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2 years ago

Given: p(a) =0.70, p(b)=0.60, p(a or b) =0.88. are a and b independent? (use your answer from question

Mathematics

1 answer:

allsm [11]2 years ago

4 0

Answer:

Step-by-step explanation:

P(a or b) = P(a) + P(b) - P(a and b)

0.88 = 0.70 + 0.60 - P(a and b)

P(a and b) = 1.30 - 0.88 = 0.42.

So the events a and b are not mutually exclusive.

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Help me out! (URGENT) I need help with numbers 1-6 please

Alex Ar [27]

Answer:

Step-by-step explanation:

1. Blank= 5

2. Blank= 10 (I believe)

3. Blank= 8

4. Blank= 13 (I believe)

5. Blank= 12

6. Blank= 7

Hope it helps!

OwO

4 0

2 years ago

Plz.. Help me.. True or false?

zimovet [89]

Answer:

$\boxed{\mathrm{False}}$

Step-by-step explanation:

$(p-q)^2$

$(p-q)(p-q)$

Use FOIL method.

$p^2-pq-pq +q^2$

$p^2-2pq +q^2$

6 0

2 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

Evaluate (-5z)^3 . Write your answer using only positive exponents. Evaluate any numerical powers.

Ludmilka [50]

Answer:

-125z³

Step-by-step explanation:

-5 times - 5 = 25 canceling negative

125 times -5 = -125

zXzXz = z to power of 3 =z³

7 0

3 years ago

57.42

Fofino [41]

Hi there!

$\large\boxed{a = 6 m/s^{2}}$

Use the kinematic equation to solve for acceleration:

$a = \frac{v_{f}-v_{i}}{t}$

Where:

vf = final velocity

vi = initial velocity

t = time

Plug in the given values:

$a = \frac{65-35}{5}\\\\a = 30 / 5 \\\\=6 m/s^{2}$

7 0

2 years ago