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Liula [17]
3 years ago
13

A civics teacher asked her students to indicate whether they believed each of two headlines. One headline was false and the othe

r was true, but the students did not know this. The probability that a student selected at random believed the true headline was 90\%90%90, percent and the probability that the student believed the false headline was 82\%82%82, percent. She found that 75\%75%75, percent of the students believed both headlines.
Mathematics
1 answer:
dusya [7]3 years ago
8 0

Answer:

The events are not mutually exclusive.

Step-by-step explanation:

To determine that in the sample, are the events "believed the false headline" and "believed the true headline" mutually exclusive.

For two events, let say E₁ and E₂ are said to be mutually exclusive if they cannot both occur simultaneously. In set-theoretical notation, we can say that the two sets  E₁ and E₂ are disjoint. i.e.  E₁ ∩  E₂ = ∅ and the probability of them occurring at the same time is zero. i.e. Pr(  E₁ ∩  E₂ ) = 0

So, we can say;

Let P(true) = E₁  

P(false) = E₂

Thus; statistically:

Pr(  E₁ ∩  E₂ ) = Pr(E₁) + Pr(E₂) - P( E₁  ∪ E₂ )

So;

the probability of students that believed the true headline P(E₁) = 0.90

the probability of students that believed the false headline P(E₁) = 0.82

the probability of students that believed both P( E₁  ∪ E₂ ) = 0.75

∴

Pr(  E₁ ∩  E₂ ) = 0.90 + 0.82 - 0.75

Pr(  E₁ ∩  E₂ ) = 0.97

Thus, the events are not mutually exclusive.

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From a box containing 10 cards numbered 1 to 10, four cards are drawn together. The probability that their sum is even is 21 21
ankoles [38]

Answer:

Step-by-step explanation:

We know that between 1 to 10 there are 5 even and 5 odd numbers.

We could get 4 even cards , 4 odd cards or 2 odd and 2 even cards

Let´s check all this combinations

Case 1: When all 4 numbers are even:  

We are going to take 4 of the 5 even numbers in the box so we have

5C4=5

Case 2: When all 4 numbers are odd:  

We are going to take 4 of the 5 odd numbers in the box, so we have

5C4=5

Case 3: When 2 are even and 2 are odd:

We are giong to take 2 from 5 even and odd cards in the box so we have

 

5C2 * 5C2

Remember that we obtain the probability from

\frac{Number-of-favourable-Outcome}{Total-number-of-outcomes}

So we have the number of favourable outcomes but we need the Total cases for drawing four cards, so we have that:  

We are taking 4 of the 10 cards:

10C_4=210

Hence we have that the probability that their sum is even

\frac{5+5+100}{210}=\frac{11}{21}

8 0
3 years ago
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konstantin123 [22]

Answer:

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Step-by-step explanation:

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What is a system of equations for the following situation? A group of 12 people went to see a movie. The cost to go to the movie
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First define the variables:

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Next create the equations to satisfy the conditions:

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Answer:

(x-1)^2 + (y+4)^2 = 100\\\\

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

Work Shown:

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You could expand terms out and simplify, but I think it's more handy to leave it in this form so you can easily spot the center and radius from a glance.

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Can someone do my last question and actually try it​
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Answer:

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Step-by-step explanation:

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