Three boys can sit on the sled.

The graph of a function is shown. Which function is represented by the graph?

Westkost [7]

Answer: D

Step-by-step explanation:

d should represent the graph

3 0

3 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:

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

2 years ago

A local bakery sells cookies for 5.50 per pound and brownies for 7.50

uysha [10]

Answer:

I may be wrong but i think the answer is D

Step-by-step explanation:

We know that y=10, so if we were to multiply 10 by 7.50, it would be 75, not 63, so A and B wouldn't be correct. and we don't know the value of x, so I think D is the best answer.

6 0

3 years ago

3 intersecting lines are shown. A line contains points E, D, A, and B. Another line intersects the line at point A and contains

GREYUIT [131]

Answer:

The one with arrows are the answers

->Line segment E B is bisected by Line segment D F .

->A is the midpoint of Line segment F C .

Line segment F C bisects Line segment D B.

->Line segment E B is a segment bisector.

->FA = One-halfFC.

Line segment D A is congruent to Line segment A B .

Step-by-step explanation:

I did it on edge and got it right

5 0

2 years ago

Read 2 more answers

15 is 23% of what number

rosijanka [135]

15 = 0.23x
x = 65.2

therefore 15 is 23% of 65.2

4 0

2 years ago