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4 years ago

(Just #6) answer both Number of solutions: Explanation:

Mathematics

1 answer:

Slav-nsk [51]4 years ago

8 0

◆ Linear equations in two Variables ◆

Hey !!

Check the attachment.
Hope it helps you :)

P.s. As you can see , above expressions have only one unique solution satisfying both equations

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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

A sports writer wished to see if a football filled with helium travels farther, on average, than a football filled with air. To

kondor19780726 [428]

Answer: According to recentest research no well actually it'd float away

Step-by-step explanation:

6 0

3 years ago

Ok I need help is this correct? I tried but I gave up.

Bas_tet [7]

X+y=12
8x+10y=102

8x+8y=96
8x+10y=102 _
-2y= -6
y=3

x+y=12
x=12- 3
x=9

or
8x+10y=102
8x+30=102
8x=102-30
8x=72
x=9

4 0

3 years ago

A pair of sneakers is on sale for 20% off. The original price is $100. What is the price of the sneakers with the discount?

kirill115 [55]

Answer:

it will be $80.00 and your saving $20.00

Step-by-step explanation:

8 0

4 years ago

PLEASE HELP

nalin [4]

Answer:

3.14159...

Step-by-step explanation:

pi is not rational number

4 0

4 years ago