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laiz [17]
1 year ago
5

What was the problem at kenya airways described in this case? what management, organization, and technology factors contributed

to this problem?
Mathematics
1 answer:
MatroZZZ [7]1 year ago
8 0

The management, organisation, and technology factors contributed to this problem are listed below.

<h3>What was the problem at Kenya Airways ?</h3>

The  problem in the airways was that corporation didn't know its customers, the airline hasn't been able to take use of its market opportunity in recent years.

Airways was unable to evaluate and keep track of its marketing efforts.

The technology factor that contributed were:

  • No reliable systems for tracking and accounting.
  • The technology used was neither accurate nor consistent.

The Organisation factors that contributed were

  • No communication between the organisation and the customers
  • No track record of the online campaigns and advertisement output
  • Customer Relations Needed to be improved.

The Management factors that contributed were

  • The management never gave reviews to the organisation about the failing system
  • The management even didn't take reviews from the customers and from the people working.

To know more about  problem at Kenya Airways

brainly.com/question/27974672

#SPJ1

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Answer:

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Step-by-step explanation:

3 0
2 years ago
Dividing polynomials. Aka the worst math subject ever<br><br> (X^3+729) / (x+8)
Deffense [45]

Answer:

x^2 - 8x + 64 remainder 217 or you could write it as

(x^2 - 8x + 64)+ 217/(x + 8)

Step-by-step explanation:

You first need to write the polynomial in descending order of degree with zero coefficients for x^2 and x terms, so here we have:

x^3 + 729  =   x^3 + 0x^2 + 0x + 729

So dividing by x+8:-

          x^2 - 8x + 64    <----------the quotient.

x + 8 ) x^3 + 0x^2 + 0x + 729

          x^3 + 8x^2

                 - 8x^2 + 0x

                 -8x^2 - 64x

                              64x + 729

                               64x + 512

                                          217 = Remainder.

6 0
3 years ago
The Math Club is sponsoring a bake sale. If their goal is to raise at least $300, how many pies must they sell at $6.00 each in
soldi70 [24.7K]
Given:
bake sale at least $300
price of each pie is $6.00
let the number of pies be represented by x.

Write the inequality:
6x <u>></u> 300  

Solve the inequality:
6x <u>></u> 300
<u>÷6       ÷6</u>
  x <u>></u> 50

Graph the inequality.
y = 6x

x is the number of pies sold; x at least 50 and gradually increases.
y is the total sales


4 0
3 years ago
Read 2 more answers
Please solve it <br>help!!
Mashcka [7]

Its upsidedown, witch one do you need help with?

4 0
3 years ago
In the isosceles △ABC m∠ACB=120° and AD is an altitude to leg BC . What is the distance from D to base AB , if CD=4cm?
snow_lady [41]

If ΔACB is an isosceles triangle, then ∠A ≅ ∠B and AC ≅ CB

Since ∠C = 120° and ∠A + ∠B + ∠C = 180°, then ∠A = 30° and ∠B = 30°

Next, look at ΔADB.  ∠A + ∠D + ∠B = 180°, so ∠A + 90° + 30° = 180° ⇒ ∠A = 30°

Now look at ΔADC.  Since ∠A = 30° in ΔACB, and ∠A = 60° in ΔADB, then ∠A = 30° in ΔADC <em>per angle addition postulate.</em>

Now that we have shown that ΔADB and ΔADC are 30-60-90 triangles, we can use that formula to calculate the side lengths.

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Since AC ≅ BC, then BC = 8 cm. Therefore, BD = 4 + 8 = 12 <em>by segment addition postulate.</em>

Lastly, look at ΔBHD.  Since ∠B = 30° and ∠H = 90°, then ∠D = 60°. So, ΔBHD is also a 30-60-90 triangle.

BD = 12 cm, so HD = \frac{12}{2}cm = 6 cm

Answer: 6 cm



8 0
3 years ago
Read 2 more answers
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