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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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solution:

The closing price (in dollars) per share of stock of Tempco Electronics on the tth day it was  

traded is approximated by  

P(t) = 20 + 12 sin πt/30 − 6 sin πt/15 + 4 sin πt/10 − 3 sin 2πt/15 (0 ≤ t ≤ 24)  

where t = 0 corresponds to the time the stock was first listed on a major stock exchange.  

What was the rate of change of the stock's price at the close of the 15th day of trading?  

P'(t) = 12(cos πt/30) - 6 (cos πt/15) + 4(cos πt/10) - 3(cos 2πt/15)  

t = 15  

P'(t) = 12(cos 15π/30) - 6 (cos 15π/15) + 4(cos 15π/10) - 3(cos 30π/15)  

P'(t) = 12(cos π/2) - 6 (cos π) + 4(cos 3π/2) - 3(cos 2π)  

P'(t) = 12(0) - 6(-1) + 4(0) - 3(1) = 6 - 3  

P'(t) = $3 per day RATE OF CHANGE  

the closing price on that day

P(t) = 20 + 12 sin πt/30 − 6 sin πt/15 + 4 sin πt/10 − 3 sin 2πt/15  

t = 15  

P(t) = 20 + 12 sin 15π/30 − 6 sin 15π/15 + 4 sin 15π/10 − 3 sin 30πt/15  

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2 years ago
Joe and Nate went for a walk. The system representing each boy’s progress is graphed below. A graph titled Joe and Nate's progre
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Answer:

A,D E

Step-by-step explanation:

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Es urgente ¡¡¡ si la MH de a y 4 es 6 y la MH de 8 y b es 12 calcula la MH de a y b
TiliK225 [7]

Answer:

Sabemos que:

MH(x, y) = \frac{2xy}{x+y}

y tenemos que:

MH (a,4) = \frac{2*a*4}{a+4} = \frac{8*a}{a+4} = 6

Con esto podemos encontrar el valor de a:

8*a/(a+ 4) = 6

8*a = 6*(a + 4) = 6*a + 24

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Tambien sabemos que:

MH(8,b) = \frac{2*8*b}{8+b} =\frac{16*b}{8+b} = 12

Y de ahí podemos despejar b:

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MH(12,24) = 2*12*24/(12 + 24) = 24*24/36 = 16

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