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WGU QTT2 Finite Mathematics Task 2 |Passed on First Attempt |Latest Update with Complete Solution

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WGU QTT2 Finite Mathematics Task 3 |Passed on First
Attempt |Latest Update with Complete Solution


Jeffrey Martin
Finite Mathematics – QTT2
Task 3

Part 1: Set Theory
Use your assigned problem from “Part 1” of the “Finite Math Assignment Form: Task 3”
supporting document to do the following:

The quality control department at a widget factory pulls a sample of 3,250 widgets to check
for defective parts.
Let the following subsets be defined.
Group 1: widgets that have a defective bearing
Group 2: widgets that have surface flaws
612 widgets in the sample are found to have a defective bearing, and 535 are found to have
surface flaws. Among the widgets with a defective bearing, 112 did not have surface flaws.

A. Construct a Venn diagram to represent your assigned problem. Include capital letter labels
for all sets and indicate what each label represents.




B: Group 1, widgets with defective bearings
S: Group 2, widgets with flawed surfaces
G: Universal set, includes all the widgets checked for defects

B. Use the set notation symbols (∪,',∩) and set labels from part A to express each of the
following sets:
• elements in both Group 1 and Group 2
B∩S
• elements in Group 2 but not Group 1

, B’ ∩ S
• elements in Group 1 or Group 2
B∪S
• elements in neither Group 1 nor Group 2
(B ∪ S)’

C. Determine the cardinality of each of the following sets, including an explanation of the
process for how you determined the cardinality:
• elements in both Group 1 and Group 2
B ∩ S has 500 widgets in it.
I got that number because I was told that 612 widgets had defective bearings and 112 of
those did not have surface flaws, so that meant that the rest did have surface flaws in
addition to having defective bearings. So I subtracted the 112 that did not have surface
flaws from the 612 with defective bearings, leaving 500 with both surface flaws and
defective bearings.

• elements in Group 2 but not Group 1
B’ ∩ S has 35 widgets in it.
I got that number by using the number of widgets in set B ∩ S which is 500 (the previous
question) and then subtracting that number from the total number of widgets in Group 2
(widgets with surface flaws) which was given to me as being 535. That leaves a total of
35 widgets in group 2 but not group 1.

• elements in Group 1 or Group 2
B ∪ S has 647 widgets in it.
I got that number by adding the total number of widgets in Group 1 (612) to the total
number of widgets in Group 2 but since 500 of those have already been determined to be
in both groups, I had to subtract those 500 since we don’t count them twice, and that left
me with 612+35 which equals 647.

• elements in neither Group 1 nor Group 2
(B ∪ S)’ equals 2603.
I got that number by taking the total number of widgets checked for
defective parts (given to me in the task information) which was 3250,
then I subtracted the elements that were in group 1 or group 2 (B ∪ S)
which was 647 and that gave me a total of 2603 widgets not in group 1
or group 2.

Part 2: Logic
Use your assigned simple and compound statements from “Part 2” of the “Finite Math
Assignment Form: Task 3” supporting document to do the following:

D. Use your assigned compound statement p to do the following:

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