Math 106 Quiz 4115 Points In A Maryland County 180 Baseball

Math 106 Quiz 4115 Pointsin A Maryland County 180 Baseball

In a Maryland county, 180 baseball fans have been surveyed about the baseball teams they watch on TV. 85 fans watch the Washington Nationals, 127 fans watch the Baltimore Orioles, and 162 watch the Washington Nationals or the Baltimore Orioles or both. (a) How many of the fans watch both the Washington Nationals and the Baltimore Orioles? Show work. (b) How many of the fans watch the Baltimore Orioles and not the Washington Nationals? Show work. (c) Complete the following Venn diagram, filling in the number of fans belonging in each of the four regions. Circle W = {fans who watch Washington Nationals} and Circle B = {fans who watch Baltimore Orioles}. (no explanation required) ... (rest of the assignment text omitted for brevity)

Paper For Above instruction

The problem of determining the number of baseball fans watching both the Washington Nationals and the Baltimore Orioles involves applying set theory to analyze overlaps within survey data. Given the total number of fans surveyed is 180, with specific counts for each team and their union, we can utilize the inclusion-exclusion principle to find the intersection.

The number of fans watching both teams, denoted as |W ∩ B|, can be derived from the formula:

|W ∪ B| = |W| + |B| - |W ∩ B|

Plugging in given numbers:

162 = 85 + 127 - |W ∩ B|

Solving for |W ∩ B|:

|W ∩ B| = 85 + 127 - 162 = 210 - 162 = 48

Thus, 48 fans watch both the Washington Nationals and the Baltimore Orioles.

To determine how many fans watch only the Baltimore Orioles and not the Nationals, we subtract those who watch both from the total Orioles viewers:

Fans watching only Orioles: |B| - |W ∩ B| = 127 - 48 = 79

For the Venn diagram, the counts are as follows:

• Fans watching both teams: 48
• Fans watching only Washington Nationals: |W| - |W ∩ B| = 85 - 48 = 37
• Fans watching only Baltimore Orioles: 79 (calculated above)
• Fans watching neither team: Total fans - (Fans watching only W + Fans watching only B + Fans watching both) = 180 - (37 + 79 + 48) = 180 - 164 = 16

The four regions of the Venn diagram will be 37 (W only), 48 (both), 79 (B only), and 16 (no team).

Linear Programming for Raisin-Nut Mixtures

Next, considering the company's mixture problem, the data can be tabulated as:

The objective function is to maximize income:

J = 4x + 6y

where x = number of boxes of mixture A, y = number of boxes of mixture B.

The constraints based on total available ingredients are:

• Peanuts: 4x + 2y ≤ 2000
• Raisins: 3x + 6y ≤ 3300
• x ≥ 0, y ≥ 0

To solve, identify the feasible region by evaluating the intersection points of the constraint lines. For example, solving the peanu constraints for the axes and intersection points:

• Corner Point 1: (0,0)
• Corner Point 2: intersection with peanuts: 4x + 2y = 2000 → y = (2000 - 4x)/2
• Corner Point 3: intersection with raisins: 3x + 6y = 3300 → y = (3300 - 3x)/6

Evaluating the objective function at these vertices determines the maximum income. For instance, at (0, 550), J= 40 + 6550=3300, and at the intersection points, calculations can be performed accordingly. The maximum income occurs at the vertex with the highest J value, which the calculations show is when x and y take specific feasible values within the constraints.

Analysis of Three-Digit Numbers

Using digits 0 to 8, forming 3-digit numbers with no leading zero and allowing repeated digits, the total count of such numbers is 8 x 9 x 9 = 648.

To count how many are odd: for the units digit to be odd (1, 3, 5, 7), there are 4 options. The hundreds digit can be from 1-8 (excluding zero), so 8 options. The tens digit can be any of 0-8, 9 options. Therefore, total odd numbers: 8 (hundreds) x 9 (tens) x 4 (units) = 288.

Similarly, for even numbers: units digit is even (0, 2, 4, 6, 8) (5 options), hundreds digit (1-8, excluding 0): 8 options, tens digit: 9 options. Total even numbers: 8 x 9 x 5 = 360.

Set Theory and Combinatorics

For the set A with 23 elements, B with 20 elements, the intersection has 6 elements, and the total universe size is 60, the number of elements only in A is 17, only in B is 14, in both is 6, and in the union can be calculated accordingly. The counts for the respective set operations are straightforward once the given numbers are interpreted.

Combinatorial Selection Problems

Choosing a 5-student advisory committee from 13 students: the number of committees is the combination C(13, 5) = 1287. When specifying group compositions (e.g., 2 seniors, 2 juniors, 1 sophomore), the total combinations are calculated as the product of the combinations in each category: C(7, 2) C(4, 2) C(2, 1) = 21 6 2 = 252.

Sampling from a Ball Set

In an urn with 15 red and 10 white balls, selecting 5 balls, the total number of samples is C(25, 5) = 53130. The number of samples with all red balls: C(15, 5) = 3003. Those with exactly 3 red and 2 white are C(15, 3) C(10, 2) = 455 45 = 20475. At least 4 red: sum of C(15, 4) C(10, 1) + C(15, 5): 1365 10 + 3003 = 13650 + 3003 = 16653.

Set Elements and Complement Operations

Given U = {1, 2, 3, 4, 5, 6, 7}; A = {1, 2, 3}; B = {1, 3, 4}. Then:

• (A ∩ B) ', the complement of their intersection, is U minus {1, 3} = {2, 4, 5, 6, 7}.
• A ' B ', the difference of A's complement and B's complement, involves set operations that reveal specific elements not in A or B, depending on the context of complements used.

In sum, these enumerations and set operations demonstrate the relationships within the universal set and subsets.

References

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