and y = x + 25

B. Number of time she spend running everyday is: 15 minutes.

C. It is not possible.

A system of equations consist of linear equations that have unknown variables whose values make the linear equations in the system true.

Part A: To write a pair of linear equations, do the following,

Let x represent the time Jackie runs

Let y represent the time Jackie dances

We are given that, Jackie runs and dances for a total of 55 minutes every day, therefore, the first linear equation would be:

x + y = 55 --> equation 1

We are also given that Jackie dances for 25 minutes longer than she runs, therefore, the second linear equation would be:

y = x + 25 --> equation 2

Therefore, the two pair of linear equations that shows the relationship between the number of minutes Jackie runs and dances everyday are:

x + y = 55

y = x + 25

Part B: Substitute y = (x + 25) into equation 1

x + y = 55 --> equation 1

x + x + 25 = 55

2x + 25 = 55

2x = 55 - 25

2x = 30

x = 15

Therefore, the number of time she spend running everyday is: 15 minutes.

Part C:

If she spent 45 minutes dancing (y = 45), and x = 15, substitute the values into x + y = 55 to find out if it is possible.

15 + 45 = 55

60 = 55 (not true)

Therefore, it is not possible.

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According to the adjustable-rate mortgage The new monthly payment in year 8 of the loan is $1292.

According o the statement

we have given that the

Remy obtains a 30-year mortgage in the amount of $625,000 And

She secures a 7/1 ARM at an initial interest rate of 3%. Her initial monthly payment is $2,635.03.

And we have to find the new monthly payment when its interest rate increased to the 4.925%.

So, For this purpose we know that the

An adjustable-rate mortgage (ARM) is a loan with an interest rate that changes.

The amount in which she bought mortgage = 625,000

Interest rate = 3%

Initial amount = 2635.03

From this it clear that the

The amount she pays in 7 years = 2635.03*7

The amount she pays in 7 years = 18445.21

The left amount = 625,000 - 18445.21

The left amount = 606554

It means the rate of interest increased on the amount 606554$

So, According to the 4.925% interest the monthly payment will become

amount on interest rate 4.925% = 4.925% *606554/23 years

amount on interest rate 4.925% = $1292.

So, According to the adjustable-rate mortgage The new monthly payment in year 8 of the loan is $1292.

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A)  Rotate by 90° counterclockwise.

B) Reflection transformation about the line y = 5.

C) Parallel Lines are C₂D₂ and A₂B₂; EF and E₁F₁.

D) Perpendicular lines are;  D₁F₁ and  E₁F₁; DF and EF; DC and AB.

E) Line segments with slope of 0 are; AB, C₂D₂, A₂B₂, EF and E₁F₁

F) Line segments on the boat that have an undefined slope are; Lines A₁B₁, DF and DC.

G) Slope of Line ED = 1

A) The blue segment CD is seen on the graph as a perpendicular line with 2 units while the line segment C₂D₂ is seen as a horizontal line. Thus, to match CD with C₂D₂, we will rotate by 90° counterclockwise.

B) The transformations that would be used on the blue triangle to get it to match with the red triangle is a reflection transformation about the line y = 5.

C) The line segments that are parallel to each other are; C₂D₂ and A₂B₂; EF and E₁F₁.

D) The line segments that are perpendicular are;  D₁F₁ and  E₁F₁; DF and EF; DC and AB.

E) Horizontal lines that are parallel to the x-axis have zero slope. Thus, AB, C₂D₂, A₂B₂, EF and E₁F₁ all have zero slopes.

F) Undefined slope is the slope of a vertical line. Thus, Lines A₁B₁, DF and DC have undefined slopes.

G) Slope of ED = (5 - 4)/(-10 - (-11))

Slope of ED = 1/1

Slope = 1

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y=2

-5(2)²-3(2)-2

-5(4)-3(2)-2

-20-6-2

=12

If we compare the given values then we can find that the database B is more likely to have smaller standard deviation.

Given that the values in database A are 0 from 100 and has mean of 50 and Database B has entries from 49 to 51 and also has mean of 50.

We are required to find the database whose standard deviation is lower.

Standard deviation measures the variation of values. It is calculated after finding mean. The square of a standard deviation is known as variance.

Database A has values from 0 to 100 and has mean of 50. Because the values are somewhat very larger than 50 and in database B has values from 49 to 51,there are more chances that the standard deviation of database B will have smaller value than from standard deviation of database A.

Hence if we compare the given values then we can find that the database B is more likely to have smaller standard deviation.

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The y- coordinate that divides the directed line segment from J to K into a ratio of 2:3 is 5. Option C

Let's the point that divides the line segment as point S.

We have that,

Point S divides the line segment into ratio 2:3

The ratio 2:3 means that we are to divide the line segment into;

= 2+3

= 5 equal parts.

We then have that the horizontal distance between the two coordinates is 5

The vertical distance between the two coordinates is 10

Now, let's divide both vertical and horizontal distance into five equal parts,

Horizontal distance = 5/5 = 1

Vertical distance = 10/ 5 = 2

The horizontal distance is 1

The vertical distance is 2

For every one unit move to the left from point J and two units up, we are dividing the line segment into five equal parts as shown in the picture.

The coordinate of point S that divides the line segment into 2 parts and 3 parts is (-5,5)

Thus, the y- coordinate that divides the directed line segment from J to K into a ratio of 2:3 is 5. Option C

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Answer:

See below

Step-by-step explanation:

[tex]A = \{2, 4, 6\}[/tex]

[tex]A^c =\{1, 3, 5\}\\B = \{4, 5, 6\}\\B^c = \{ 1,2, 3\}[/tex]

To prove 1 we have

[tex](A \bigcup B) = \{2, 4, 6\} \bigcup \{4, 5, 6\} = \{2, 4, 5, 6\}\\(A \bigcup B) ^c =\bold{\{1,3\}}\\A^c \bigcap B^c = {\{1, 3, 5\} \bigcap \{1, 2, 3\} =\bold{\{1,3\}}\\[/tex]

Hence proved

To prove 2

[tex](A \bigcap B) = \{2, 4, 6\} \bigcap \{4, 5, 6\} = \{4, 6\}\\\\(A \bigcap B)^c = \bold{\{1,2, 3, 5\}}\\A^c \bigcup B^c = \{1, 3, 5\} \bigcup \{1, 2, 3\} = \bold{\{1,2,3,5\}}\[/tex]

Hence proved

Notes

The complement of a set is the set of all elements not in the set

Here the set of all outcomes is {1, 2, 3,4,5, 6}

So if A = {2, 4, 6} then the complement of A s=is the set of outcomes not in A ie the set {1, 3, 5} which is the set of odd numbers on a die throw

The union of two sets is the set of all elements in both sets without duplication

The intersection of two sets is the set of all elements common to both sets

1529.3 square unit is the area of the cheetah area at a zoo is designed in a triangular fashion, surrounded on all three sides by sidewalks, given that the property has 67 feet of frontage on one sidewalk, and 48 feet of frontage on another; these two sidewalks intersect at a 72° angle. This can obtained by using the formula of Area of a Triangle with 2 Sides and Included Angle.

Area of a triangle is obtained using the formula of Area of a Triangle with 2 Sides and Included Angle.

If in a triangle ΔABC,

Here it is given that,

67 feet and 48 feet are the sides of the triangular space and angle 72° is the included angle.

By using the formula of Area of a Triangle with 2 Sides and Included Angle,

Area of the cheetah area = 1/2 × bc × sin(A)

Area of the cheetah area = 1/2 × (67)(48) × sin(72°)

sin(72°) = 0.951056516

Area of the cheetah area = 1608 × 0.951056516

Area of the cheetah area = 1529.29888 ≈ 1529.3 square unit

Hence 1529.3 square unit is the area of the cheetah area at a zoo is designed in a triangular fashion, surrounded on all three sides by sidewalks, given that the property has 67 feet of frontage on one sidewalk, and 48 feet of frontage on another; these two sidewalks intersect at a 72° angle.

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The number of ways of constructing questions from the pool of 28 questions if her test is to have 3 difficult, 4 average and 3 easy questions is  924, 000 ways

Note that the formula for combination is given as;

Combination = [tex]\frac{n!}{r!(n-r)!}[/tex]

From the information we have that;

There are 28 questions in the pool

The test should have a total of 10 questions;

6 difficult , 10 average and 12 easy questions

We are asked to determine the combination of;

3 difficult questions

4 average questions

3 easy questions

6C3 = [tex]\frac{6!}{3!(6-3)!}[/tex]

6C3 = [tex]\frac{720}{36}[/tex]

6C3 = 20

10C4 = [tex]\frac{10!}{4!(10-4)!}[/tex]

10C4 = [tex]\frac{3628800}{17280}[/tex]

10C4 = 210

12C4 = [tex]\frac{12!}{4!(12-4)!}[/tex]

12C4 = 220

The number of ways of constructing the questions is

= 20 × 210 × 220

= 924, 000 ways

Thus, the number of ways of constructing questions from the pool of 28 questions if her test is to have 3 difficult, 4 average and 3 easy questions is  924, 000 ways

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Answer: Choice 3

Step-by-step explanation:

Oof, this is a tough one. But anyways

Angle 19 + Angle 20 = 180

So Angle 20 = 180 - Angle 19

As l and m are parallel, Angle 20 = Angle 6 from the z rule

So Angle 6 = 180 - Angle 19

Substitute the information given at the start into the equation

2x - 4 = 180 - (5x - 8)

180 - 5x + 8 = 2x - 4

7x = 192

x = 27.42

From what we have found earlier, Angle 20 = Angle 6 so

Angle 20 = 2x - 4 = 2 * 27.42 - 4 = 50.85 rounding it to 50.9

So the answer is 50.9 or Choice 3

Answer:

d = [tex]\sqrt{65}[/tex]

Step-by-step explanation:

The distance between two points is found by

d = [tex]\sqrt{( x2-x1)^2 + ( y2-y1)^2}[/tex] where (x1,y1) and (x2,y2) are the two points

d = [tex]\sqrt{(-2 - -1)^2+(-6 - -14)^2 }[/tex]

d =[tex]\sqrt{( -2+1)^2 + (-6+14)^2}\\[/tex]

d = [tex]\sqrt{(-1)^2 +( 8)^2 }[/tex]

d = [tex]\sqrt{1+64}[/tex]

d = [tex]\sqrt{65}[/tex]

Answer:

x=3/2 y=1/2

Step-by-step explanation:

The Area of the shaded region of the circle = 60.75π m²

The Length of arc ADB = ∅/360 × 2πr = 13.5π m.

The area of the shaded region in the circle = area of the sector of a circle = ∅/360 × πr², where r is the radius of the circle and the central angle is ∅.

The length of the arc on a circle = ∅/360 × 2πr, where r is the radius of the circle and the central angle is ∅.

Given the following:

Central angle (∅) = 360 - 90 = 270°

Radius (r) = 9 m.

Area of the shaded region of the circle = ∅/360 × πr² = 270/360 × π(9²)

Area of the shaded region of the circle = 270/360 × π81

Area of the shaded region of the circle = 60.75π m²

Length of arc ADB = ∅/360 × 2πr = 270/360 × 2π(9)

Length of arc ADB = ∅/360 × 2πr = 270/360 × 18π

Length of arc ADB = ∅/360 × 2πr = 13.5π m.

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Answer:

Step-by-step explanation:

D because the quadratic formula is basically [tex]\frac{-b+\sqrt{b^{2}-4ac } }{2a}[/tex]±

so if we place the equation D into the formula

a = 1

b = +3

c = +24

we can clearly say that D eq is correctly used and can be.

Answer:

1.

Step-by-step explanation:

10C0 = 1.

Answer:

there are 8 million different phone numbers whose phone numbers whose digits aren't 0 or 1

The simplified forms of the given expressions are

a. 32x

b. 0

c. -26t + 14x

d. 3t

From the question, we are to simplify the given expressions by adding or subtracting

a. (10x) + (22x)

= 32x

b. (-6x) - (-6x)

= -6x + 6x

= 0

c. (-13t) + (14x) - (+13t)

= -13t + 14x - 13t

Collect like terms

= -13t -13t + 14x

= -26t + 14x

d. (-4t) + (+4t) - (-3t)

= -4t + 4t + 3t

= 3t

Hence, the simplified forms of the given expressions are

a. 32x

b. 0

c. -26t + 14x

d. 3t

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Answer:

(x-5)(x+1)

Step-by-step explanation:

x² - 4x + 5 = (x-5)(x+1)

Answer:

750 - y(y-5) = 0

Step-by-step explanation:

Area is Length times width.  In a rectangle we have two lengths and 2 widths.

We can use y to represent the length.  The width would be y-5.

Area = LW

750 = y(y-5) or another way to write this is

750 - Y(y-5) = 0

Answer:

3.2

Step-by-step explanation:

4 x 8 / 8 + 2

32/10

=3.2

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Based on the number of years that Peter and Blair have till retirement, the total amount that they need to save to completely fund their income shortfall is $65,545.72.

First, find the income shortfall:

= 75,017 - 54,000

= $21,017

The amount they need to save to cover this shortfall is the future value of this amount at retirement. The rate will be the real rate which is:

= 6% - 2% inflation

= 4%

The future value is:

= Income shortfall x future value interest factor, 4%, 29 years

= 21,017 x 3.1187

= $65,545.72

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The equation that represents the distance, in kilometers, Seb travels (d) based on how many hours he bikes (t) at this rate is d = 20t

The given parameter is

constant rate of 20 kilometers per hour.

This means that

Rate = 20km per hour

When Seb bikes for t hours, the distance is

d = rate * t

This gives

d = 20t

Hence, the equation that represents the distance, in kilometers, Seb travels (d) based on how many hours he bikes (t) at this rate is d = 20t

This means that

t = 5

So, we have:

d = 20 * 5

Evaluate

d = 100

Hence, Seb will travel 100 kilometers if he bikes at this rate for 5 hours

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Answer:

1. d=20t

2. 50

Answer:

(1,2)

Step-by-step explanation:

Substitution:

x + 2y = 5  Solve for x

x = -2y + 5  Substitute -2y + 5 in for x in the second equation

4x - 12y = -20

4(-2y + 5) - 12y = -20  Distribute the 4

-8y + 20 - 12 y = -20  Combine the y term

-20y + 20 = -20  Subtract 20 from both sides

-20y = -40  Divide both sides by -20

y = 2

Plug y into either of the 2 original equations to get x.

x + 2y = 5

x + 2(2) = 5

x + 4 = 5

x = 1

The answer is (2,1).

Elimination:

x + 2y = 5      4x - 12y = -20.  We want to eliminate with the x or the y.  I am going to eliminate the x's that means that I have to multiply the first equation all the way through by -4

(-4)(x + 2y) = (5) (-4)  That makes the equivalent expression

-4x - 8y = -20  I will add that to 4x - 12y = -20

4x - 12y = -20

0x -20y = -40

-20y = -40

y = 2.  Plug 2 into either the 2 original equation to find x.  This time I will select the second original equation to find x.

4x -12y = -20

4x - 12(2) = -20

4x - 24 = -20

4x = 4

x = 1

Answer:

Step-by-step explanation:

I think you are trying to use synthetic division for

[tex]x^{4} -2x^{3} -2x^{2} +3x-5[/tex] divided by x-3

First  

x -3 = 0 , make it equal to 0 and add 3 to both sides

x = 3

Then we write all the coefficients and continue with

a pattern of multiply by 3 and add to the next coefficient.

See attachment.

Using the normal distribution, there is a 0.1894 = 18.94% probability that the sample average will exceed $75.

The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The parameters for this problem are given as follows:

[tex]\mu = 70, \sigma = 40, n = 50, s = \frac{40}{\sqrt{50}} = 5.66[/tex]

The probability that the sample average will exceed $75 is one subtracted by the p-value of Z when X = 75, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

Z = (75 - 70)/5.66

Z = 0.88

Z = 0.88 has a p-value of 0.8106.

1 - 0.8106 = 0.1894.

0.1894 = 18.94% probability that the sample average will exceed $75.

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a) The preparation of a traditional income statement for Carbon Company is as follows:

Carbon Company

Sales revenue                         $600,000

Costs:

Materials               $ 101,300

Energy                       21,590

Setups                      24,360

Purchasing              26,200

Customer service   20,200

Long-term labor     29,400

Administrative        30,700     $253,750

Operating Income                  $346,250

b) The preparation of an activity-based income statement for Carbon Company is as follows:

Carbon Company

Sales revenue                                                                     $600,000

                   Resources supplied    Unused Resources   Used Resources

Materials               $ 101,300                     $ 2,300              $99,000

Energy                       21,590                         2,210                  19,380

Setups                      24,360                                0                 24,360

Purchasing              26,200                         7,800                  18,400

Customer service   20,200                         9,700                 10,500

Long-term labor     29,400                         4,650                 24,750

Administrative        30,700                         5,950                 24,750        

Total                   $253,750                      $32,610              $221,140

Operating Income                                                            $378,860

                          Unused Resources Capacity     Cost Driver Volume

Resources used

Materials                        $ 2,300                                 9,000 pounds

Energy                               2,210                        380 machine-hours

Setups                                     0                                         87 setups

Purchasing                      7,800                         80 purchase orders

Customer service           9,700                                         70 returns

Long-term labor              4,650                              330 labor-hours

Administrative                5,950                              450 labor-hours

Materials $ 101,300

Energy 21,590

Setups 24,360

Purchasing 26,200

Customer service 20,200

Long-term labor 29,400

Administrative 30,700

Sales revenue  $600,000

                   Resources supplied       Unused Resources   Used Resources

Materials               $ 101,300                         $ 2,300              $99,000

Energy                       21,590                             2,210                  19,380

Setups                      24,360                                    0                 24,360

Purchasing              26,200                             7,800                  18,400

Customer service   20,200                             9,700                 10,500

Long-term labor     29,400                             4,650                 24,750

Administrative        30,700                             5,950                 24,750        

Total                   $253,750                          $32,610              $221,140

Thus, the traditional income statement for Carbon Company shows an operating income of $346,250, while the activity-based income statement shows an operating income of $378,860.

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