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Aliun [14]
3 years ago
15

What common fraction is equivalent to 3/12

Mathematics
1 answer:
777dan777 [17]3 years ago
3 0

Answer:

1/4 is the common fraction

You might be interested in
Please help me solve this
Oduvanchick [21]
\dfrac{x}{ - 3} < - 4 \\\\ x > 12

P. S. The inequality sign changed or flipped when there's a negative figure involved.

Hope this helps. - M
3 0
3 years ago
Anwser: <br> 36÷(1-9×1-4)+1+2
kkurt [141]
So this is a BIDMAS question so first you do what’s in the brackets so you have (1-9x1-4) and you have to do the multiplication first out of that to get (1-9-4) which is -12 so then you have overall 36 / -12 + 1 + 2 so you have to do the division first so do 36/-12 to get -3 and you have -3 + 1 + 2 which is 0 so your answer is 0.
5 0
3 years ago
Read 2 more answers
Does the following table show a proportional relationship between the variables x and y? x 5/6 25/6 125/6 y 2 10 50
ki77a [65]

Answer: A:Yes

Step-by-step explanation:

This table shows a proportional relationship between the variables x and y.

3 0
3 years ago
I need help! Will mark Brainliest for full answer!!!
Softa [21]

<u><em>Answer:</em></u>

Part a .............> x = 11

Part b .............> k = 57.2

Part c .............> y = 9.2

<u><em>Explanation:</em></u>

The three problems deal with inverse variation between two variables

An inverse variation relation between two variables means that when one of the variables increases, the other will decrease (and vice versa)

<u>Mathematically, an inverse variation relation is represented as follows:</u>

y = \frac{k}{x}

where x and y are the two variables and k is the constant of variation

<u><em>Now, let's check the givens:</em></u>

<u>Part a:</u>

We are given that y = 3 and k = 33

<u>Substitute in the original relation and solve for x as follows:</u>

y = \frac{k}{x}\\ \\3 = \frac{33}{x}\\ \\x=\frac{33}{3}=11

<u>Part b:</u>

We are given that y = 11 and x = 5.2

<u>Substitute in the original relation and solve for k as follows:</u>

y=\frac{k}{x}\\ \\11=\frac{k}{5.2}\\ \\k=11*5.2=57.2

<u>Part c:</u>

We are given that x=7.8 and k=72

<u>Substitute in the original relation and solve for y as follows:</u>

y=\frac{k}{x}=\frac{72}{7.8}=9.2 to the nearest tenth

Hope this helps :)

6 0
3 years ago
Recall that a 6-bit string is a bit strings of length 6, and a bit string of weight 3, say, is one with exactly three 1's. How m
strojnjashka [21]

Answer:

1.. Total number of 6 bit strings is 64

2. Number of 6-bit strings with weight of 0 is 1

3. Number of 6-bit strings with weight of 1 is 6

4. Number of 6-bit strings with weight of 3 is 20

5. Number of 6-bit strings with weight of 5 is 6

6. Number of 6-bit strings with weight of 6 is 1

7. Number of 6-bit strings with weight of 7 is 0

Step-by-step explanation:

A bit string is a string that contains 0 and 1 only

1. Total number of 6 bit strings is 2^6 = 64

2. Number of 6 bit strings with weight 0 is 1

Explanation

Weight 0 means a string with no occurrence of 1

Here, we are only interested in occurrence and not order of occurrence

We apply combination formula for this

nCr = n!/(n-r)!r!

n = 6 and r = 0 i.e. no occurrence of 1

6C0 = 6!/(6-0)!0!

6C0 = 6!/6!0!

6C0 = 1

Hence, the number of string with weight 0 (i.e. no occurrence of 1 ) is 1

3. Number of string with weight 1 is 6

Explanation

Weight 0 means a string with exactly 1 occurrence of '1'

Here, we are only interested in occurrence and not order of occurrence

We apply combination formula for this

nCr = n!/(n-r)!r!

n = 6 and r = 1

6C1 = 6!/(6-1)!1!

6C1 = 6!/5!1!

6C1 = 6

Hence, the number of string with weight 6

4. Number of string with weight 3 is 20

Explanation

n = 6 and r = 3

6C3 = 6!/(6-3)!3!

6C3 = 6!/3!3!

6C3 = 20

Hence, the number of string with weight 3 is 20

5. Number of string with weight 5 is 6

Explanation

n = 6 and r = 5

6C5 = 6!/(6-5)!5!

6C5 = 6!/1!5!

6C5 = 6

Hence, the number of string with weight 5 is 6

6. Number of string with weight 6 is 1

Explanation

n = 6 and r = 6

6C6 = 6!/(6-6)!6!

6C6 = 6!/0!6!

6C6 = 1

Hence, the number of string with weight 6 is 1

7. Number of string with weight 7 is 0

Weight of 7 means that a string that has 7 occurrence of 1

The total length of a 6 bit is 6

Since 6 is less than 7, there's no way a bit of weight 7 can occur.

So, the right answer for this is 0.

8 0
3 years ago
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