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enot [183]
3 years ago
15

Please answer question now

Mathematics
1 answer:
Radda [10]3 years ago
5 0

Answer:

TU and TS (with arrows on top)

Step-by-step explanation:

The sides of any angle are rays. This are half lines, which in your case both start at the point T, and one goes towards point U, while the other one goes from toward point S. So I would write TU (with a little arrow symbol on top). and in the other box I would write TS also with a little arrow symbol on top.

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I just took the review so this is a verified answer.

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2 years ago
Suppose you are given either a fair dice or an unfair dice (6-sided). You have no basis for considering either dice more likely
hoa [83]

Answer: Our required probability is 0.83.

Step-by-step explanation:

Since we have given that

Number of dices = 2

Number of fair dice = 1

Probability of getting a fair dice P(E₁) = \dfrac{1}{2}

Number of unfair dice = 1

Probability of getting a unfair dice  P(E₂) = \dfrac{1}{2}

Probability of getting a 3 for the fair dice P(A|E₁)= \dfrac{1}{6}

Probability of getting a 3 for the unfair dice P(A|E₂) = \dfrac{1}{3}

So, we need to find the probability that the die he rolled is fair given that the outcome is 3.

So, we will use "Bayes theorem":

P(E_1|A)=\dfrac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)}\\\\(E_1|A)=\dfrac{0.5\times 0.16}{0.5\times 0.16+0.5\times 0.34}\\\\P(E_1|A)=0.83

Hence, our required probability is 0.83.

8 0
3 years ago
In a large statistics course, 74% of the students passed the first exam, 72% of the students pass the second exam, and 58% of th
11111nata11111 [884]

Answer:

Required probability is 0.784 .

Step-by-step explanation:

We are given that in a large statistics course, 74% of the students passed the first exam, 72% of the students pass the second exam, and 58% of the students passed both exams.

Let Probability that the students passed the first exam = P(F) = 0.74

     Probability that the students passed the second exam = P(S) = 0.72

     Probability that the students passed both exams = P(F \bigcap S) = 0.58

Now, if the student passed the first exam, probability that he passed the second exam is given by the conditional probability of P(S/F) ;

As we know that conditional probability, P(A/B) = \frac{P(A\bigcap B)}{P(B) }

Similarly, P(S/F) = \frac{P(S\bigcap F)}{P(F) } = \frac{P(F\bigcap S)}{P(F) }  {As P(F \bigcap S) is same as P(S \bigcap F) }

                          = \frac{0.58}{0.74} = 0.784

Therefore, probability that he passed the second exam is 0.784 .

5 0
3 years ago
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