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NemiM [27]
3 years ago
13

What is the greatest common factor of 8 and 18?

Mathematics
1 answer:
SCORPION-xisa [38]3 years ago
4 0

Answer: 2

Step-by-step explanation: To find the greatest common factor or GCF of 8 and 18, we begin by finding all of the factors of each number.

To find the factors of 8, we know that 8 ÷ 1 is 8 so 1 and 8 are factors.

8 ÷ 2 is 4 so 2 and 4 are factors.

However, if we continue to divide 8 by 3, 4, 5, and so on, we won't find any new factors. So the factors of 8 are 1, 2, 4, and 8.

Now let's find the factors of 18.

18 ÷ 1 is 18 so 1 and 18 are factors.

18 ÷ 2 is 9 so 2 and 9 are factors.

18 ÷ 3 is 6 so 3 and 6 are factors.

However, if we continue to divide by 4, 5, 6, and so on, we won't find any new factors. So the factors of 18 are 1, 2, 3, 6, 9, and 18.

Now that we have our list of factors, to find the greatest common factor, we simply find the largest number that is shared by the two lists.

Notice that our lists share a 1 but the largest number they share is a <em>2</em><em>.</em>

<em />

So the greatest common factor or <em>gcf</em> of 8 and 18 is 2.

I have also shown my work on the whiteboard.

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Answer:

Almost about 4500 mm

Step-by-step explanation:

Measure the pennies width and estimate. Multilpy by 2000 and there you go!

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2 years ago
Can someone please help me with math.
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neither positive nor negative

Step-by-step explanation:

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2 years ago
Solve for<br> Help me please
Delvig [45]

Answer:  The first one: 2 square roots of 3

Step-by-step explanation:

4/2 would give you the base= 2

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2 square roots of 3

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Question 54 of 98
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submitdjememendixodme ejej

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3 years ago
student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te
alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = \frac{1}{4}.

Thus, the probability that the student does not receive version A = 1-\frac{1}{4} = \frac{3}{4}.

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}

= (\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5

= (\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]

= (\frac{3}{4^4})[1+\frac{1}{16}]

= (\frac{3}{256})[\frac{17}{16}]

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0
3 years ago
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