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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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tester [92]
The correct answer would be 128.
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3 years ago
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Colin was thinking of a number. Colin subtracts 5 from it and gets an answer of 10.7. What was the original number?
Gemiola [76]

Answer:

15.7

Step-by-step explanation:

10.7 + 5 = 15.7

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Answer:

The copy machine makes 87.5 copies in one minute.

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3 0
3 years ago
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natta225 [31]
The sum of the opposite interior angles is equal to the exterior angle.

So,

∠CAB + ∠ACB = ∠CBD

x + 40 + 3x + 10 = 6x

4x + 50 = 6x

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x = 25°

∠CAB = x + 40 = 25 + 40 = 65°

Hence, the answer is D.


4 0
3 years ago
16) Please help with question. WILL MARK BRAINLIEST + 10 POINTS.
Katyanochek1 [597]
We will use the sine and cosine of the sum of two angles, the sine and consine of \frac{\pi}{2}, and the relation of the tangent with the sine and cosine:

\sin (\alpha+\beta)=\sin \alpha\cdot\cos\beta + \cos\alpha\cdot\sin\beta

\cos(\alpha+\beta)=\cos\alpha\cdot\cos\beta-\sin\alpha\cdot\sin\beta

\sin\dfrac{\pi}{2}=1,\ \cos\dfrac{\pi}{2}=0

\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha}

If you use those identities, for \alpha=x,\ \beta=\dfrac{\pi}{2}, you get:

\sin\left(x+\dfrac{\pi}{2}\right) = \sin x\cdot\cos\dfrac{\pi}{2} + \cos x\cdot\sin\dfrac{\pi}{2} = \sin x\cdot0 + \cos x \cdot 1 = \cos x

\cos\left(x+\dfrac{\pi}{2}\right) = \cos x \cdot \cos\dfrac{\pi}{2} - \sin x\cdot\sin\dfrac{\pi}{2} = \cos x \cdot 0 - \sin x \cdot 1 = -\sin x

Hence:

\tan \left(x+\dfrac{\pi}{2}\right) = \dfrac{\sin\left(x+\dfrac{\pi}{2}\right)}{\cos\left(x+\dfrac{\pi}{2}\right)} = \dfrac{\cos x}{-\sin x} = -\cot x
3 0
3 years ago
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