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lina2011 [118]
2 years ago
14

A university offers 3 calculus classes: Math 2A, 2B, and 2C. A set of students have each taken at least one of the three classes

. 51 have taken Math 2A, 80 have taken Math 2B, and 70 have taken Math 2C. 15 students have taken Math 2A and 2B, 20 have taken Math 2A and 2C, and 13 have taken Math 2B and 2C. Only 4 have taken all three classes. How many students are there in the set?

Mathematics
1 answer:
lyudmila [28]2 years ago
4 0

Answer:

157

Step-by-step explanation:

We are given that

Number of students in maths 2A=n(2A)=51

Number of students in math 2B=n(2B)=80

Number of students in math 2C=n(2C)=70

Number of students in maths 2A and 2B=n(A\cap B)=15

Number of students in maths 2B and 2C=n(B\cap C)=13

Number of students in maths 2A and 2C=n(A\cap C)=20

Number of students in maths 2A ,2B and 2C=4

We have to find the number of students in a set.

Formula:n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)

Substitute the values in the given formula

n(2A\cup 2B\cup 2C)=51+80+70-15-13-20+4=157

Hence, total number of students in a set=157

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Step-by-step explanation:

The question is just subtraction of fractions. First, you subtract the whole numbers. 10-2=8. Then you subtract the fractions. To do this, you have to make them equivalent. Just multiply the 1/4 by 2 and you get 2/8. Now, 7/8 - 2/8 = 5/8. Your final answer is 8 and 5/8

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3/4 x 1/8 in simples form
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Step-by-step explanation:

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Construct the discrete probability distribution for the random variable described. Express the probabilities as simplified fract
lisov135 [29]

Answer:

P(X = 0) = 0.03125

P(X = 1) = 0.15625

P(X = 2) = 0.3125

P(X = 3) = 0.3125

P(X = 4) = 0.15625

P(X = 5) = 0.03125

Step-by-step explanation:

For each toss, there are only two possible outcomes. Either it is tails, or it is not. The probability of a toss resulting in tails is independent of any other toss, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

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Equally as likely to be heads or tails, so p = 0.5

5 tosses:

This means that n = 5

Probability distribution:

Probability of each outcome, so:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{5,0}.(0.5)^{0}.(0.5)^{5} = 0.03125

P(X = 1) = C_{5,1}.(0.5)^{1}.(0.5)^{4} = 0.15625

P(X = 2) = C_{5,2}.(0.5)^{2}.(0.5)^{3} = 0.3125

P(X = 3) = C_{5,3}.(0.5)^{3}.(0.5)^{2} = 0.3125

P(X = 4) = C_{5,4}.(0.5)^{4}.(0.5)^{1} = 0.15625

P(X = 5) = C_{5,5}.(0.5)^{5}.(0.5)^{0} = 0.03125

5 0
2 years ago
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