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3 years ago

Solve for X. HELP PLEASE!!! ASAP

Mathematics

2 answers:

HACTEHA [7]3 years ago

5 0

Answer:

5, 6, 9

Step-by-step explanation:

1. x= -5

2.x= 6

3. x=9

gizmo_the_mogwai [7]3 years ago

3 0

1. C) -5
2. D) 6
thats what i know

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Write an inequality that represents the graph below:

Ray Of Light [21]

Answer:

X is greater than 3

X > 3

Step-by-step explanation:

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3 years ago

F(c)=x²-3x find f (-8)

aliina [53]

Answer:

put value of x in x

Step-by-step explanation:

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2 years ago

Read 2 more answers

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

Length=x^2 -8; W= -x+12 PLEASE HELP!

ASHA 777 [7]

L×W
x^2+8x+12
am not sure

4 0

3 years ago

Which number has the same value as 3 X 1/10?

irakobra [83]

To find the same value as 3 x 1/10 you first have to solve 3 x 1/10

so how to do this is you have to make 3 into a fraction

so to make 3 into a fraction you can put 3 as the numerator and 1 as the denominator

so you multiply
$\frac{3}{1} \times \frac{1}{10} = \frac{3}{10}$
so 3/10 has the same value as 3x1/10

HOPE THIS HELPS!!!

5 0

3 years ago