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frutty [35]
4 years ago
6

What is the justification for step 2 in the solution process?

Mathematics
2 answers:
gulaghasi [49]4 years ago
7 0

Answer: OPTION B.

Step-by-step explanation:

Given the following equation:

10x - 25 - 3x = 4x- 1

These are the steps to solve it:

<em>Step 1</em>

Add like terms on the left side of the equation:

7x -25= 4x - 1

<em>Step 2</em>

Apply Addition property of equality, which states that: If\ a=b\ then\ a+c=b+c

Then, we can add 25 to both sides of the equation:

7x -25+(25)= 4x - 1+(25)\\\\7x=4x+24

<em>Step 3</em>

Applying the Subtraction property of equality, we can subtract 4x from both sides:

7x-4x=4x+24-(4x)\\\\3x=24

<em>Step 4</em>

Applying the Division property of equality, we can both sides of the equation by 3. Then:

\frac{3x}{3}=\frac{24}{3}\\\\x=8

Dmitrij [34]4 years ago
7 0

Answer:

B. the addition property of equality

Step-by-step explanation:

we have:

Step 1: 7x − 25= 4x − 1

we apply the addition property of equality:

7x − 25+25= 4x − 1 +25

finally we have

Step 2: 7x = 4x + 24

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4 years ago
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.

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