The width of a rectangle is 3 feet longer than it's length. What's the dimensions of the rectangle such that the perimeter of th
e rectangle is 238 feet.
1 answer:
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Answer:
CS = 4
Step-by-step explanation:
From the figure below;
Triangle ABC is similar to triangle RSC
Therefore, the ratio of the corresponding sides in the two triangles is equal;
That is;
In this case, CR = 2 , RA = 3 and BC = 10 we are required to determine CS
But;
AC=CR + RA
= 5
Assuming CS is y, then
Therefore, CS is 4
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.