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4 years ago

Please helpppppppppp me

Mathematics

1 answer:

Naily [24]4 years ago

5 0

Answer:

Please read explanation below.

Step-by-step explanation:

Let's go over what some of the symbols of inequalities represent:

$>$ : greater than

: less than

$\ge$ : greater than or equal to

$\le$ : less than or equal to

The symbol in the equation that is given to you is $\ge$ . It seems as though each of the answer choices have everything written as the same thing except for the description of the inequalities. Check the one that applies.

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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

What is the least common multiple of 2, 6, and 15?

VLD [36.1K]

Answer:

Step-by-step explanation:

Prime factorization of the numbers:

2 = 2 × 1

6 = 2 × 3

15 = 3 × 5

LCM(2, 6, 15)

= 2 × 3 × 5

= 30

5 0

3 years ago

Which three angles with the given measures can be used to construct at least one triangle? 40°,20°,40° 45°,45°,45° 23°,23°,134°

goldenfox [79]

D is the answer :) hope this helped

3 0

3 years ago

Read 2 more answers

Find the algebraic expression for the word phrase 7 less than p

True [87]

C. p-7 because 7 less than p implies that we would subtract 7 from p

5 0

3 years ago

A bookmark is shaped like a rectangle with a semicircle attached at both ends. The rectangle is 11 cm long and 4 cm wide. The di

marin [14]

Figure=1rectangle+2 semicircles
1semicircle=1/2circle
2 (1/2 circles)=1 circle
figure=1 rectangle+1 circle
area of rectangle=legnth times width
area of circle=pir^2

so the width of the rectangle=diameter of the circle
widht=4

diameter=2r
d/2=r
width=d=4
4/2=r=2

area of circle=3.14 times 2^2
are of circle=3.14 times 4
aera of circle=12.56

area of rectangle=11 times 4
area=44

total=add them up
12.56+44=56.56 cm^2

8 0

3 years ago