The chances that the student was merely guessing is 1/3.
Bayes Theorem determines the conditional probability of an event A given that event B has already occurred.
denoted by

let A be the event that the student knows the answer .
B be the event that the student does not knows the answer .
and
E be the event he gets answer correct .
According to the given question

Probability that the answer is correct ,given that he knows the answer is

Probability that the answer is correct ,given that he guesses it is
[as the MCQ has 3 options and only one is correct]
We need to find the probability that he guesses the answer given that it is correct.
Required probability 
Substituting the values we get


Therefore , the chances that the student was merely guessing is 1/3.
Learn more about Probability here brainly.com/question/13140147
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Answer:
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Answer:
Part A:

Part B:
The closure property describes cases when mathematical operations are CLOSED. It means that if you apply certain mathematical operations in a polynomial it will still be a polynomial. Polynomials are closed for sum, subtraction, and multiplication.
It means:
But when it is about division:
<u>Example of subtraction of polynomials:</u>
<u />
<u />
<u />
<u />
Step-by-step explanation:
<u>First, it is very important to define what is a polynomial in standard form: </u>
It is when the terms are ordered from the highest degree to the lowest degree.
Therefore I can give:

but,
is not in standard form.
For this question, I can simply give the answer:
and it is correct.
But I will create a fifth-degree polynomial using this formula

Also, note that

For 


Sorry but I will not type every step for each value of 
The first one is enough.
For 

Doing that for
values:

Answer:
70.5°
Step-by-step explanation:
To solve for the ladder's angle of elevation, we solve using the Trigonometric function of Cosine
= cos θ = Adjacent/Hypotenuse
Where
Adjacent = Distance of the base from the house = 10ft
Hypotenuse = Length of the ladder = 30 ft
cos θ = 10/30
cos θ = 1/3
θ = arc cos 1/3
θ = 70.528779366
Approximately = 70.5°
Therefore, the ladder's angle of elevation is 70.5°