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NARA [144]
2 years ago
14

At a certain college, 30% of the students major in engineering, 20% play club sports, and 10% both major in engineering and play

club sports. A student is selected at random. Navidi, William,Navidi, William. Statistics for Engineers and Scientists (p. 85). McGraw-Hill Higher Education. Kindle Edition.

Engineering
1 answer:
hodyreva [135]2 years ago
5 0

At a certain college, 30% of the students major in engineering, 20% play club sports, and 10% both major in engineering and play club sports. A student is selected at random.

a) What is the probability that the student is majoring in engineering?

b) What is the probability that the student plays club sports?

c) Given that the student is majoring in engineering, what is the probability that the

student plays club sports?

d) Given that the student plays club sports, what is the probability that the student is

majoring in engineering?

e) Given that the student is majoring in engineering, what is the probability that the

student does not play club sports?

f) Given that the student plays club sports, what is the probability that the student is not

majoring in engineering?

Answer and Explanation

The venn diagram for the question is in the attachment.

Percentage majoring in engineering = 30% = 0.3

Percentage that plays club sport = 20% = 0.2

Percentage that major in engineering and play sport = 10% = 0.1

Percentage majoring in engineering & do not play club sport = 30 - 10 = 20% = 0.2

Percentage that plays club sport & do not major in engineering = 20 - 10 = 10% = 0.1

Total percentage = 100%

a) probability that student is majoring in Engineering P(E) = 30/100 = 0.3

b) probability that student plays club sport = 20/100 P(S) = 0.2

c) probability that the

student plays club sport given that the student is majoring in engineering, P(S|E) = (P(E and S))/(P(E))

P(E and S) = 10/100 = 0.1, P(E) = 0.3

P(S|E) = 0.1/0.3 = 0.3333

d) probability that the

student is majoring in engineering given that the student plays club sport, P(E|S) = (P(S and E))/(P(S))

P(S and E) = 10/100 = 0.1, P(S) = 0.2

P(E|S) = 0.1/0.2 = 0.5

e) probability that the

student does not play club sport given that the student is majoring in engineering, P(S'|E) = (P(E and S'))/(P(E))

P(E and S') = 20/100 = 0.2, P(E) = 0.3

P(S'|E) = 0.2/0.3 = 0.667

f) probability that the

student is not majoring in engineering given that the student plays club sport, P(E'|S) = (P(S and E'))/(P(S))

P(S and E') = 10/100 = 0.1, P(S) = 0.2

P(E|S) = 0.1/0.2 = 0.5

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Answer:

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Write multiple if statements: If carYear is before 1967, print "Probably has few safety features." (without quotes). If after 19
Free_Kalibri [48]

Answer:

The solution code is written in Python 3.

  1. carYear = 1995
  2. if(carYear < 1967):
  3.    print("Probably has few safety features.\n")
  4. if(carYear > 1970):
  5.    print("Probably has head rests. \n")
  6. if(carYear > 1991):
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Explanation:

Firstly, create a variable, <em>carYear</em> to hold the value of year of the car make. (Line 1)

Next, create multiple if statements as required by the question (Line 3-13). The operator "<" denotes "smaller" and therefore <em>carYear < 1967</em> means any year before 1967. On another hand, the operator ">" denotes "bigger" and therefore <em>carYear > 1970 </em>means any year after 1970.

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2 years ago
The rectangular frame is composed of four perimeter two-force members and two cables AC and BD which are incapable of supporting
Rama09 [41]

Answer:

Your question is lacking some information attached is the missing part and the solution

A) AB = AD = BD = 0, BC = LC

    AC = \frac{5L}{3}T, CD = \frac{4L}{3} C

B) AB = AD = BC = BD = 0

   AC = \frac{5L}{3} T, CD = \frac{4L}{3} C

Explanation:

A) Forces in all members due to the load L in position A

assuming that BD goes slack from an inspection of Joint B

AB = 0 and BC = LC from Joint D, AD = 0 and CD = 4L/3 C

B) steps to arrive to the answer is attached below

AB = AD = BC = BD = 0

AC = \frac{5L}{3} T,  CD = \frac{4L}{3}C

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2 years ago
A platinum resistance temperature sensor has a resistance of 120 Ω at 0℃ and forms one arm of a Wheatstone bridge. At this tempe
oksian1 [2.3K]

Answer : 9.36ohms/ temperature

Explanation:

Expression for the variation of resistance of platinum with temperature

Rt= Ro(1+*t)

Rt= resistance @ t°C

Ro= resistance @ 0°C

*= temperature coefficient of resistance

Calculate the change in resistance by putting 120ohms for Ro,

0.0039/K for *

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Using this formula:

Rt = Ro(1+*t)

Rt- Ro = Ro*t

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Answer:

// The method is defined with a void return type

// It takes a parameter of integer called numCycles

// It is declared static so that it can be called from a static method

public static void printShampooInstructions(int numCycles){

// if numCycles is less than 1, it display "Too few"

   if (numCycles < 1){

       System.out.println("Too few.");

   }

// else if numCycles is less than 1, it display "Too many"

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// else it uses for loop to print the number of times to display

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}

Explanation:

The code snippet is written in Java. The method is declared static so that it can be called from another static method. It has a return type of void. It takes an integer as parameter.

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