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2 years ago

HELP Combining like terms solve all please :) c

Mathematics

2 answers:

In-s [12.5K]2 years ago

6 0

Answer:

11. x = -16

12. k = 6

13. x = -19

14. x = -6

15. x = -20

16. Combining like terms isn't to be used on this type of problem. I'm sorry, can you guess on this one?

17. x = 19

18. n = -10

19. b = 11

20. n = 4

21. r = -6

22. n = -4

Again super sorry about question 16 :(

stealth61 [152]2 years ago

5 0

Answer:

Step-by-step explanation:

11) -1= x/16

x= -16(because when dividing a negative and positive number the answer is always a negative.

12) 7=1+k

k=7-1

k=6

13) x-6=-25

x=-25+6

x=-19

14) -8x=48

x=48/-8

x= -8

15) x/10=-2

x=10*-2

x=-20

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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.

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We can do simultaneous equations either by substitute method or elimination. In this case I'm using substitute.
We can already obtain the value of y (in terms of x) in euqation no. 2, so all we gonna do is to put y into equation 1.
X + x + 19 = 43
Solve this by algebra.
2x = 43 - 19
X = 12
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