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Masteriza [31]
3 years ago
10

10. There are 40 students at Oostburg College majoring in computer science. Computer science is not considered an engineering ma

jor.
Calculate an estimate of the number of computer science majors you think are in the marching band. Explain how you calculated your estimate.

Mathematics
1 answer:
Luden [163]3 years ago
3 0

Answer:

x= 0.1905*40 = 7.62 \approx 8

So then we would expect between 7 or 8 computer science majors are in the marching band.

Step-by-step explanation:

For this case we assume the two frequency chart attached on the figure.

For this case we are interested on this question "If a randomly selected student is not majoring in engineering. What is the  probability that this student is in the marching band? "

And in order to calculate this probability we need to look at the row "Not engineering Major" and we see that we have a total of 630 people and from these we are interested in who are in the marching band and as we can see 120 satisfy this condition, so then we can find the probability like this:

p = \frac{120}{630}=0.1905

Now since we expect 0.1905 or 19.05% of students are from computer science mejors that are in the marching band and we have a sample selected of 40 students, then the expected value would be:

x= 0.1905*40 = 7.62 \approx 8

So then we would expect between 7 or 8 computer science majors are in the marching band.

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Answer:

6r+7=13+7r

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−r+7=13

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\frac{-r}{-1} = \frac{6}{-1}

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Step-by-step explanation:

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6 0
3 years ago
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DanielleElmas [232]

Question:  The probability that s student owns a car is 0.65, and the probability that a student owns a computer is 0.82.

a. If the probability that a student owns both is 0.55, what is the probability that a randomly selected student owns a car or computer?

b. What is the probability that a randomly selected student does not own a car or computer?

Answer:

(a) 0.92

(b) 0.08

Step-by-step explanation:

(a)

Applying

Pr(A or B) = Pr(A) + Pr(B) – Pr(A and B)................. Equation 1

Where A represent Car, B represent Computer.

From the question,

Pr(A) = 0.65, Pr(B) = 0.82, Pr(A and B) = 0.55

Substitute these values into equation 1

Pr(A or B) = 0.65+0.82-0.55

Pr(A or B) = 1.47-0.55

Pr(A or B) = 0.92.

Hence the probability that a student selected randomly owns a house or a car is 0.92

(b)

Applying

Pr(A or B) = 1 – Pr(not-A and not-B)

Pr(not-A and not-B) = 1-Pr(A or B) ..................... Equation 2

Given: Pr(A or B)  = 0.92

Substitute these value into equation 2

Pr(not-A and not-B) = 1-0.92

Pr(not-A and not-B) = 0.08

Hence the probability that a student selected randomly does not own a car or a computer is 0.08

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3 years ago
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Katyanochek1 [597]

Answer:

4/663

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Probability = Expected outcome/Total outcome

Since we are to draw from a pack of card, the total outcome will be 52

Since there are 4 hearts;

Pr(selecting heart) = 4/52 = 1/13

If a club is then selected without replacement, the total number of card remaining will be 51;

Pr(selecting a heart) = 4/51

probability of drawing a heart and a club in that order = 4/52 * 4/51

probability of drawing a heart and a club in that order = 16/2652

probability of drawing a heart and a club in that order = 4/663

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Use o fato de que o determinante de qualquer matriz quadrada é o mesmo da sua transposta.

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4 years ago
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lara [203]

Answer:

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Step-by-step explanation:

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\frac{6B+4}{R} = T  (divide R on both sides)

thus

T = \frac{6b+4}{R}

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