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sertanlavr [38]
3 years ago
12

Compare the functions represented by the ordered pairs:

Mathematics
2 answers:
mel-nik [20]3 years ago
6 0

We are given

Set A: {(5,1), (4,4), (3,9), (2,16), (1,25)}

We can see that for each x value of the set is increasing exponentially.

<h3>We can set up a rule for it y = (6-x)^2</h3>

y=(6-5)^2 = (1)^2 = 1

y=(6-4)^2 = (2)^2 = 4

y=(6-3) = (3)^2 = 9

y=(6-2)^2 = (4)^2 = 16

y=(6-1)^2 = (5)^2 = 25.

For set B:

Set B: {(1,-5), (2,-3), (3,-1), (4,1), (5,3)}

It's simply linear function as diffrences of y values are constant that is 2.

<h3>We can set a rule for it : y = 2x - 7.</h3>

y = 2(1) -7 = 2-7 = -5

y = 2(2) -7 = 4-7 = -3

y = 2(3) -7 = 6-7 = 1

y = 2(4) -7 = 8-7 = 1

y = 2(5) -7 = 10-7 = 3.




Korolek [52]3 years ago
3 0

Answer:

Set A is a quadratic function : y=x^2-12x+36

Set B is a linear function: y=2x-7

Step-by-step explanation:

Here  set A : {(1,25), (2,16), (3,9), (4, 4) , (5, 1)}

This is a quadratic function

Let

y=ax^2+bx+c

For (1,25) we get 25=a+b+c     ....(1)

For (2,16) we get 16=4a+2b+c   ....(2)

For (3,9) we get 9=9a+3b+c     ....(3)

Subtracting (1) from (2) we get

-9=3a+b   .....(4)

Subtracting (2) from (3) we get

-7=5a+b    ....(5)

Subtracting (4) from (5)

2a=2

a=1

Substituting a= 1 in (4), we get

-9=3(1) + b

b=-12

Substituting a = 1 & b = -12 in (1) we get

c=25-1+12= 36

The function is

y=x^2-12x+36

Set B: {(1,-5), (2,-3), (3,-1), (4,1),(5,3)}

This is a linear function

Slope = \frac{-3+5}{2-1}=2

The equation is

y=2x+c

For (1,-5) we get

-5=2+c

c=-7

The equation is

y=2x-7

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Return to the credit card scenario of Exercise 12 (Section 2.2), and let C be the event that the selected student has an America
Nadya [2.5K]

Answer:

A. P = 0.73

B. P(A∩B∩C') = 0.22

C. P(B/A) = 0.5

   P(A/B) = 0.75

D. P(A∩B/C) = 0.4

E. P(A∪B/C) = 0.85

Step-by-step explanation:

Let's call A the event that a student has a Visa card, B the event that a student has a MasterCard and C the event that a student has a American Express card. Additionally, let's call A' the event that a student hasn't a Visa card, B' the event that a student hasn't a MasterCard and C the event that a student hasn't a American Express card.

Then, with the given probabilities we can find the following probabilities:

P(A∩B∩C') = P(A∩B) - P(A∩B∩C) = 0.3 - 0.08 = 0.22

Where P(A∩B∩C') is the probability that a student has a Visa card and a Master Card but doesn't have a American Express, P(A∩B) is the probability that a student has a has a Visa card and a MasterCard and P(A∩B∩C) is the probability that a student has a Visa card, a MasterCard and a American Express card. At the same way, we can find:

P(A∩C∩B') = P(A∩C) - P(A∩B∩C) = 0.15 - 0.08 = 0.07

P(B∩C∩A') = P(B∩C) - P(A∩B∩C) = 0.1 - 0.08 = 0.02

P(A∩B'∩C') = P(A) - P(A∩B∩C') - P(A∩C∩B') - P(A∩B∩C)

                   = 0.6 - 0.22 - 0.07 - 0.08 = 0.23

P(B∩A'∩C') = P(B) - P(A∩B∩C') - P(B∩C∩A') - P(A∩B∩C)

                   = 0.4 - 0.22 - 0.02 - 0.08 = 0.08

P(C∩A'∩A') = P(C) - P(A∩C∩B') - P(B∩C∩A') - P(A∩B∩C)

                   = 0.2 - 0.07 - 0.02 - 0.08 = 0.03

A. the probability that the selected student has at least one of the three types of cards is calculated as:

P = P(A∩B∩C) + P(A∩B∩C') + P(A∩C∩B') + P(B∩C∩A') + P(A∩B'∩C') +              

     P(B∩A'∩C') + P(C∩A'∩A')

P = 0.08 + 0.22 + 0.07 + 0.02 + 0.23 + 0.08 + 0.03 = 0.73

B. The probability that the selected student has both a Visa card and a MasterCard but not an American Express card can be written as P(A∩B∩C') and it is equal to 0.22

C. P(B/A) is the probability that a student has a MasterCard given that he has a Visa Card. it is calculated as:

P(B/A) = P(A∩B)/P(A)

So, replacing values, we get:

P(B/A) = 0.3/0.6 = 0.5

At the same way, P(A/B) is the probability that a  student has a Visa Card given that he has a MasterCard. it is calculated as:

P(A/B) = P(A∩B)/P(B) = 0.3/0.4 = 0.75

D. If a selected student has an American Express card, the probability that she or he also has both a Visa card and a MasterCard is  written as P(A∩B/C), so it is calculated as:

P(A∩B/C) = P(A∩B∩C)/P(C) = 0.08/0.2 = 0.4

E. If a the selected student has an American Express card, the probability that she or he has at least one of the other two types of cards is written as P(A∪B/C) and it is calculated as:

P(A∪B/C) = P(A∪B∩C)/P(C)

Where P(A∪B∩C) = P(A∩B∩C)+P(B∩C∩A')+P(A∩C∩B')

So, P(A∪B∩C) = 0.08 + 0.07 + 0.02 = 0.17

Finally, P(A∪B/C) is:

P(A∪B/C) = 0.17/0.2 =0.85

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