McCoy Brothers manufactures and sells two products, A and Z in the ratio of 5:2. Product A sells for $86; Z sells for $124. Variable costs for product A are $45; for Z $49. Fixed costs are $429,500. Compute the contribution margin per composite unit.

Answer:

your pics a little blurry but if i am reading it right

the first one

f(x)=log x+5

Answer:

1) 19.0km

2) 23.9mi

3) 53.9yd

4) 30.0mi

(all answers rounded off to the nearest tenth)

Step-by-step explanation:

Please see the attached pictures for full solution.

Answer:

[tex] \sqrt{2 \times 2 \times 2 \times 3 } = \sqrt{24} = \sqrt{4 \times 6} = 2 \sqrt{6} [/tex]

Answer:

Step-by-step explanation:

Answer:

4

Step-by-step explanation:

Answer:

3y-2

Step-by-step explanation:

y+y+y-2

3y-2

The perimeter of the isosceles triangle with base length y - 2 and length of legs y is 3y - 2. legs.

Here,

Base length of triangle is y - 2.

Length of legs of triangle is y.

What is Perimeter of triangle?

Perimeter of triangle is sum of all three sides of a triangle.

Now,

Base length of triangle is y - 2.

Length of legs of triangle is y.

Perimeter = y - 2 + y + y

               = 3y - 2

Hence, The perimeter of the isosceles triangle with base length y - 2 and legs of length y is 3y - 2.

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18m

Step-by-step explanation:

5x3=15 +3 =18

Answer:

Here Ratio=1:2:4

total money =$70

Step-by-step explanation:

let the ratio number be 1x,2x and4x

Now,

1x+2x+4x =$70

or, 7x. =$70

or, x. =$70÷7=$10

again,

1x=1×$10=$10,

2x=2×$10=$20 and

4x=4×$10=$40

Answer:

Seth

Step-by-step explanation:

3/12 = 0.25

0.25*100 = 25%

Seth scored at 25%

2/10 = 0.2

0.2*100 = 0.2

0.2*100 = 20%

Zak scored at 20%

then:

25>20

then:

Seth scored on a greater fraction of his shots.

Zak scored on a greater fraction of his shots.

Given,

In a hockey game, Seth took 12 shots and scored 3 times.

Zak took 10 shots and scored twice.

We need to find who scored on a greater fraction of his shots.

We can compare fractions by finding their decimal values a nd comparing them.

Example: 1/2 and 2/5

1/2 = 0.5

2/5 = 0.4

0.5 is greater than 0.4.

Find the number of score Seth shots out of 12 shots.

= 3

We can write in fractions as:

3/12 _____(1)

Find the number of score Zak shots out of 10 shots.

= 2 x 3

= 6

We can write in fractions as:

6/10 _____(2)

Compare (1) and (2)

3/12 = 1/4 = 0.25
6/10 = 3/5 = 0.6

0.25 is less than 0.6

6/10 is a greater fraction than 3/12

Thus Zak scored on a greater fraction of his shots.

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Answer:

Yes, provided they are all the same size.

Step-by-step explanation:

4/5=0.8

1/2=0.5

3/4=0.75

0.8+0.5+0.75=n

8+5=13

0.8+0.5=1.3

1.3+0.75=2.05

Answer:

AB = 4.5

Step-by-step explanation:

Distance between A(-1, 2) and B(1, -2):

[tex] AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} [/tex]

Let,

[tex] A(-1, 2) = (x_1, y_1) [/tex]

[tex] B(1, -2) = (x_2, y_2) [/tex]

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

[tex] AB = \sqrt{(2)^2 + (-4)^2} [/tex]

[tex] AB = \sqrt{4 + 16} = \sqrt{20} [/tex]

[tex] AB = 4.5 [/tex] (nearest tenth)

Answer:

12, 12 and 12 so 36

Step-by-step explanation:

w²+7w>293 is the inequality which we use to determine what lengths would make the area of the base of the tree house greater than 293 square Inches

The area of Rectangle is length times of width.

Given that Adam built a tree house with a rectangular base.

The length of the base is 7 inches more than its width.

Length=W+7

W is the width

The area of the base of the tree house greater than 293

We need to find the inequality to represent the area of the base of tree

Area of rectangle=Length×width

293=(w+7)w

293=w²+7w

As area is greater we have

w²+7w>293

Hence, w²+7w>293 is the inequality which we use to determine what lengths would make the area of the base of the tree house greater than 293 square Inches

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Answer:

a.  y= e raise to power y

c. y = e^ky

Step-by-step explanation:

The first derivative is obtained by making the exponent the coefficient and decreasing the exponent by 1 . In simple form the first derivative of

x³ would be 2x³-² or 2x².

But when we take the first derivative of  y= e raise to power y

we get y= e raise to power y. This is because the derivative of e raise to power is equal to e raise to power y.

On simplification

y= e^y

Applying ln to both sides

lny= ln (e^y)

lny= 1

Now we can apply chain rule to solve ln of y

lny = 1

1/y y~= 1

y`= y

therefore

derivative of e^y = e^y

The chain rule states that when we have a function having one variable and one exponent then we first take the derivative w.r.t to the exponent and then with respect to the function.

Similarly when we take the first derivative of  y= e raise to power ky

we get y=k multiplied with e raise to power ky. This is because the derivative of e raise to a  constant and power is equal to constant multiplied  with e raise to power y.

On simplification

y= k e^ky

Applying ln to both sides

lny=k ln (e^y)

lny=ln k

Now we can apply chain rule to solve ln of y ( ln of constant would give a constant)

lny = ln k

1/y y~= k

y`=k y

therefore

derivative of e^ky = ke^ky

Answer:

Step-by-step explanation:

Answer:

B) -2

Step-by-step explanation:

[tex]\frac{25-(-15)}{-12-8} =\frac{40}{-20} =-2[/tex]

Answer:

(a) 0.16759

(b)  0.9649

Step-by-step explanation:

Given that:

n = 10 , p = 0.3 and  [tex]\hat p = 0.4[/tex]

[tex]P(|\hat p - p| > 0.2 ) = 1 - P ( |\hat p -p| \leq 0.2)[/tex]

= [tex]1 - P(-0.2 \leq \hat p-p \leq 0.2)[/tex]

= [tex]1 - P \begin {pmatrix} \dfrac{-0.2}{\sqrt{\dfrac{pq}{n}}} \leq \dfrac{\hat p -p}{\sqrt{\dfrac{pq}{n}}} \leq \dfrac{0.2}{\sqrt{\dfrac{pq}{n}}} \end {pmatrix}[/tex]

[tex]=1 - P \begin {pmatrix} \dfrac{-0.2}{\sqrt{\dfrac{(0.3)(1-0.3)}{10}}} \leq Z \leq \dfrac{0.2}{\sqrt{\dfrac{(0.3)(1-0.3)}{10}}} \end {pmatrix}[/tex]

[tex]=1 - P \begin {pmatrix} \dfrac{-0.2}{\sqrt{\dfrac{(0.3)(0.7)}{10}}} \leq Z \leq \dfrac{0.2}{\sqrt{\dfrac{(0.3)(0.7)}{10}}} \end {pmatrix}[/tex]

[tex]=1 - P \begin {pmatrix} \dfrac{-0.2}{\sqrt{0.021}} \leq Z \leq \dfrac{0.2}{\sqrt{0.021}} \end {pmatrix}[/tex]

[tex]=1 - P \begin {pmatrix} -1.380 \leq Z \leq 1.380 \end {pmatrix}[/tex]

= 1 - P( Z ≤ 1.380) - P(-1.380)

= 1 - ( 0.91620 - 0.08379 )

= 1 - 0.83241

= 0.16759

b) when n = 400; p =0.3 , q = 1 - p = 1 - 0.3 = 0.7

[tex]P( |\hat p - p | > 0.001) = 1- P ( |\hat p - p | < 0.001 )[/tex]

[tex]= 1- P ( -0.001 < \hat p - p < 0.001 )[/tex]

[tex]= 1- P ( \dfrac{-0.001}{\sqrt{\dfrac{pq}{n}}} < \dfrac{ \hat p - p}{\dfrac{pq}{n}} < \dfrac{0.001}{\sqrt{\dfrac{pq}{n}}} )[/tex]

[tex]= 1- P ( \dfrac{-0.001}{\sqrt{\dfrac{0.3\times 0.7}{400}}} < Z < \dfrac{0.001}{\sqrt{\dfrac{0.3 \times 0.7}{400}}} )[/tex]

[tex]= 1- P ( \dfrac{-0.001}{\sqrt{5.25 \times 10^{-4}}} < Z < \dfrac{0.001}{\sqrt{5.25 \times 10^{-4}}} )[/tex]

[tex]= 1- P ( -0.0436< Z < 0.0436)[/tex]

= 1 - P ( Z < 0.0436) - P ( -0.0436)

= 1 - (0.5176 - 0.4825)

= 1 -  0.0351

= 0.9649

Answer:

i believe it's 39

Step-by-step explanation:

29 -39 = -10

ans is 25

Step-by-step explanation:

12(10-5)-40+(4+1)

=12 of 5 - 40 + 5

= 60 - 40 +5

=60 +5 - 40

=65-40

=25

Answer:

-25

Step-by-step explanation:

We start off by using PEMDAS. 10-5 then 4+1.

12(5)-40+5

we multiply, then add and subtract giving us -25

Answer:

[tex]\frac{x}{5} + 6y[/tex]

Step-by-step explanation:

Answer:

C. .024

Step-by-step explanation:

You multiply 25 by 4 to get 100 and then you multiply 6 by 4 to get twenty four. After that you divide both numbers by one hundred and you get C

Answer: Area of Polygon Q is  [tex]\dfrac14[/tex]  of  Area of Polygon

Step-by-step explanation:

Given: Polygon Q is a scaled copy of Polygon P using a scale factor of [tex]\dfrac12[/tex] .

According to the property of scale factor ,

Area of the image = (scale factor )²x (Area of the original figure)

So, Area of Polygon Q = [tex](\dfrac12)^2[/tex]  x (Area of Polygon P)

⇒ Area of Polygon Q = [tex]\dfrac14[/tex]  x (Area of Polygon P)

Hence,  Area of Polygon Q is   [tex]\dfrac14[/tex]  of  Area of Polygon P.

Answer:

1/4

I did on khan

Step-by-step explanation:

Answer:

The saying money breeds money means that with money, you have the possibility to create more money. Simple interest is the idea that you have something, and you obtain a percentage of that every designated time period. That being said, simple interest = money breeds money

Answer:

7 and 9

Step-by-step explanation:

So we want two consecutive odd numbers whose product is 63.

Let's write an equation.

Let's let n be a random integer: doesn't matter what it is. Therefore, the first integer must be 2n+1.

This is because we're letting n be whatever it wants to be. If we multiply that whatever number by 2, then it will turn even. If we add 1 to an even number, it becomes odd.

Therefore, our first odd number is (2n+1). Our second, then, must be (2n+3).

Multiply them together. They equal 63. Thus:

[tex](2n+1)(2n+3)=63[/tex]

Expand:

[tex]4n^2+6n+2n+3=63[/tex]

Combine like terms:

[tex]4n^2+8n+3=63[/tex]

Subtract 63 from both sides:

[tex]4n^2+8n-60=0[/tex]

Divide both sides by 4:

[tex]n^2+2n-15=0[/tex]

And now, factor:

[tex](n+5)(n-3)=0[/tex]

Zero Product Property:

[tex]n+5=0\text{ or } x-3=0[/tex]

Find n:

[tex]n=-5\text{ or } n=3[/tex]

So, we've found n.

Then the first integer is either:

[tex]2(-5)+1 \text{ or } 2(3)+1[/tex]

Evaluate:

[tex]-9 \text{ or } 7[/tex]

However, we want two consecutive odd natural numbers. So, ignore the -9.

Therefore, our first odd integer is 7.

And our second one would be 9.

So, our answer is 7 and 9.

And we're done!

Answer:

3m ≤ ∫ f(x) dx ≤ 3M, at limit of b, a

Step-by-step explanation:

Like the question asked, which property of integral was used.

Property 8 of integrals was the basis upon which the question was solved.

The property of the integral is used to solve the problem. The function is given below.

[tex]3m & \leq \int_a^b f(x)dx \leq 3M \end{aligned}[/tex]

The function is an expression, rule, or law that defines the relationship between one variable to another variable. Functions are ubiquitous in mathematics and are essential for formulating physical relationships.

Suppose f has absolute minimum value m and absolute maximum value M.

We know that the property of the integral can be used.

If m ≤ f(x) ≤ M ∀ a ≤ x ≤ b. Then we have

[tex]m (b-a) \leq \int_a^b f(x) dx \leq M(b-a)\\[/tex]

Then we have

[tex]m \leq f(x) \leq M \ \ and \ \ 4 \leq x \leq 7[/tex]

This means that

[tex]\begin{aligned} m(7-4) & \leq \int_a^b f(x)dx \leq M(7-4)\\\\3m & \leq \int_a^b f(x)dx \leq 3M \end{aligned}[/tex]

More about the function link is given below.

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step 1: open up the brackets ( rmb to flip the signs )

step 2: bring 'a' to one side and the numbers to the other

6a + 34 = -3(-2a + 8)

6a + 34 = 6a - 24

6a - 6a = -24 - 34

0 = - 58

the final ans is kind of weird but here's my solution :))

Answer:

47.53

Step-by-step explanation:

First you add all of the numbers together.

66.9+5.6+70.1=142.6

Then you divide the sum of numbers by how many number there are.

142.6/3=47.53

Answer:

Step-by-step explanation:

3x^2 - 4x^2 - 6mn + 5mn

-x^2 - mn