The 10th grade is having a picnic this Friday. There will be 182 students and 274 adults. Each table seats 12 people. How many t
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<h3>Answer: (a - 1)(2a² + a + 2)</h3>
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Explanation:
Use the rational root theorem to determine this list of possible rational roots: 1, -1, 1/2, -1/2
Plug each possible root one at a time into the original expression given. If the simplified result is 0, then that possible root is an actual root.
If we tried say a = -1, then,
2a³-a²+a-2 = 2(-1)³-(-1)²+(-1)-2 = -6
The result is not zero, so a = -1 is not an actual root.
But if we tried say a = 1, then,
2a³-a²+a-2 = 2(1)³-1²+1-2 = 0
We get 0 so a = 1 is an actual root. I'll let you try the other values, but you should find that a = 1 is the only rational root.
Since a = 1 is a root, this makes (a-1) to be a factor.
From here, use either synthetic or polynomial long division to determine the other factor. Refer to the diagram below for each method.
Regardless of which method you pick, the quotient is 2a² + a + 2 which is the other factor needed. The remainder of 0 tells us we have (a-1) as a factor. For more information, check out the remainder theorem.
Answer:
(i) The probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.
(ii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.
(iii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.
Step-by-step explanation:
Denote the events as follows:
<em>B</em> = bugs are present in a computer program.
<em>D</em> = a de-bugging program detects the bug.
The information provided is:
(i)
The probability that there is a bug in the program given that the de-bugging program has detected the bug is, P (B | D).
The Bayes' theorem states that the conditional probability of an event <em>E </em>given that another event <em>X</em> has already occurred is:
Use the Bayes' theorem to compute the value of P (B | D) as follows:
Thus, the probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.
(ii)
The probability that a bug is actually present given that the de-bugging program claims that bug is present is:
P (B|D) = 0.3333
Now it is provided that two tests are performed on the program A.
Both the test are independent of each other.
The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is:
P (Bugs are actually present | Detects on both test) = P (B|D) × P (B|D)
Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.
(iii)
Now it is provided that three tests are performed on the program A.
All the three tests are independent of each other.
The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is:
P (Bugs are actually present | Detects on all 3 test)
= P (B|D) × P (B|D) × P (B|D)
Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.