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Anna11 [10]
3 years ago
10

Figure ABCDE has vertices A(−2, 3), B(2, 3), C(5, −2), D(0, −4), and E(−2, −2). Plot the points on your own coordinate grid and

connect the points in alphabetical order. Decompose Figure ABCDE into rectangles and triangles.
please help :)
Part A: How many triangles and rectangles did you make? (1 point)

Part B: Use Figure ABCDE created on your coordinate grid to find the lengths, in units, of Sides AB and AE. (4 points)

Part C: What is the area of Figure ABCDE? Show your work. (5 points)

Mathematics
1 answer:
Anastasy [175]3 years ago
5 0

There's a right angle that they want you to make a rectangle out of.  I'm ignoring that and using each of the five sides to make a triangle with the origin.

A.  Five triangles, zero rectangles.

B.

AB=4, the difference of the x coordinates since the y coordinates are the same

AE = 5, the difference in the y coordinates since the x coordinates are the same

C.  We'll use the Shoelace Formula, adding up all those triangles I made.  A triangle with vertices (0,0), (a,b), (c,d) has signed area A = (1/2)(ad-bc).  We go around counterclockwise so they all have positive areas, starting with AE, then ED, DC, CB, BA.

Area = (1/2) (  (-2)(-2) - 3(-2)  +  (-2)(-4) - (-2)(0)  +  0(-2) - (-4)(5)  + 5(3) - (-2)(2) + 2(3) - (3)(-2))

Area = 69/2

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Dr. Miriam Johnson has been teaching accounting for over 20 years. From her experience, she knows that 60% of her students do ho
oksano4ka [1.4K]

Answer:

a) The probability that a student will do homework regularly and also pass the course = P(H n P) = 0.57

b) The probability that a student will neither do homework regularly nor will pass the course = P(H' n P') = 0.12

c) The two events, pass the course and do homework regularly, aren't mutually exclusive. Check Explanation for reasons why.

d) The two events, pass the course and do homework regularly, aren't independent. Check Explanation for reasons why.

Step-by-step explanation:

Let the event that a student does homework regularly be H.

The event that a student passes the course be P.

- 60% of her students do homework regularly

P(H) = 60% = 0.60

- 95% of the students who do their homework regularly generally pass the course

P(P|H) = 95% = 0.95

- She also knows that 85% of her students pass the course.

P(P) = 85% = 0.85

a) The probability that a student will do homework regularly and also pass the course = P(H n P)

The conditional probability of A occurring given that B has occurred, P(A|B), is given as

P(A|B) = P(A n B) ÷ P(B)

And we can write that

P(A n B) = P(A|B) × P(B)

Hence,

P(H n P) = P(P n H) = P(P|H) × P(H) = 0.95 × 0.60 = 0.57

b) The probability that a student will neither do homework regularly nor will pass the course = P(H' n P')

From Sets Theory,

P(H n P') + P(H' n P) + P(H n P) + P(H' n P') = 1

P(H n P) = 0.57 (from (a))

Note also that

P(H) = P(H n P') + P(H n P) (since the events P and P' are mutually exclusive)

0.60 = P(H n P') + 0.57

P(H n P') = 0.60 - 0.57

Also

P(P) = P(H' n P) + P(H n P) (since the events H and H' are mutually exclusive)

0.85 = P(H' n P) + 0.57

P(H' n P) = 0.85 - 0.57 = 0.28

So,

P(H n P') + P(H' n P) + P(H n P) + P(H' n P') = 1

Becomes

0.03 + 0.28 + 0.57 + P(H' n P') = 1

P(H' n P') = 1 - 0.03 - 0.57 - 0.28 = 0.12

c) Are the events "pass the course" and "do homework regularly" mutually exclusive? Explain.

Two events are said to be mutually exclusive if the two events cannot take place at the same time. The mathematical statement used to confirm the mutual exclusivity of two events A and B is that if A and B are mutually exclusive,

P(A n B) = 0.

But, P(H n P) has been calculated to be 0.57, P(H n P) = 0.57 ≠ 0.

Hence, the two events aren't mutually exclusive.

d. Are the events "pass the course" and "do homework regularly" independent? Explain

Two events are said to be independent of the probabilty of one occurring dowant depend on the probability of the other one occurring. It sis proven mathematically that two events A and B are independent when

P(A|B) = P(A)

P(B|A) = P(B)

P(A n B) = P(A) × P(B)

To check if the events pass the course and do homework regularly are mutually exclusive now.

P(P|H) = 0.95

P(P) = 0.85

P(H|P) = P(P n H) ÷ P(P) = 0.57 ÷ 0.85 = 0.671

P(H) = 0.60

P(H n P) = P(P n H)

P(P|H) = 0.95 ≠ 0.85 = P(P)

P(H|P) = 0.671 ≠ 0.60 = P(H)

P(P)×P(H) = 0.85 × 0.60 = 0.51 ≠ 0.57 = P(P n H)

None of the conditions is satisfied, hence, we can conclude that the two events are not independent.

Hope this Helps!!!

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Answer:

2

Step-by-step explanation:

subtract 2x from both sides then 3 from both. this gives you 5x = 10. divide by 5 and you get 2.

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Answer:

See below in bold.

Step-by-step explanation:

The median will be the middle number - that is the 11th.

The first quartile will be the middle value of the first ten numbers.

That is the mean of the 5th and 6th number.

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4 years ago
Use the recursive formula f(n) = 0.4 ⋅ f(n − 1) + 12 to determine the 2nd term if f(1) = 4.
Maru [420]

Answer:

f(2) = 13.6

Step-by-step explanation:

Using the recursive formula and f(1) = 4 , then

f(2) = 0.4 f(1) + 12 = (0.4) × 4) + 12 = 1.6 + 12 = 13.6

7 0
3 years ago
Using a pencil, pair of compasses and ruler, construct a triangle with sides 8cm, 9cm and 6cm
Sonja [21]

Answer:

The nearest degree of the angle between the 8cm and 9cm sides is:

  • <u>41°</u>

Step-by-step explanation:

To obtain the correct angle in a small triangle like the triangle formed with the measures in the exercise is a little bit hard, in a change of this, I used a triangle calculator, with this I obtained the right angle formed by the sides with 8 cm and 9 cm, which I'm gonna add like an attachment.

From this image, you can guide yourself to draw the triangle without mistakes and to identify the right angle, which is 40,8° but the compass is not so defined, by this reason you can take it like 41° or 40°.

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