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Dovator [93]
2 years ago
10

I need help to find the lcm of 81 and 3

Mathematics
2 answers:
velikii [3]2 years ago
8 0

Answer:

81 is the LCM of 81 and 3

Step-by-step explanation:

81 is the smallest (least) number that both (common) 81 and 3 "go into" (multiple)

81 goes into 81.

3 goes into 81.

81 is the smallest number that they both go into.

Lina20 [59]2 years ago
4 0

Hi there,

please see below for solution steps :

________

<u></u>\ast<u> Key moment to solving this problem</u> :

  • the LCM stands for least common multiple

Okay. Now, what is the lcm of 3 and 81?

One way to find the lcm is : list several multiples of each number.

However, it's not a really efficient way, because listing all the multiples of 3 until we get to 81 is time-consuming!

So we can ask ourselves one question : <em>Is 81 divisible by 3?</em>

Sure! with this in mind, let's ask ourselves another question : <em>what is the lcm of 81 and 3?</em>

The answer to that is \boldsymbol{81}.

Hope the answer - and explanation - made sense to you,

happy studying !!                                                                 \bold{Frozen \ melody}

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Step-by-step explanation:

the width of the first frame is 12cm because 120/10 = 12

the length of the second frame is 24cm because 288/12 = 24

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We want to determine if the probability that a student enrolled in an accelerated math pathway is independent of whether the stu
Molodets [167]

Answer:

The correct option is (c).

Step-by-step explanation:

The complete question is:

The data for the student enrollment at a college in Southern California is:

                    Traditional          Accelerated            Total

                  Math-pathway     Math-pathway

Female              1244                       116                   1360

Male                  1054                       54                    1108

Total                  2298                     170                  2468

We want to determine if the probability that a student enrolled in an accelerated math pathway is independent of whether the student is female. Which of the following pairs of probabilities is not a useful comparison?

a. 1360/2468 and 116/170

b. 170/2468 and 116/1360

c. 1360/2468 and 170/2468

Solution:

If two events <em>A</em> and <em>B</em> are independent then:

P(A|B)=P(A)\\\\\&\\\\P(B|A)=P(B)

In this case we need to determine whether a student enrolled in an accelerated math pathway is independent of the student being a female.

Consider the following probabilities:

P (F|A) = \farc{116}{170}\\\\P(A|F)=\frac{116}{1360}\\\\P(A)=\frac{170}{2468}\\\\P(F)=\frac{1360}{2468}

If the two events are independent then:

P (F|A) = P(F)

&

P (A|F) = P (A)

But what would not be a valid comparison is:

P (A) = P(F)

Thus, the correct option is (c).

4 0
3 years ago
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