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3 years ago

The question is.....

Mathematics

1 answer:

yaroslaw [1]3 years ago

8 0

Pi x 10 2/4
Pi x 10 x 10 + 4
Pi x 100 + 4
Pi x 25
3.14 x 25
= 78.5. Hope this helps. :)

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Which of the following is a correct name for a side of the given angle?

Zigmanuir [339]

Answer:

C. ED

Step-by-step explanation:

The center of the angle is the point E, so it would be written first.

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1 year ago

Question 14 (1 point)

valina [46]

Answer:

C(x) = 25 + 2x

Step-by-step explanation:

He already has 25 cards.

He is collecting 2 cards per month. In x months, he will have collected:

2 * x = 2x cards

That means that the number of cards he will have after x months will be the sum of the cards he already has and the cards he collects per month:

C(x) = 25 + 2x

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2 years ago

Naval intelligence reports that 4 enemy vessels in a fleet of 17 are carrying nuclear weapons. If 9 vessels are randomly targete

icang [17]

Answer:

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

Step-by-step explanation:

The vessels are destroyed without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question:

Fleet of 17 means that $N = 17$

4 are carrying nucleas weapons, which means that $k = 4$

9 are destroyed, which means that $n = 9$

What is the probability that more than 1 vessel transporting nuclear weapons was destroyed?

This is:

$P(X > 1) = 1 - P(X \leq 1)$

In which

$P(X \leq 1) = P(X = 0) + P(X = 1)$

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,17,9,4) = \frac{C_{4,0}*C_{13,9}}{C_{17,9}} = 0.0294$

$P(X = 1) = h(1,17,9,4) = \frac{C_{4,1}*C_{13,8}}{C_{17,9}} = 0.2118$

Then

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0294 + 0.2118 = 0.2412$

$P(X > 1) = 1 - P(X \leq 1) = 1 - 0.2412 = 0.7588$

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

8 0

2 years ago

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gladu [14]

2x - 8y = 32
-8y = -2x + 32
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