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bogdanovich [222]

2 years ago

8

selena has developed A function that models the amount of time it will take to feel like concert hall as a function of the numbe

r of people who are attending a show at the hall

1 answer:

Volgvan2 years ago

3 0

Sorry I do not know what your asking

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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:

<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.

4 0

3 years ago

A cube has and edge length of 2 centimeters. What is it’s volume, in cubic centimeters?

77julia77 [94]

Answer:

8

Step-by-step explanation:

2x2 is 4 and x 2 is 8. I think im right?

5 0

3 years ago

Find the coordinates of the vertices of a triangle after a 90 degree rotation counterclockwise about the origin.

Oksana_A [137]

Step-by-step explanation:

The rule for a 90 degree counterclockwise is

(x,y) -> (-y,x)

so

C(-1,2) -> C'(-2,-1)

D(3,5) -> D'(-5,3)

E(1,2) -> E'(-2,1)

7 0

3 years ago

Increase 120 centimetres by 24%

Eva8 [605]

It is a simple problem where 120 centimeters need to be increased by 24%. The increased length can be found by:
120 * (24/100)
= 12 * (24/10)
= 288/10
= 28.8 centimeters
Then the total length of increase = 28.8 cm
Then the increased length = (120 + 28.8) cm
= 148.8 cm
So the length becomes 148.8 cm after it is increased by 24%.

3 0

3 years ago

You have a 1-gallon paint can in the shape of a cylinder. One gallon is 231 cubic inches. The radius of the can is 3 inches. Wha

aleksandr82 [10.1K]

Answer:

The approximate height of the paint can is 8.2 in

Step-by-step explanation:

we know that

The volume of the cylinder ( can of paint) is equal to

$V=\pi r^{2} h$

we have

$V=231\ in^{3}$

$r=3\ in$

$\pi=3.14$

Substitute the values and solve for h

231=(3.14)(3)^{2} h

h=231/(3.14*9)=8.2 in

3 0

2 years ago

Read 2 more answers