What decimal is halfway between 2.7 and 2.8?

Identify a pattern in the given list of number 1,2,1,4,1,8,1 what’s the next number ?

Ipatiy [6.2K]

10
it looks like the pattern is 1, (x+2) x is the number before it

3 0

4 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

3 years ago

Let f(X)=3√x

SOVA2 [1]

g(x) = 3√(x-5) -1

The process of altering a graph to produce a different version of the preceding graph is known as graph transformation. The graphs can be moved about the x-y plane or translated. They may also be stretched, or they may undergo a mix of these changes.

Horizontal stretching: It means the graph is elongated or shrink in x direction.

Vertical stretching : It means the graph is elongated or shrink in y direction

Vertical translation : It means moving the base of the graph in y direction

Horizontal translation : It means moving the base of the graph in x direction

According to rules of transformation f(x)+c shift c units up and f(x)-c shift c units down.

Therefore, in order to move the graph down 1 units, we need to subtract given function by 1 , we get

g(x) = 3√x -1

According to rules of transformation f(x+c) shift c units left and f(x-c ) shift c units right.

Therefore, in order to move the graph left by 5 units, we need to add given function by 5 , we get

g(x) = 3√(x-5) -1

To learn more about graphical transformation, refer to brainly.com/question/4025726

#SPJ9

4 0

1 year ago

A rectangular school banner has a length of 44 inches, a perimeter of 156 inches, and an area of 1,496 square inches. the cheerl

STatiana [176]

A rectangular school banner has a length of 44 inches, a perimeter of 156 inches, and an area of 1,496 square inches. the cheerleaders make signs similar to the banner. the length of a sign is 11 inches. First solve the width of the rectangle:

1496 sq in/ 44 = 34 in

So the sign has also a width of 34 in and a length of 11 so the area is 34*11 =374 sq in

The perimeter is (34*2) +(11*2) =90 in

3 0

3 years ago

Help??? It’s pretty complicated, well for me.

masha68 [24]

Answer:

The first box is 10.2

The second box is 15.

The last box is 76.

8 0

3 years ago