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3 years ago

15

How to solve please help.

a1" title="7 x^{2} 8x\alpha \sqrt{x} 7" alt="7 x^{2} 8x\alpha \sqrt{x} 7" align="absmiddle" class="latex-formula">

1 answer:

DedPeter [7]3 years ago

5 0

The answer should be 56V7 ax^3 V7

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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

How do I solve this problem? The green dots can be moved

olganol [36]

Answer :

Points at (1, 3) and (3, 3) and (2, 1)

Explanation:

The original graph (dashed) is an absolute value function meaning f(x) = |x|

<u>Given:</u> y = 2f(x - 2) + 1

f(x - 2) means you plug in x - 2 for x in f(x) function

the new graph equation is: y = 2|x - 2| + 1

The graph will shift up 1 because +1
The graph will shift right because -2
The graph has a scale factor of 2
The graph is V shaped because it is an absolute value graph

Learn more about Absolute Value here: brainly.com/question/729268

<u>Visual:</u>

8 0

2 years ago

Calculate a four-day moving average for the price of the stock for the end of day 7. day 1 - 62.00; day 2 - 56.00; day 3 - 50.00

Aneli [31]

At day 7, the four-day moving average for the price of the stock would be $58.25.

<h3>What is the four-day moving average at day 7?</h3>

This can be found as:

= (Day 7 price + Day 6 + Day 5 + Day 4) / Number of days

Solving gives:

= (59 + 55 + 59 + 60) / 4

= 233 / 4

= $58.25

Find out more on moving averages at brainly.com/question/15188858.

#SPJ1

8 0

2 years ago

bazaltina [42]

Answer:

6

Step-by-step explanation:

8(1/2) + 2(4)

/2

4 + 8

/2

12/2

6

6 0

3 years ago

Read 2 more answers

What is 18/48 simplify

Y_Kistochka [10]

Hmm they're both even numbers so maybe we can start by cutting each number in half.
$\rm \dfrac{18}{48}=\dfrac{9\cdot2}{24\cdot2}=\dfrac{9}{24}$
18 and 48 had 2 as a common factor.
So factoring a 2 out of each number was the same as cutting each number in half. Try to do something similar with the 9 and 24. They each have something in common.

4 0

3 years ago

Read 2 more answers