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Hunter-Best [27]
2 years ago
8

Weber's Mail Manufacturing is a company that produces chain mail. Their most popular product is saw-cut metal links that hobbyis

ts can use to make their own creations. The company recently spent $600 on advertising, which they hope to make back quickly. The wire used to produce one ounce of links costs $1, but the finished links sell for $4 per ounce. Eventually, the company will make back its investment and break even. What will their expenses and sales both equal?
Mathematics
1 answer:
mariarad [96]2 years ago
8 0

Answer:

Their expenses and sales will both equal at 200 ounces.

Step-by-step explanation:

This can be determined using the break-even point in ounce formula as follows:

Break even point in ounce = Fixed cost / Contribution per ounce

Where;

Fixed cost = Advertising = $600

Contribution per ounce = Selling price per ounce - Cost per ounce = $4 - $1 = $3

Substituting the values into equation (1), we have:

Break even point in ounce = $600 / $3 = 200 ounces

Therefore, their expenses and sales will both equal at 200 ounces.

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Which expressions are equivalent to 64’1Check all that apply.
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3 years ago
The base of an aquarium with given volume V is made of slate and the sides are made of glass. If the slate costs seven times as
Olin [163]

Answer:

x = ∛(2V/7)

y = ∛(2V/7)

z = 3.5 [∛(2V/7)]

{x,y,z} = { ∛(2V/7), ∛(2V/7), 3.5[∛(2V/7)] }

Step-by-step explanation:

The aquarium is a cuboid open at the top.

Let the dimensions of the base of the aquarium be x and y.

The height of the aquarium is then z.

The volume of the aquarium is then

V = xyz

Area of the base of the aquarium = xy

Area of the other faces = 2xz + 2yz

The problem is to now minimize the value of the cost function.

The cost of the area of the base per area is seven times the cost of any other face per area.

With the right assumption that the cost of the other faces per area is 1 currency units, then, the cost of the base of the aquarium per area would then be 7 currency units.

Cost of the base of the aquarium = 7xy

cost of the other faces = 2xz + 2yz

Total cost function = 7xy + 2xz + 2yz

C(x,y,z) = 7xy + 2xz + 2yz

We're to minimize this function subject to the constraint that

xyz = V

The constraint can be rewritten as

xyz - V = 0

Using Lagrange multiplier, we then write the equation in Lagrange form

Lagrange function = Function - λ(constraint)

where λ = Lagrange factor, which can be a function of x, y and z

L(x,y,z) = 7xy + 2xz + 2yz - λ(xyz - V)

We then take the partial derivatives of the Lagrange function with respect to x, y, z and λ. Because these are turning points and at the turning point, each of the partial derivatives is equal to 0.

(∂L/∂x) = 7y + 2z - λyz = 0

λ = (7y + 2z)/yz = (7/z) + (2/y) (eqn 1)

(∂L/∂y) = 7x + 2z - λxz = 0

λ = (7x + 2z)/xz = (7/z) + (2/x) (eqn 2)

(∂L/∂z) = 2x + 2y - λxy = 0

λ = (2x + 2y)/xy = (2/y) + (2/x) (eqn 3)

(∂L/∂λ) = xyz - V = 0

We can then equate the values of λ from the first 3 partial derivatives and solve for the values of x, y and z

(eqn 1) = (eqn 2)

(7/z) + (2/y) = (7/z) + (2/x)

(2/y) = (2/x)

y = x

Also,

(eqn 1) = (eqn 3)

(7/z) + (2/x) = (2/y) + (2/x)

(7/z) = (2/y)

z = (7y/2)

Hence, at the point where the box has minimal area,

y = x,

z = (7y/2) = (7x/2)

We can then substitute those into the constraint equation for y and z

xyz = V

x(x)(7x/2) = V

(7x³/2) = V

x³ = (2V/7)

x = ∛(2V/7)

y = x = ∛(2V/7)

z = (7x/2) = 3.5 [∛(2V/7)]

The values of x, y and z in terms of the volume that minimizes the cost function are

{x,y,z} = {∛(2V/7), ∛(2V/7), 3.5[∛(2V/7)]}

Hope this Helps!!!

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Answer:

5.7

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Answer:

C) There is not sufficient evidence to support the claim that the mean attendance is greater than 523.

Step-by-step explanation:

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Null and alternative hypotheses are:

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This means that <em>test statistic</em> is lower than <em>critical value</em>.  Therefore it is not significant, there is no significant evidence to accept the <em>alternative</em> hypothesis.

That is no significant evidence that the average attendance at games of the football team is greater than 523.

7 0
2 years ago
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