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bixtya [17]
3 years ago
15

Suppose that we have a sample space S5 {E1, E2, E3, E4E4E, E5, E6, E7}, where E1, E2, . . ., E7 denote the sample points. The fo

llowing probability assignments apply: P(E1) 5 .05, P(E2) 5 .20, P(E3) 5 .20, P(E4) 5 .25, P(E5) 5.15, P(E6) 5 .10, and P(E7) 5 .05. LetA5{E1,E4E4E,E6}B5{E2E2E,E4E4E,E7E7E}C5{E2E2E,E3,E5,E7E7E}a.Find P(A(A(), P(B), and P(C).b.Find AB and P(A(A(>B).d.Are events A and C mutually exclusive?C mutually exclusive?Ce.Find Bc and P(Bc).
Mathematics
1 answer:
VashaNatasha [74]3 years ago
5 0

Answer:

P(A) = 0.4 ; P(B) = 0.50 ; P(C) = 0.60 ; P(A u B) = 0.65 ; P(A n B) = 0.25 ;

A and C are mutually exclusive ;

0.5

Step-by-step explanation:

S = {E1, E2, E3, E4, E5, E6, E7}

P(E1) = .05, P(E2) = .20, P(E3) = .20, P(E4) = .25, P(E5) = .15, P(E6) = .10, and P(E7) = .05

Let:

A{E1,E4,E6}

B{E2,E4,E7}

C{E2,E3,E5,E7}

P(A) = 0.05 + 0.25 + 0.10 = 0.40

P(B) = 0.20 + 0.25 + 0.05 = 0.50

P(C) = 0.20 + 0.20 + 0.15 + 0.05 = 0.60

AUB = {E1, E2, E4, E6, E7}

P(A u B) = 0.05 + 0.20 + 0.25 + 0.10 + 0.05 = 0.65

AnB = {E4}

P(A n B) = 0.25

A and C are mutually exclusive if AnC = ∅

A n C = ∅

Hence, A and C are mutually exclusive.

B complement = B' = {E1, E3, E5, E6}

P(B') = 0.05 + 0.20 + 0.15 + 0.10 = 0.5

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Ivan used coordinate geometry to prove that quadrilateral EFGH is a square.
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Answer:

(A)Segment EF, segment FG, segment GH, and segment EH are congruent

Step-by-step explanation:

<u>Step 1</u>

Quadrilateral EFGH with points E(-2,3), F(1,6), G(4,3), H(1,0)

<u>Step 2</u>

Using the distance formula

Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Given E(-2,3), F(1,6)

|EF|=\sqrt{(6-3)^2+(1-(-2))^2}=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt{2}

Given F(1,6), G(4,3)

|FG|=\sqrt{(3-6)^2+(4-1)^2}=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt{2}

Given G(4,3), H(1,0)

|GH|=\sqrt{(0-3)^2+(1-4)^2}=\sqrt{(-3)^2+(-3)^2}=\sqrt{18}=3\sqrt{2}

Given E (−2, 3), H (1, 0)

|EH|=\sqrt{(0-3)^2+(1-(-2))^2}=\sqrt{(-3)^2+(3)^2}=\sqrt{18}=3\sqrt{2}

<u>Step 3</u>

Segment EF ,E (−2, 3), F (1, 6)

Slope of |EF|=\frac{6-3}{1+2} =\frac{3}{3}=1

Segment GH, G (4, 3), H (1, 0)

Slope of |GH|= \frac{0-3}{1-4} =\frac{-3}{-3}=1

<u>Step 4</u>

Segment EH, E(−2, 3), H (1, 0)

Slope of |EH|= \frac{0-3}{1+2} =\frac{-3}{3}=-1

Segment FG, F (1, 6,) G (4, 3)

Slope of |EH| =\frac{3-6}{4-1} =\frac{-3}{3}=-1

<u>Step 5</u>

Segment EF and segment GH are perpendicular to segment FG.

The slope of segment EF and segment GH is 1. The slope of segment FG is −1.

<u>Step 6</u>

<u>Segment EF, segment FG, segment GH, and segment EH are congruent. </u>

The slope of segment FG and segment EH is −1. The slope of segment GH is 1.

<u>Step 7</u>

All sides are congruent, opposite sides are parallel, and adjacent sides are perpendicular. Quadrilateral EFGH is a square

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Answer:

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