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3 years ago

(x2 + 4)(x2 - 4) please help

Mathematics

1 answer:

VikaD [51]3 years ago

4 0

( $x^{2}$ + 4)( $x^{2}$ - 4)

To solve this question you must FOIL (First, Outside, Inside, Last) like so

First:

(x^2 + 4)(x^2 - 4)

x^2 * x^2

x^4

Outside:

(x^2 + 4)(x^2 - 4)

x^2 * -4

-4x^2

Inside:

(x^2 + 4)(x^2 - 4)

4 * x^2

4x^2

Last:

(x^2 + 4)(x^2 - 4)

4 * -4

-16

Now combine all the products of the FOIL together like so...

x^4 - 4x^2 +4x^2 - 16

Combine like terms:

x^4 - 4x^2 +4x^2 - 16

- 4x^2 +4x^2 = 0

x^4 - 16 <<<This is your answer

Hope this helped!

~Just a girl in love with Shawn Mendes

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3 years ago

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Answer:

Step-by-step explanation:

Remember that the sum of the interior angles of a triangle will always total 180.

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Let’s solve for x:

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3 years ago

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2 years ago

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3 years ago

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

4 0

3 years ago