Consider the following information about passengers on a cruise ship on vacation: 41% check work e-mail, 31% use a cell phone to stay connected to work, 25% bring a laptop with them on vacation, 22% both check work e-mail and use a cell phone to stay connected, and 50% neither check work e-mail nor use a cell phone to stay connected nor bring a laptop. In addition, 89% of those who bring a laptop also check work e-mail and 71% of those who use a cell phone to stay connected also bring a laptop. With

E = event that a traveler on vacation checks work e-mail

C = event that a traveler on vacation uses a cell phone to stay connected

L = event that a traveler on vacation brought a laptop use the given information to determine the following probabilities.

A Venn diagram may help. (Round all answers to four decimal places.)

(a) P(E) =

(b) P(C) =

(c) P(L) =

(d) P(E and C) =

(e) P(Ec and Cc and Lc ) =

(f) P(E or C or L) =

(g) P(E|L) = (h) P(L|C) =

(i) P(E and C and L) =

(j) P(E and L) =

(k) P(C and L) =

(l) P(C|(E and L)) =

Step-by-step explanation:

Let E = events that a traveler on vacation checks works email

C = events that a traveler on vacation uses a cell phone to stay connected

L = event that a traveler on vacation broughgt a laptop

from the information given ; P(E) = 0.41, P(C) = 0.31, P(L)= 0.25, P(EnC) = 0.22

50% neither check work e-mail nor use a cell phone to stay connected nor bring a laptop= P(EUCUL)' = 0.50

9% of those who bring a laptop also check work e-mail = P(E|L) = 0.89

71% of those who use a cell phone to stay connected also bring a laptop = P(L|C) = 0.71

a) P(E) = 0.41

(b) P(C) = 0.31

(c) P(L) = 0.25

(d) P(EnC) = 0.22

(e) P(E'nC'nL' ) = 0.50

(f) P(EUCUL) = 1 - P(EUCUL)' = 0.50

(g) P(E|L) = 0.89

Also P(EnL) = P(E|L)P(L) = 0.25 X 0.89 = 0.2225

(h) P(L|C) = 0.71

(I) P(EnCnL) = P(EUCUL) + P(E) + P(C) + P(L) - P(EnC) -P(EnL) -P(CnL)

But P(LnC) = P(C)P(L|C) = 0.31 X 0.71 = 0.2201 and P(EnL) = 0.2225

P(EnCnL) = 0.1926

(j) P(EnL) = 0.2225

(k) P(CnL) = 0.2201

(l) P(C|(EnL)) = P(CnEnL)/P(EnL)

= 0.8656

Step-by-step explanation: Let P( E and C and L)' = 0.50 and P(Ec and Cc and Lc) = x

P(E) = 0.41 - {( 0.89 - x) + (0.22 - x) + x} = 0.41 - 1.11 - x = - 0.70 - x

P(C) = 0.31 - {(0.71 - x) + (0.22 - x) + x} = 0.31 - 0.92 - x = - 0.61 - x

P(L) = 0.25 - {(0.71 - x) + (0.89 - x) + x} = 0.25 - 1.60 - x = - 1.35 - x

1 = (-0.70 - x) + (-0.61 - x) + (-1.35 - x) + (0.71 - x) + (0.22 - x) + (0.89 - x) + x + 0.50

1 = -0.24 - 5x

5x = -1 - 0.24 = -1.24

x = -1.24/5 = -0.248

a) P(E) = -0.70 - (-0.248) = -0.70 + 0.248 = -0.452

b) P(C) = -0.61 - (-0.248) = -0.61 + 0.248 = -0.362

c) P(L) = -1.35 - (-0.248) = -1.35 + 0.248 = -1.102

d) P(E and C) = 0.22 - (-0.248) = 0.22 + 0.248 = 0.468

e) P(Ec and Cc and Lc) = -0.248

f)  P(E or C or L) = P(E and C and L)'= 0.50

g)

i) P(E and C and L) = -0.452 - 0.362 - 1.102 = -1.916

j) P(E and L) = 0.89 - (-0.248) = 0.89 + 0.248 = 1.138

k) P(C and L) = 0.71  - (-0.248) = 0.71 + 0.248 = 0.958

l) P(C\(E and L) = 0.468 + 0.958 - 1.1386 - 0.248 = 1.427 - 1.386 = 0.041

answer:   the paragraph proof is mentioned below.

step-by-step explanation:

given: in a δxwy where

such that, ∠wzx ≅ ∠yzx and zw≅zy

we have to prove that: segment zx is a perpendicular bisector of segment wy.

prove: since, zw≅zy ⇒ ∠wzy is a supplementary angle.

then, ∠wzy= 180°

⇒∠wzx + ∠yzx = 180° ( because xz divides ∠wzy into two parts)

but, ∠wzx ≅ ∠yzx ( given)

⇒∠wzy+∠wzy=180° ( by substitution)

⇒2∠wzy=180°

⇒∠wzy=90°

⇒∠wzy=∠yzx =90°

therefore, xz⊥wy where wz=zy

⇒zx is a perpendicular bisector of wy.