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2 years ago

Please help me I really need this

Mathematics

2 answers:

Alenkasestr [34]2 years ago

8 0

Answer:

Step-by-step explanation:

8C5 = 56

Serjik [45]2 years ago

7 0

Answer:

8C5 = 56

Step-by-step explanation:

The formula for nCk is ...

nCk = n!/(k!(n-k)!)

Then the value of 8C5 is ...

8!/(5!(8-5)!) = 8·7·6/(3·2·1) = 8·7

8C5 = 56

_____

Additional comment

The larger of the denominator factorials will cancel most of the factors of the numerator. The remaining denominator factors can be factored from the remaining numerator values to simplify the problem of doing the math by hand. Most graphing calculators and all spreadsheets can compute this function for you.

$\dfrac{8!}{5!\cdot3!}=\dfrac{8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{(5\cdot4\cdot3\cdot2\cdot1)(3\cdot2\cdot1)}=\dfrac{8\cdot7\cdot6}{3\cdot2\cdot1}=\dfrac{8\cdot7\cdot6}{6}=8\cdot7$

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2 years ago

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3 years ago

Suppose that bugs are present in 1% of all computer programs. A computer de-bugging program detects an actual bug with probabili

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Answer:

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(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.

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2 years ago

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Rufina [12.5K]

Answer:

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Step-by-step explanation:

First multiply -13 with (q+4)

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Now you have to change the order before = the numbers with q and after = the numbers without anything

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Remember that when you change positions the symbol too

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Now divide by the greater divisor in this case is 4

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And organize it

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Never forget to do q=(answer)

Hopefully it helps.

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3 years ago