Sunita bought a 10 kg box of grapes from the market. She gave 3 1/2 kg of her grapes to her friend Reena and 2 1/4 kg to Anita.

Answer:

1. Addition Property of Equality

2. Segment addition

3. Substitution Property of Equality.

4. Transitive Property of Equality.

Step-by-step explanation:

Given: CD = EF and AB = CE

To Show: AB = DF

Following the steps outlined:

1. CD + DE = EF + DE by the (addition) Property of Equality.

This shows that the same value has been added to both sides of the equation.

2. CE = CD + DE and DF = EF + DE through (segment addition).

This indicates that CE and DF are segments, and the length of any segment equals the total of its parts.

3. CE = DF according to the (Addition, subtraction, substitution, transitive) Property of Equality. Recognizing that CE = CD + DE and using the fact that CD = EF allows us to replace CD with EF.

So, CE = EF + ED = FD (via substitution) Property of Equality.

4. Since AB = CE and CE = DF, we can conclude AB = DF by the (transitive) Property of Equality.

This follows the rule that if x = y and y = z, then x = z.

Response:

a. 0.76

b. 0.23

c. 0.5

d. p(B/A) signifies the likelihood that a student with a visa card also possesses a MasterCard.

p(A/B) indicates the probability that a student with a MasterCard also has a visa card.

e. 0.35

f. 0.31

Detailed explanation:

a. p(AUBUC) = P(A) + P(B) + P(C) - P(AnB) - P(AnC) - P(BnC) + P(AnBnC)

        = 0.6 + 0.4 + 0.2 - 0.3 - 0.11 - 0.1 + 0.07 = 0.76

b. P(AnBnC') = P(AnB) - P(AnBnC)

        = 0.3 - 0.07 = 0.23

c. P(B/A) = P(AnB)/P(A)

        = 0.3/0.6 = 0.5

e. P((AnB)/C) = P((AnB)nC)/P(C)

        = P(AnBnC)/P(C)

        = 0.07/0.2 = 0.35

f. P((AUB)/C) = P((AUB)nC)/P(C)

        = (P(AnC) U P(BnC))/P(C)

        = (0.11 + 0.1)/0.2

        = 0.21/0.2 = 0.31