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Citrus2011 [14]
3 years ago
12

In a freshman class of 80 students,22 students take Consumer Education,20 students take French,and 4 students take both.Which eq

uation can be used to find the probability,P, that a randomly selected student from this class takes Consumer Education, French,or both?
A: P = 11/40 + 1/4 + 1/20

B: P = 11/40 + 1/4

C: P = 11/40 + 1/4 - 1/20

D: P = 11/40 + 1/4 - 1/10

​
Mathematics
1 answer:
MAXImum [283]3 years ago
6 0
<h3>Answer: Choice C</h3>

P = 11/40 + 1/4 - 1/20

=========================================================

Explanation:

The formula we use is

P(A or B) = P(A) + P(B) - P(A and B)

In this case,

  • P(A) = 22/80 = 11/40 = probability of picking someone from consumer education
  • P(B) = 20/80 = 1/4 = probability of picking someone taking French
  • P(A and B) = 4/80 = 1/20 = probability of picking someone taking both classes

So,

P(A or B) = P(A) + P(B) - P(A and B)

P(A or B) = 11/40 + 1/4 - 1/20

which is why choice C is the answer

----------------

Note: P(A and B) = 1/20 which is nonzero, so events A and B are not mutually exclusive.

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6 0
3 years ago
A 3 ft by 5 ft wooden board has three equally sized circles cut out of it. A penny will be randomly tossed onto the board. What
Arte-miy333 [17]

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Let's begin by listing out the data given unto us, we have:

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Area of board = Length * Width = 3 * 5 = 15 ft²,

Diameter of each circle (d) = 1 ft; r = d ÷ 2

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7 0
3 years ago
Let sets A and B be defined as follows.
erica [24]

There are 4 elements in the set A and 8 elements in set B that is n(A) = 4 and n(B) = 8

<h3>Cardinality of a set</h3>

This is the total number of elements in a set. Given the following sets;

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Since there are 4 elements in the set A and 8 elements in set B hence n(A) = 4 and n(B) = 8

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6 0
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