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Maksim231197 [3]
3 years ago
7

Lcm of 6 and 13 what is it?

Mathematics
2 answers:
andriy [413]3 years ago
7 0

Answer:

78

Step-by-step explanation:

Since 13 is a prime, the least common multiple of it and 6 is their product. 6*13=78. Hope this helps!

Marianna [84]3 years ago
4 0

Answer:

78

Step-by-step explanation:

Least Common Multiple of 6 and 13. Least common multiple (LCM) of 6 and 13 is 78.

6- 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78

13- 13, 26, 39, 52, 65, 78

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Type the correct answer in each box. If necessary, use / for the fraction bar. A bag contains 5 blue marbles, 2 black marbles, a
konstantin123 [22]

Answer:

Step-by-step explanation:

Problem One

Blue =   5

Black  =2

Red    = 3

First of all there are 10 marbles, 2 of which are black.

That means that 8 others are not black

You can draw any one of the 8.

P(not black) = 8/10 = 4/5

Problem Two

There are 10 marbles in all

3 of them are red.

P(Red) = 3/10  

7 0
3 years ago
What is the solution to the equation t minus 15 = 76?
Phoenix [80]

Answer:

4. / D.

Step-by-step explanation:

t-15=76\\t=76-15\\t=61

8 0
2 years ago
I forget how to turn fractions into percent
Illusion [34]
<span>Divide the top of the fraction by the bottom,t<span>hen multiply your answer by 100</span></span>
3 0
3 years ago
Urn 1 contains 3 blue tokens and 2 red tokens; urn 2 contains 2 blue tokens and 4 red tokens. All tokens are indistinguishable.
Dmitry_Shevchenko [17]

Answer:

RR = 0.4

RB = 0.3

BB = 0.22

BR = 0.30

Step-by-step explanation:

P( Urn 1 ) = 2/6 = 1/3

P( Urn 2 ) = 1 - 1/3 = 2/3

Urn 1 contains : 3 blue and 2 red

P( blue | urn 1 ) = 3/5 ( with replacement ) , P( blue | urn 1 ) = 3/4 ( without replacement )

P( red | urn 1 ) = 2 / 5 ( with replacement ) , P(red | urn 1 ) = 1/2 ( without replacement )

Urn 2 contains : 2 blue and 4 red

P ( blue | urn 2 ) = 1/3 ( with replacement ) , P( blue | urn 2 ) = 2/5 ( without replacement )

P ( red | urn 2 ) = 2/3 ( with replacement) , P( red | urn 2 ) = 4/5 ( without replacement )

Determine

<u>i) Possible outcomes when two tokens are drawn from either Urn without replacement </u>

RR = [[ ( 2/5 * 1/3 ) + ( 2/3 * 2/3 ) ] * [( 1/2 * 1/3 ) + ( 4/5 * 2/3 ) ]] = 0.4

RB = [[ (2/5 * 1/3 ) + ( 2/3 * 2/3 ) ] * [ ( 3/4 *1/3 ) + ( 2/5 * 2/3 ) ]] ≈ 0.3

BB = [[ ( 3/5 * 1/3 ) + ( 1/3 * 2/3 ) ] * [ ( 3/4 *1/3 ) + ( 2/5 * 2/3 ) ]] ≈ 0.22

BR = [[ ( 3/5 * 1/3 ) + ( 1/3 * 2/3 ) ] * [ ( 1/2 * 1/3 ) + ( 4/5 * 2/3 ) ]] ≈ 0.30

<u />

8 0
3 years ago
HELP HELP PLEEEAASSSEE!
Nitella [24]

Answer:

What's the question?

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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