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Whitepunk [10]
3 years ago
10

What is the least common multiple of 2, 3, and 20?

Mathematics
2 answers:
beks73 [17]3 years ago
5 0

Answer:

60

Step-by-step explanation:

20x3=60

2x30=60

3x20=60

60 is the least common multiple.

tia_tia [17]3 years ago
3 0

Answer:

60

Step-by-step explanation:

The least common multiple of 2 and 20 is 20.

The only factor that is shared with 20 and 3 is 1. Therefore, the least common multiple will have to be 20*3, which is 60.

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Zane needs 20 yards of fencing to go around his rectangular yard. What are possible dimensions for Zane's yard? Select all that
stiv31 [10]

Answer:

1 yard by 9 yards

2 yards by 8 yards

3 yards by 7 yards

4 yards by 6 yards

Step-by-step explanation:

the perimeter of the rectangular yard needs to be 20 yards

Perimeter of a rectangle = 2 x ( length + width)

20 = 2 x ( length + width)

( length + width) = 20 /2

( length + width) = 10

The sum of the length and width should equal 10

Possible dimensions that would equal 10 are :

1 yard by 9 yards

2 yards by 8 yards

3 yards by 7 yards

4 yards by 6 yards

7 0
3 years ago
What does 5 + 5 equal
bixtya [17]

Answer:

10 Stupid. You're in college

Step-by-step explanation:

Bruh

7 0
4 years ago
Calculate the Area of the square
Naddika [18.5K]

Answer:

208 squared

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Sara left a bin outside in her garden to collect rainwater. She notices that 1/8 gallon of water fills 2/3 of the bin. Write and
Tpy6a [65]

Answer: 3/16 gallons of water

Step-by-step explanation:

With any fraction, if you multiply it by its reciprocal, it equals 1.  (ex: 3/4*4/3=1, 1/2*2/1=1, 71/256*256/71=1)  Thus, 2/3*3/2=1.  Because 2/3 of the bin=1/8 gallon, simply multiply both sides of the equation by 3/2 and you will get that 1 bin of water=0.1875 gallons of water, or 3/16 gallons of water.

Hope it helps <3

8 0
3 years ago
Discuss the validity of the following statement. If the statement is always​ true, explain why. If​ not, give a counterexample.
Zarrin [17]

Correction:

Because F is not present in the statement, instead of working on​P(E)P(F) = P(E∩F), I worked on

P(E∩E') = P(E)P(E').

Answer:

The case is not always true.

Step-by-step explanation:

Given that the odds for E equals the odds against E', then it is correct to say that the E and E' do not intersect.

And for any two mutually exclusive events, E and E',

P(E∩E') = 0

Suppose P(E) is not equal to zero, and P(E') is not equal to zero, then

P(E)P(E') cannot be equal to zero.

So

P(E)P(E') ≠ 0

This makes P(E∩E') different from P(E)P(E')

Therefore,

P(E∩E') ≠ P(E)P(E') in this case.

8 0
3 years ago
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