1.6y + y - 4/15 y + 1 1/6y = 2 1/3

HELP HELP HELP FIRST RIGHT ANSWER GETS BRAINLIEST

Sati [7]

Answer: B

Explanation

3 0

2 years ago

HELP!...................................................

DaniilM [7]

Answer:

1.6 hrs remaining, or 3.12 km/h

Step-by-step explanation:

7 0

2 years ago

Help pls :) 10 points

Alexeev081 [22]

Answer:

sq root of 64 = 8, sq root of 81 = 9, 8 and 9

Step-by-step explanation:

Since Sq root of 64 = 8 and Sq root of 81 = 9 it is known that sq root of 66 is between 8 and 9

7 0

2 years ago

2. Which fraction is in its simplest form? 2/4 2/8 2/6 2/5

wariber [46]

Answer:

2/5

Step-by-step explanation:

2 / 4 could be divided by two in both sides.

2 ÷ 2 = 1

4 ÷ 2 = 2

2/4 = 1/2

Same goes for 2/8 and 2/6

2 ÷ 2 = 1

8 ÷ 2 = 4

2/8 = 1/4

2 ÷ 2 = 1

6 ÷ 2 = 3

2/6 = 1/3

2/5 however, could not be divided by two on both sides because there is 5 and it is not a factor of 2. So it is in its simplest form

7 0

3 years ago

Read 2 more answers

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