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3 years ago

Hhelp me pleaseeeeee

Mathematics

1 answer:

svetlana [45]3 years ago

4 0

A. Area = ½b × h

= ½ × 16 × 9

= 72

B. Area = b × h

= 26 × 18

= 468

C. Area = a²

= 11²

= 121

D. Area = ½(a+b) × h

= ½(6+21) × 8

= 108

Hope this helps :)

Please consider marking my answer the brainliest!

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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

4 0

3 years ago

1/2y^2(6x+2y+12x-2y) is equivalent

vazorg [7]

The answer is 9y^2x(the 2 goes on top were the arrow is ) Hope this was helpful!

3 0

3 years ago

Two dice are numbered 1,2,3,4,5,6 and 1,1,2,2,3,3 respectively. They are rolled and

Alexxx [7]

Answer:

P(2) = 1/18

P(3) = 1/9

P(4) = 1/6

P(5) = 1/6

P(6) = 1/6

P(7) = 1/6

P(8) = 1/9

P(9) = 1/18

Step-by-step explanation:

Each die has six possible outcomes, so both dice together have 6*6 = 36 outcomes:

(1,1), (1,1), (1,2), (1,2), (1,3), (1,3),

(2,1), (2,1), (2,2), (2,2), (2,3), (2,3),

(3,1), (3,1), (3,2), (3,2), (3,3), (3,3),

(4,1), (4,1), (4,2), (4,2), (4,3), (4,3),

(5,1), (5,1), (5,2), (5,2), (5,3), (5,3),

(6,1), (6,1), (6,2), (6,2), (6,3), (6,3).

The sum of the values for each outcome is:

2, 2, 3, 3, 4, 4,

3, 3, 4, 4, 5, 5,

4, 4, 5, 5, 6, 6,

5, 5, 6, 6, 7, 7,

6, 6, 7, 7, 8, 8,

7, 7, 8, 8, 9, 9.

To find the probability of each sum, we just need to divide the number of times this sum appears over the total number of possibilities (36), so we have:

P(2) = 2/36 = 1/18

P(3) = 4/36 = 1/9

P(4) = 6/36 = 1/6

P(5) = 6/36 = 1/6

P(6) = 6/36 = 1/6

P(7) = 6/36 = 1/6

P(8) = 4/36 = 1/9

P(9) = 2/36 = 1/18

8 0

3 years ago

$3 is what percent of $4?

ozzi

Answer:

The answer is 3/4 = 75%

Step-by-step explanation:

3/4 = 75%

It is: 3/4 = 75%

4 0

3 years ago

What is the median value of the data set shown on the line plot? Enter your answer in the box.

frozen [14]

1/2 of 14 is 7 so 7th x plus 8th x = 30+30 / 2 = the median is 30

3 0

3 years ago

Read 2 more answers