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Pavlova-9 [17]
4 years ago
14

Ajay Analyzed two companies and has formulated tables to model each company growth, where x represents the age of the companies

in years and y represents growth in terms of number of employees

Mathematics
2 answers:
loris [4]4 years ago
5 0

If the number of employees is an indicator of a successful business, the company that Ajay should invest is company 1 because it is adding six employees each year while company 2 is multiplying its number of employees by six each year. You can also see form the table above that as the number of years increased, the number of employees also increases. Company 1 and 2 is directly proportional with the number of years and the number of employees.

Hope this helps!

tensa zangetsu [6.8K]4 years ago
5 0

Answer:

Option D.

Step-by-step explanation:

In the given question the number of employees is an indicator of a successful business.

Now we will analyze how the employees are being added in two companies.

Ajay will invest his money in the company which adds the employees more.

Company 1.

6 employees are being added so addition of employees are in the arithmetic progression.

Company 2.

There is a common ratio of 6 in each term of y which indicates the sequence formed is an exponential sequence.

And it's a fact that exponential growth is always greater than the linear growth.

Therefore, company 2 should be the choice of Ajay to invest his money.

Option D. is the answer.

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A large but sparsely populated county has two small hospitals, one at the south end of the county and one at the north end. The
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Step-by-step explanation:

Probability distribution of this type is a vicariate distribution because it specifies two random variable X and Y. We represent the probability X takes x and Y takes y by

f(x,y) = P((X=x,Y=y) ,

and given that the random variables are independent the joint pmf isbgiven by:

f(x,y) = fX(x) .fY(y) and this gives the table (see attachment)

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f(x,y) = sum{(P(X ≤ 1, Y ≤ 1) }

= [0.1 +0.2 ] × [0.1 + 0.3]

= 0.3 × 0.4

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Now we find

fX(x) = sum{[f(x.y)]} for y =0, 1 and similarly for fY(y)= sum{[f(x,y)]} for x=0, 1

fX(x) = sum{[f(x,y)]}= 0.1+0.3

= 0.4

fY(y) = sum{[f(x.y)]} = 0.1 + 0.2

= 0.3

Since fY(y) × fX(x) = 0.4 × 0.3

= 0.12

Hence f(x,y) = fX(x) .fY(y), the events are independent

(c) the required event is give by: [P(X<=1, PY<=1)]

= P(X<=1) . P(Y<=1)

= [sum{[f(x.y]} over Y=0,1] × [sum{[f(x.y)]}, over X= 0, 1

= 0.4 × 0.3

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(d) the required event is given by the idea that: either The South has no bed and the North has or the the North has no bed and the South has. Let this event be A

A =sum[(P(X = 1<= x<= 4, Y =0)] + sum[P(X =0 Y= 1 <= x <= 3)]

A = [0.02 + 0.03 + 0.02 + 0.02] + [0.03 + 0.04 + 0.02]

A = [0.09] + [ 0.09]

A = 0.18

Pls note the sum of the vertical column X=2 is 0.3

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