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Roman55 [17]
2 years ago
8

At a college, 69% of the courses have final exams and 42% of courses require research papers. Suppose that 29% of courses have a

research paper and a final exam. Let F be the even that a course has a final exam. Let R be the event that a course requires a research paper. (a) Find the probability that a course has a final exam or a research paper. Your answer is : (b) Find the probability that a course has NEITHER of these two requirements. Your answer is :
Mathematics
1 answer:
laila [671]2 years ago
3 0

Answer:

a) 0.82

b) 0.18

Step-by-step explanation:

We are given that

P(F)=0.69

P(R)=0.42

P(F and R)=0.29.

a)

P(course has a final exam or a research paper)=P(F or R)=?

P(F or R)=P(F)+P(R)- P(F and R)

P(F or R)=0.69+0.42-0.29

P(F or R)=1.11-0.29

P(F or R)=0.82.

Thus, the the probability that a course has a final exam or a research paper is 0.82.

b)

P( NEITHER of two requirements)=P(F' and R')=?

According to De Morgan's law

P(A' and B')=[P(A or B)]'

P(A' and B')=1-P(A or B)

P(A' and B')=1-0.82

P(A' and B')=0.18

Thus, the probability that a course has NEITHER of these two requirements is 0.18.

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Consider the points A(5, 3t+2, 2), B(1, 3t, 2), and C(1, 4t, 3). Find the angle ∠ABC given that the dot product of the vectors B
Vilka [71]

Answer:

66.42°

Step-by-step explanation:

<u>Given:</u>

A(5, 3t+2, 2)

B(1, 3t, 2)

C(1, 4t, 3)

BA • BC = 4

Step 1: Find t.

First we have to find vectors BA and BC. We do that by subtracting the coordinates of the initial point from the coordinates of the terminal point.

In vector BA B is the initial point and A is the terminal point.

BA = OA - OB = (5-1, 3t+2-3t, 2-2) = (4, 2, 0)

BC = OC - OB = (1-1, 4t-3t, 3-2) = (0, t, 1)

Now we can find t because we know that BA • BC = 4

BA • BC = 4

To find dot product we calculate the sum of the produts of the corresponding components.

BA • BC = (4)(0) + (2)(t) + (0)(1)

4 = (4)(0) + (2)(t) + (0)(1)

4 = 0 + 2t + 0

4 = 2t

2 = t

t = 2

Now we know that:

BA = (4, 2, 0)

BC = (0, 2, 1)

Step 2: Find the angle ∠ABC.

Dot product: a • b = |a| |b| cos(angle)

BA • BC = 4

|BA| |BC| cos(angle) = 4

To get magnitudes we square each compoment of the vector and sum them together. Then square root.

|BA| = \sqrt{4^2 + 2^2 + 0^2} = \sqrt{20} = 2\sqrt{5}

|BC| = \sqrt{0^2 + 2^2 + 1^2} = \sqrt{5}

2\sqrt{5}\sqrt{5}\cos{(m\angle{ABC})} = 4

10\cos{(m\angle{ABC})} = 4

\cos(m\angle{ABC}) = \frac{4}{10}=\frac{2}{5}

m\angle{ABC} = cos^{-1}{(\frac{2}{5})}

m\angle{ABC} = 66.4218^{\circ}

Rounded to two decimal places:

m\angle{ABC} = 66.42^\circ

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36 divided by 3 = 12

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Step-by-step explanation:

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HTH :)

6 0
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There are 276 calories in 6 ounces of chicken. How many calories are in 1 pound of chicken
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Just divide 276 and 6.

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If C= (-10, -9, -8, -7 ,-6 , -5) and D= ( -6, -5, -4, -3, -2, -1, ) , what is C n D ?
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