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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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The probability that two people have the same birthday in a room of 20 people is about 41.1%. It turns out that
salantis [7]

Answer:

a) Let X the random variable of interest, on this case we know that:

X \sim Binom(n=20, p=0.411)

This random variable represent that two people have the same birthday in just one classroom

b) We can find first the probability that one or more pairs of people share a birthday in ONE class. And we can do this:

P(X\geq 1 ) = 1-P(X

And we can find the individual probability:

P(X=0) = (20C0) (0.411)^0 (1-0.411)^{20-0}=0.0000253

And then:

P(X\geq 1 ) = 1-P(X

And since we want the probability in the 3 classes we can assume independence and we got:

P= 0.99997^3 = 0.9992

So then the probability that one or more pairs of people share a birthday in your three classes is approximately 0.9992

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:  

P(X)=(nCx)(p)^x (1-p)^{n-x}  

Where (nCx) means combinatory and it's given by this formula:  

nCx=\frac{n!}{(n-x)! x!}  

Solution to the problem

Part a

Let X the random variable of interest, on this case we know that:

X \sim Binom(n=20, p=0.411)

This random variable represent that two people have the same birthday in just one classroom

Part b

We can find first the probability that one or more pairs of people share a birthday in ONE class. And we can do this:

P(X\geq 1 ) = 1-P(X

And we can find the individual probability:

P(X=0) = (20C0) (0.411)^0 (1-0.411)^{20-0}=0.0000253

And then:

P(X\geq 1 ) = 1-P(X

And since we want the probability in the 3 classes we can assume independence and we got:

P= 0.99997^3 = 0.9992

So then the probability that one or more pairs of people share a birthday in your three classes is approximately 0.9992

4 0
3 years ago
Henry and Natalie are writing a comic book. Henry writes 12 pages for the comic book, and Natalie writes 16 pages. They want to
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answer: 1736

Step-by-step explanation: big brain

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The combined area of two squares is 17 square meters. the size of the larger Square are four times as long as the size of the sm
7nadin3 [17]

we are given that ,

The combined area of two squares is 17 square meters.

Area of square=(side)^2

Let the side length of smaller square be x

The size of the larger Square are four times as long as the size of the smaller square

Area \ of  \ smaller \ square \ =x^2

So Area of larger square=4x^2

The combined Area is 17 square meter.

x^2+4x^2=17\\\\5x^2=17\\\\x=\sqrt{\frac{17}{5}}

Hence the smaller square has the side length=\sqrt{\frac{17}{5}}

So the larger square has side length=2\sqrt{\frac{17}{5}}

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3 years ago
The sum of 5 consecutive integers is 265.<br><br>What is the fifth number in this sequence?
tiny-mole [99]
The number missing is 234
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A recipe code lasagna calls for 3 pounds of tomatoes to serve 5 people. A caterer wants to make enough lasagna to serve 110 peop
yKpoI14uk [10]

Answer:

66 pounds of tomatoes

Step-by-step explanation:

A recipe code lasagna calls for 3 pounds of tomatoes to serve 5 people. A caterer wants to make enough lasagna to serve 110 people. How many pounds of tomatoes does he neeed

From the above question:

5 people = 3 pounds of tomatoes

110 people = x pounds of tomatoes

Cross Multiply

5 people × x pounds = 110 people × 3 pounds

x pounds = 110 people × 3 pounds/5 people

x pounds = 66 pounds of tomatoes

4 0
3 years ago
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