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True [87]
2 years ago
8

Six teachers and 12 students volunteer for a committee to discuss extra-curricular activities. How many committees of 5 people c

an be made if:
a) there must be exactly 3 students on the committee

b) there must be at least one teacher and at least one student on the committee
Mathematics
1 answer:
mariarad [96]2 years ago
4 0

Answer:

a) 11,880

b) 7,770

Step-by-step explanation:

The number of teachers that volunteer = 6 teachers

The number of students that volunteer = 12 students

The number of people in the committee = 5 people

a) The exact number of student on the committee = 3 student

Therefore, the number of teachers in the committee of 5 = 5 - 3 = 2

The number of teachers in the committee = 2 teachers

We get;

The number of committees that can be made = ₁₂C₃ × ₆C₂ = 792×15 = 11,880

b) When there is at least one teacher and at least one student on the committee, we have;

₁₂C₄ × ₆C₁ + ₁₂C₃ × ₆C₂ + ₁₂C₂ × ₆C₃ + ₁₂C₁ × ₆C₄ = 7,770

The number of committees having at least one teacher and at least one student = 7,770

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A (0, 2) and B (6,6) are points on the straight line ABCD.
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Answer:

(18, 14)

Step-by-step explanation:

We know that C and D lie on the line AB and BC = CD = AB. Then we need to use the distance formula and equation of the line AB to find the other two coordinates.

The distance formula states that the distance between two points (x_1,y_1) and (x_2,y_2), the distance is denoted by: \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}. Let's find the distance between A and B:

d = \sqrt{(6-0)^2+(6-2)^2}=\sqrt{6^2+4^2} =\sqrt{36+16} =\sqrt{52} =2\sqrt{13}

Now say the coordinates of D are (a, b). Then the distance between D and B will be twice of 2√13, which is 4√13:

4√13 = \sqrt{(6-a)^2+(6-b)^2}

Square both sides:

208 = (6 - a)² + (6 - b)²

Let's also find the equation of the line AB. The y-intercept we know is 2, so in y = mx + b, b = 2. The slope is (6 - 2) / (6 - 0) = 4/6 = 2/3. So the equation of the line is: y = (2/3)x + 2. Since (a, b) lines on this line, we can put in a for x and b for y: b = (2/3)a + 2. Substitute this expression in for b in the previous equation:

208 = (6 - a)² + (6 - b)²

208 = (6 - a)² + (6 - (2/3a + 2))² = (6 - a)² + (-2/3a + 4)²

208 = a² - 12a + 36 + 4/9a² - 16/3a + 16 = 13/9a² - 52/3a + 52

0 = 13/9a² - 52/3a - 156

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We know a can't be negative so a = 18. Plug this back in to find b:

b = 2/3a + 2 = (2/3) * 18 + 2 = 12 + 2 = 14

So point D has coordinates (18, 14).

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