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OlgaM077
1 month ago
9

Which transformations could have occurred to map △ABC to △A"B"C"? a rotation and a reflection a translation and a dilation a ref

lection and a dilation a dilation and a rotation
Mathematics
1 answer:
Inessa [12.5K]1 month ago
4 0

Response:

(B) Reflection and Rotation

Detailed explanation:

You might be interested in
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
zzz [12365]

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:

Denote A as the event of a student having a Visa card, B as the event of holding a MasterCard, and C as the event of owning an American Express card. Additionally, let A' indicate the event of not having a Visa card, B' signify not having a MasterCard, and C denote the event of not possessing an American Express card.

Thus, with the given probabilities, we can determine the following probabilities:

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

Here, P(A∩B∩C') refers to the chance that a student has both a Visa and MasterCard but does not own an American Express, P(A∩B) indicates the probability that a student possesses both a Visa and a MasterCard, and P(A∩B∩C) represents the likelihood that a student has a Visa, MasterCard, and American Express. Similarly, 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 likelihood that the selected student holds at least one of the three card types is calculated as follows:

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 chosen student possesses both a Visa and a MasterCard without an American Express card can be represented as P(A∩B∩C') equaling 0.22

C. P(B/A) represents the chance that a student holds a MasterCard provided they have a Visa. This is calculated as:

P(B/A) = P(A∩B)/P(A)

By substituting in the values, we find:

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

In a similar manner, P(A/B) represents the probability a student has a Visa given they possess a MasterCard, calculated as:

P(A/B) = P(A∩B)/P(B) = 0.3/0.4 = 0.75

D. For a student with an American Express card, the likelihood they also hold both a Visa and a MasterCard is expressed as P(A∩B/C), calculated as:

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

E. If the student has an American Express card, the probability they possess at least one of the other two card types is denoted as P(A∪B/C), computed 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')

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

Ultimately, P(A∪B/C) equals:

P(A∪B/C) = 0.17/0.2 =0.85

4 0
1 month ago
Jada plans to serve milk and healthy cookies for a book club meeting. She is preparing 12 ounces of milk and 4 cookies per perso
Zina [12379]

Jada intends to provide milk and nutritious cookies at a gathering for her book club. She has planned to offer 12 ounces of milk and 4 cookies for each participant. With her included, the total number of club members is 15. Each package of cookies has 24 cookies and is priced at $4.50. A gallon of milk has 128 ounces and costs $3. Let n denote the club members, m for the ounces of milk, c for the cookies, and b for Jada's budget in dollars. Identify all equations that could represent the relevant quantities and constraints in this scenario.

â m = 12(15)

n refers to the club members

b. 3m + 4.5c = 6

c. 4n =

m stands for ounces of milk

& 44.50) = C

c indicates the cookies

e b=2(3) + 3(4.50)

b denotes Jada's budget in

currency

The equations applicable for representing the quantities and constraints in this scenario are:

a) m = 12(15)

c)4n = c

e) b = 2(3) + 3(4.50)

Detailed explanation:

Jada is set to provide milk and nutritious cookies for her book club meeting

She plans to prepare 12 ounces of milk and 4 cookies for each individual

She has 15 members in total

A pack of cookies contains 24 cookies and costs $4.50

A gallon of milk has 128 ounces and costs $3

Designate the total members by n

Ounces of milk = m

The number of cookies = c

Jada's budget in dollars = d

y = 12(15)

Each individual will receive 4 cookies

4n = c

As one pack contains 24 cookies

She requires = 15 × 4 = 60 cookies

60 ÷ 24 = 2.5

Thus, she would need to purchase 3 packs of cookies

She needs = 15 × 12 = 180 ounces of milk

A gallon of milk contains 128 ounces

180 ÷ 128 = 1.406

She will need to acquire 2 gallons of milk

It's noted: from the prompt

Each gallon of milk costs $3

Each package of cookies costs $4.50

b serves as Jada's dollar budget

b = 2(3) + 3(4.50)

<pThe relevant equations that could signify the quantities and constraints in this scenario are:

Options a, c and e are valid

3 0
2 months ago
If we want to provide a 95% confidence interval for the mean of a population, the confidence coefficient will be
Inessa [12570]

Please clarify the question as I don't quite understand it.

5 0
2 months ago
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