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ExtremeBDS [4]
3 years ago
15

How do you solve -h/3 - 4 = 13

Mathematics
1 answer:
Phantasy [73]3 years ago
8 0
Answer is -h = -51 (download photomath)

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Suppose that bugs are present in 1% of all computer programs. A computer de-bugging program detects an actual bug with probabili
lawyer [7]

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:

P(B) =0.01\\P(D|B)=0.99\\P(D|B^{c})=0.02

(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:

P(E|X)=\frac{P(X|E)P(E)}{P(X|E)P(E)+P(X|E^{c})P(E^{c})}

Use the Bayes' theorem to compute the value of P (B | D) as follows:

P(B|D)=\frac{P(D|B)P(B)}{P(D|B)P(B)+P(D|B^{c})P(B^{c})}=\frac{(0.99\times 0.01)}{(0.99\times 0.01)+(0.02\times (1-0.01))}=0.3333

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)

                                                                                     =0.3333\times 0.3333\\=0.11108889\\\approx 0.1111

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)

=0.3333\times 0.3333\times 0.3333\\=0.037025927037\\\approx 0.037

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.

4 0
3 years ago
Help me out please ty!
Naya [18.7K]

Answer:

a5 = 2604

Step-by-step explanation:

a1 = 4

a2 = -5(4)-1 = -21

a3 =-5(-21)-1 = 104

a4 = -5(104)-1 =-521

a5 =-5(-521)-1 = 2604

5 0
2 years ago
Read 2 more answers
-5x + y = -6 <br><br>7x + 2y = 39
storchak [24]

Answer:

so i don't know if they are separate math problems but for the first question i got M=5 and the second question i got M=-7/2

Step-by-step explanation:

hope this helps please mark brainliest

6 0
3 years ago
During a sale, everything in the store was ⅕ off the ticketed price. What percent of an item's original price should you expect
tiny-mole [99]
You would pay 80% of the original price because 1/5 is 20% and you would subtract 20 from 100 to get 80 as your answer 

Hope this Helps :3
4 0
3 years ago
Adele has 6 sheets of stickers .Bea has 9 sheets of the same stickers.how many stickers do they have altogether?
Jlenok [28]
They have 15 sheets of stickers all together
7 0
3 years ago
Read 2 more answers
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