Resistors. An electrical engineer has two boxes of resistors, with four resistors in each box. The resistors in the first box are labeled 10 ohms, but in fact their resistances are 9, 10, 11, and 12 ohms. The resistors in the second box are labeled 20 ohms, but in fact haveresistances of 18, 19, 20, and 21 ohms. The engineer chooses one resistor from each box and determines the resistance of each. Answer the following: A. List all possible outcomes in the sample space.
B. List all outcomes in the event B, the event that the second resistor has a resistance less than 19.
C. List all outcomes in the event C, the event that the sum of the resistances is equal to 28.
D. List all outcomes in B∪C.
E. List all outcomes in B(complement) ∩C .

A. Sample Space = S = {(9, 18),(9, 19),(9, 20),(9, 21),(10, 18),(10, 19),(10, 20),(10, 21),(11, 18),(11, 19),(11, 20),(11, 21),(12, 18),(12, 19),(12, 20),(12, 21)}

B.  {(9, 18),(10, 18),(11, 18),(12, 18)}

C. {(9, 18),(9, 19),(10, 18)}

D.  {(9, 18),(9, 19),(10, 18),(11, 18),(12, 18)}

E. B(complement) ∩C =  {(9, 19)}

Step-by-step explanation:

The Sample Space would contain each element of the box A associated with each element of the box B.

A. Sample Space = S = {(9, 18),(9, 19),(9, 20),(9, 21),(10, 18),(10, 19),(10, 20),(10, 21),(11, 18),(11, 19),(11, 20),(11, 21),(12, 18),(12, 19),(12, 20),(12, 21)}

B.  Let the Outcomes in the event B, the event that the second resistor has a resistance less than 19 be denoted by J then

J=  {(9, 18),(10, 18),(11, 18),(12, 18)}

C. Let the outcomes in the event C, the event that the sum of the resistances is equal to 28 be denoted by L then

L =  {(9, 18),(9, 19),(10, 18)}

D. The  outcomes in B∪C contains all the elements of B and C

B∪C=J∪L= {(9, 18),(9, 19),(10, 18),(11, 18),(12, 18)}

E. B complement contains those elements of the Sample Space which are not the elements of Set B.

B complement= S-B=  {(9, 19),(9, 20),(9, 21),(10, 19),(10, 20),(10, 21),(11, 19),(11, 20),(11, 21),(12, 19),(12, 20),(12, 21)}

B(complement) ∩C contains those elements of B and C which are common to both B complement and C.

B(complement) ∩C =  {(9, 19)}

the camry sales+accord sale=828,000. to make it so the two numbers are 18 thousand apart and still equal 828,000, you first need to find out what half of 828,000 is. 828,000/2=414,000. for the two numbers to equal 828,000, you have to add the same amount as you subtract from the middle. in this case, the number is 9,000 because it's half of 18,000. to equal 828,000, you have to add 9,000 to 414,000 for the camry and subtract 9,000 from 414,000 for the accord. 414,000+9,000=423,000.

414,000-9,000=405,000.

423,000+405,000=828,000.

423,000-405,000=18,

hope it