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2 years ago

6 Answer the following questions about this figure. If mZ1= 53°, what is mZ3?

Mathematics

1 answer:

Scrat [10]2 years ago

5 0

Answer:

143 degrees

Step-by-step explanation:

Triangle EDF is a right triangle and we know that angle 1 is equal to 53 degrees. So, angle 2 is equal to 180-90-53 = 37 degrees. Now that we want to find angle 3, we subtract the measure of angle 2 (which is 37) from 180. That gives us 143 degrees.

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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

2 years ago

What is the answer to y=3.648·1.182×

Goryan [66]

Answer:

if you are solving for X then X=0

Step-by-step explanation:

7 0

3 years ago

URGENT URGENT URGENT URGENT URGENT

fiasKO [112]

Answer:

x=7

Step-by-step explanation:

To set up for this answer, we need to set it up as an equation: 2x+12=5x-9.

get the xs on one side and the whole numbers on the other.

that leaves you with 21=3x.

divide by 3 on both sides to get the answer of x=7

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2 years ago

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If -11 + N = -11, then N is the

gladu [14]

Answer:additive inverse

Step-by-step explanation:

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3 years ago

Three robots go out to lunch. They decide

jarptica [38.1K]

Answer:

Step-by-step explanation:

11.91/3 =3.97

the answer is $3.97

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2 years ago

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