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nignag [31]
1 year ago
7

Revenue management methodology was originally developed for the banking industry. group of answer choices

Mathematics
1 answer:
anygoal [31]1 year ago
8 0

The statement 'Revenue management methodology was originally developed for the banking industry.' is False.

The revenue Management is an analytics technique.

This technique is used to predict consumer behavior at the micro-level, which is ultimately useful in optimizing the product availability and pricing and maximize revenue growth.

This methodology is used by companies in certain industries, particularly those with fixed costs and capacity and products or services that expire.

It is the operational procedures and practices that maximize revenues without creating additional products or services.

Therefore, The statement 'Revenue management methodology was originally developed for the banking industry.' is False.

Learn more about the revenue management here:

brainly.com/question/14884122

#SPJ4

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How do i find the area of this composite figure?
diamong [38]

To find the area of a composite figure, separate the figure into simpler shapes whose area can be found. Then add the areas together. Be sure than none of the simpler figures have overlapping areas

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2 years ago
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A person is watching a boat from the top of a lighthouse. The boat is approaching
Lapatulllka [165]

9514 1404 393

Answer:

  445.10 feet

Step-by-step explanation:

The relation between angle of depression and distance to the boat is ...

  Tan = Opposite/Adjacent

  tan(angle of depression) = (200 ft)/(distance to boat)

Then the distance to the boat is ...

  distance to boat = (200 ft)/tan(angle of depression)

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We want the change in distance between the two angles, so ...

  change in distance = (200 ft)/tan(17°31') -(200 ft)/tan(46°41')

  = (200 ft)(cot(17°31') -cot(46°41'))

  change in distance ≈ 445.10 ft

7 0
3 years ago
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creativ13 [48]
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7 0
2 years ago
8 - (-3) + 33 : (-3)
Lynna [10]

Answer:

Step-by-step explanation:

8-(-3)= 8+3+33=42

7 0
3 years ago
Assume {v1, . . . , vn} is a basis of a vector space V , and T : V ------> W is an isomorphism where W is another vector spac
Degger [83]

Answer:

Step-by-step explanation:

To prove that w_1,\dots w_n form a basis for W, we must check that this set is a set of linearly independent vector and it generates the whole space W. We are given that T is an isomorphism. That is, T is injective and surjective. A linear transformation is injective if and only if it maps the zero of the domain vector space to the codomain's zero and that is the only vector that is mapped to 0. Also, a linear transformation is surjective if for every vector w in W there exists v in V such that T(v) =w

Recall that the set w_1,\dots w_n is linearly independent if and only if  the equation

\lambda_1w_1+\dots \lambda_n w_n=0 implies that

\lambda_1 = \cdots = \lambda_n.

Recall that w_i = T(v_i) for i=1,...,n. Consider T^{-1} to be the inverse transformation of T. Consider the equation

\lambda_1w_1+\dots \lambda_n w_n=0

If we apply T^{-1} to this equation, then, we get

T^{-1}(\lambda_1w_1+\dots \lambda_n w_n) =T^{-1}(0) = 0

Since T is linear, its inverse is also linear, hence

T^{-1}(\lambda_1w_1+\dots \lambda_n w_n) = \lambda_1T^{-1}(w_1)+\dots +  \lambda_nT^{-1}(w_n)=0

which is equivalent to the equation

\lambda_1v_1+\dots +  \lambda_nv_n =0

Since v_1,\dots,v_n are linearly independt, this implies that \lambda_1=\dots \lambda_n =0, so the set \{w_1, \dots, w_n\} is linearly independent.

Now, we will prove that this set generates W. To do so, let w be a vector in W. We must prove that there exist a_1, \dots a_n such that

w = a_1w_1+\dots+a_nw_n

Since T is surjective, there exists a vector v in V such that T(v) = w. Since v_1,\dots, v_n is a basis of v, there exist a_1,\dots a_n, such that

a_1v_1+\dots a_nv_n=v

Then, applying T on both sides, we have that

T(a_1v_1+\dots a_nv_n)=a_1T(v_1)+\dots a_n T(v_n) = a_1w_1+\dots a_n w_n= T(v) =w

which proves that w_1,\dots w_n generate the whole space W. Hence, the set \{w_1, \dots, w_n\} is a basis of W.

Consider the linear transformation T:\mathbb{R}^2\to \mathbb{R}^2, given by T(x,y) = T(x,0). This transformations fails to be injective, since T(1,2) = T(1,3) = (1,0). Consider the base of \mathbb{R}^2 given by (1,0), (0,1). We have that T(1,0) = (1,0), T(0,1) = (0,0). This set is not linearly independent, and hence cannot be a base of \mathbb{R}^2

8 0
3 years ago
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