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2 years ago

Write three to five sentences explaining which levels of production provide Alonzo’s Cycling with the maximum profit.

Mathematics

2 answers:

denis-greek [22]2 years ago

7 0

Answer:

The maximum profit can be attained when 4 bikes are produced each day.

Step-by-step explanation:

Look at the attached picture:

In the table given in the picture, the number of bikes produces varies. We cannot properly compare the profits per day. To be consistent, let us determine the profit per unit of bike produced.The relationship between cost, revenue and profit can be as:

Profit = Revenue - Cost

To find the profit per unit of bike, simply divide the profit with the number of bikes produced (1st column). After you see the results, we can see that the highest profit is $17.5 per unit of bike produced. Therefore, the maximum profit can be attained when 4 bikes are produced each day.

daser333 [38]2 years ago

4 0

Answer: Sample response:Based on the data in the graphs, Alonzo's Cycling should produce five or six bikes if it wants to maximize its profits. This is because the sixth bike is the point at which the marginal revenue matches the marginal cost. The company will make $90 in profit with the sale of its fifth or sixth bike each day. After that, the profit begins to fall with each sale.

Step-by-step explanation:

yeah this is the actual answer. Youre welcome. just took the test and got this.

You might be interested in

Deal with these relations on the set of real numbers: R₁ = {(a, b) ∈ R² | a > b}, the "greater than" relation, R₂ = {(a, b) ∈

uranmaximum [27]

Answer:

a) R1ºR1 = R1

b) R1ºR2 = R1

c) R1ºR3 = $\{ (a,b) \in R^2 \}$

d) R1ºR4 = $\{ (a,b) \in R^2 \}$

e) R1ºR5 = R1

f) R1ºR6 = $\{ (a,b) \in R^2 \}$

g) R2ºR3 = $\{ (a,b) \in R^2 \}$

h) R3ºR3 = R3

Step-by-step explanation:

R1ºR1

(a,c) is in R1ºR1 if there exists b such that (a,b) is in R1 and (b,c) is in R1. This means that a > b, and b > c. That can only happen if a > c. Therefore R1ºR1 = R1

R1ºR2

This case is similar to the previous one. (a,c) is in R1ºR2 if there exists b such that (a,b) is in R2 and (b,c) is in R1. This means that a ≥ b, and b > c. That can only happen if a > c. Hence R1ºR2 = R1

R1ºR3

(a,c) is in R1ºR3 if there exists b such that a c. Independently of which values we use for a and c, there always exist a value of b big enough so that b is bigger than both a and c, fulfilling the conditions. We conclude that any pair of real numbers are related.

R1ºR4

This is similar to the previous one. Independently of the values (a,c) we choose, there is always going to be a value b big enough such that a ≤ b and b > c. As a result any pair of real numbers are related.

R1ºR5

If a and c are related, then there exists b such that (a,b) is in R5 and (b,c) is in R1. Because of how R5 is defined, b must be equal to a. Therefore, (a,c) is in R1. This proves that R1ºR5 = R1

R1ºR6

The relation R6 is less restrictive than the relation R3, if we find 2 numbers, one smaller than the other, in particular we find 2 different numbers. If we had 2 numbers a and c, we can find a number b big enough such that ac. In particular, b is different from a, so (a,b) is in R6 and (b,c) is in R1, which implies that (a,c) is in R1ºR6. Since we took 2 arbitrary numbers, then any pair of real numbers are related.

R2ºR3

This is similar to the case R1ºR3, only with the difference that we can take b to be equal to a as long as it is bigger than c. We conclude that any pair of real numbers are related.

R3ºR3

If a and c are real numbers such that there exist b fulfilling the relations a < b and b < c, then necessarily a < c. If a < c, then we can use any number in between as our b. Therefore R3ºR3 = R3

I hope you find this answer useful!

5 0

2 years ago

Find all x in set of real numbers R Superscript 4 that are mapped into the zero vector by the transformation Bold x maps to Uppe

sukhopar [10]

Answer:

$x_3 = \left[\begin{array}{c}4&3&1\\0\end{array}\right]$

Step-by-step explanation:

According to the given situation, The computation of all x in a set of a real number is shown below:

First we have to determine the $\bar x$ so that $A \bar x = 0$

$\left[\begin{array}{cccc}1&-3&5&-5\\0&1&-3&5\\2&-4&4&-4\end{array}\right]$

Now the augmented matrix is

$\left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\2&-4&4&-4\ |\ 0\end{array}\right]$

After this, we decrease this to reduce the formation of the row echelon

$R_3 = R_3 -2R_1 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\0&2&-6&6\ |\ 0\end{array}\right]$

$R_3 = R_3 -2R_2 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\0&0&0&-4\ |\ 0\end{array}\right]$

$R_2 = 4R_2 +5R_3 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&4&-12&0\ |\ 0\\0&0&0&-4\ |\ 0\end{array}\right]$

$R_2 = \frac{R_2}{4}, R_3 = \frac{R_3}{-4} \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&1\ |\ 0\end{array}\right]$

$R_1 = R_1 +3 R_2 \rightarrow \left[\begin{array}{cccc}1&0&-4&-5\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&-1\ |\ 0\end{array}\right]$

$R_1 = R_1 +5 R_3 \rightarrow \left[\begin{array}{cccc}1&0&-4&0\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&-1\ |\ 0\end{array}\right]$

$= x_1 - 4x_3 = 0\\\\x_1 = 4x_3\\\\x_2 - 3x_3 = 0\\\\ x_2 = 3x_3\\\\x_4 = 0$

$x = \left[\begin{array}{c}4x_3&3x_3&x_3\\0\end{array}\right] \\\\ x_3 = \left[\begin{array}{c}4&3&1\\0\end{array}\right]$

By applying the above matrix, we can easily reach an answer

5 0

1 year ago

USING STRUCTURE An athletic facility is building an indoor track. The track is composed of a rectangle and two semicircles, as s

Nesterboy [21]

Answer:

Step-by-step explanation:

Find the digram attached.

Perimeter of the track = perimeter of the rectangle + perimeter of the 2semicircles

Perimter of a rectangle = 2(x+r) where:

x is the length

2r is the width of the rectangle = diameter of the semicircle

Perimeter of semicircle = 2πr/2 = πr

Perimeter of 2semicirle = 2πr

Perimeter of the track = 2(x+2r) + 2πr

r is the radius if the semicircle

Expand

Perimeter of the track = 2x+4r + 2πr

Perimeter of the track = 2(x+2r+πr)

b) Given P = 2(x+2r+πr), we are to make x the subject of the formula.

P = 2x+4r+2πr

P-4r-2πr = 2x

Divide both sides by 2.

(P-4r-2πr)/2 = 2x

x = (P-4r-2πr)/2

c) Given

P = 600fr

r = 50ft

x = (600-4(50)-2π(50))/2

x = (600-200-100(3.14))/2

x = 400-314/2

x = 86/2

x = 43ft

Hence the value of x to nearest foot is 43ft

4 0

2 years ago

Suppose you have a rectangular shipping box that measures 30 cm × 15 cm × 20 cm. What is the longest length of a 1 cm diameter w

atroni [7]

The longest length of dowel will feet in the dimension that will give us the highest length of the hypotenuse;
given that the dimension of our box is:
15 by 20 by 30:
this will be given by:
c^2=a^2+b^2+c^2
c^2=15^2+20^2+30^2
c^2=225+400+900
c^2=1525
c=√1525
c=39.05 cm
Therefore the longest dowel will be 39 cm (to the nearest whole centimeter)

7 0

1 year ago

Read 2 more answers

A student solving for the acceleration of an object has applied appropriate physics principles and obtained the expression a=a1+

egoroff_w [7]

Answer:

$a=\frac{3\times 7}{7}+\frac{12}{7}$ (First step for obtaining a common denominator for the two fraction)