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3 years ago

What is average of 46.66% and 60%

1 answer:

5 0

The average for the 2 percents added together is 0.5333 I believe this is the answer. And to double check look up average percentage calculator and Average Calculator - Omni will pop up and use that calculator to be sure. Hope this helped.

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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:

B = bugs are present in a computer program.

D = 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 E given that another event X 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

PLEASE HELP! WILL GIVE BRAINLIEST!

irinina [24]

Answer:

1: y intercept = 5

2: y interpcept = -2

3: y intercept = -3

Step-by-step explanation:

So for the 1st and 3rd ones (the graphs), all you have to do is find which y matches the x value of 0

The 2nd one is a bit harder. I’ll explain it the best I can

So in the y section, you see that -4 goes to -3 when the x is raised by 1. There for everytime the x raises by 1, so does the y. That means when x = 0, the y = -2

3 0

3 years ago

Read 2 more answers

What is the output of this function when the input is 2

nalin [4]

Easy question.

So, the slope is -x and the y-intercept is at -4.

y = -x - 4

Use this equation and you will find:

y = -2 - 4

y = -6

Answer: (2, -6)

5 0

3 years ago

For the transformation to be reflection which statement mist be true

Fiesta28 [93]

Are there any options?

5 0

3 years ago

Friction causes vehicles, such as this truck, to use more what?

ziro4ka [17]

Answer:

A. Gasoline

Step-by-step explanation:

7 0

3 years ago