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3 years ago

Which pair of numbers has a GCF of 3? Group of answer choices (A) 3 and 18 (B) 8 and 24 (C)12 and 18 (D) 1 and 3

Mathematics

1 answer:

Arte-miy333 [17]3 years ago

5 0

Answer:

A, B and C

Step-by-step explanation:

A) 3 and 18 = has 3 as factor

B) 8 and 24 = has 3 as factor

C) 12 and 18 = has 3 as factor

D) 1 and 3 = has as factor of 1 and 3

so the answer is A, B and C

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Answer:

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Step-by-step explanation:

Given that

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And, the commission rate is 25%

We need to find out the sales commission

So, the sales commission is

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= $900 × 0.25

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3 years ago

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3 years ago

If 1 inch on a map represents 20 miles, what is the mileage represented by 2.5 inches?

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Step-by-step explanation:

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5 0

3 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

3 years ago