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mart [117]
3 years ago
6

Are 2/3 and 6/8 equivalent?

Mathematics
1 answer:
miskamm [114]3 years ago
7 0
No they are not, 6/8 could also be represented as 3/4. Therefore they are not equivalent. You would divide each side by the factor they have in common such as 2.
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To transmit information on the internet, large files are broken into packets of smaller sizes. Each packet has 1,500 bytes of in
erastova [34]

Answer: x= 187.5p

Step-by-step explanation:

Given : An equation relating packets to bytes of information : b = 1,500p where<em> p</em> represents the number of packets and <em>b</em> represents the number of bytes of information.

1 byte = 8 bits

Let x= Number of bits such that 1 b = 8x

Put b = 8x in given equation , we get

8x=1500p

Divide both sides by 8 , we get

x= 187.5p

Hence, the equation represent the relationship between the number of packets and the number of bits : x= 187.5p

3 0
3 years ago
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What fraction is equivalent to eight tentHs
aleksklad [387]
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2 years ago
The product of 162 and m is the same as 17
elena-s [515]

Answer:

No

Step-by-step explanation:

Just took a test

5 0
3 years ago
What is the solution to the equations represented by these two lines
salantis [7]

Look at the picture.

A.) (3, 1)

6 0
3 years ago
A boy has 3 red , 2 yellow and 3 green marbles. In how many ways can the boy arrange the marbles in a line if: a) Marbles of the
Ilya [14]

Answer:

The total number of different arrangements is 560.

Step-by-step explanation:

A multiset is a collection of objects, just like a set, but can contain an object more than once.

The multiplicity of a particular type of object is the number of times objects of that type appear in a multiset.

Permutations of Multisets Theorem.

The number of ordered n-tuples (or permutations with repetition) on a collection or multiset of n objects, where there are k kinds of objects and object kind 1 occurs with multiplicity n_1, object kind 2 occurs with multiplicity n_2, ... , and object kind k occurs with multiplicity n_k is:

                                                 \begin{equation*}\frac{n!}{n_1!*n_2!*\dots * n_k!}\end{equation*}

We know that a boy has 3 red, 2 yellow and 3 green marbles. In this case we have n = 8.

If marbles of the same color are indistinguishable, then the total number of different arrangements is

{8 \choose 3, 2, 3}  = \frac{8 !}{3 ! 2 ! 3 !} = \frac{8\cdot \:7\cdot \:6\cdot \:5\cdot \:4}{2!\cdot \:3!}=\frac{6720}{2!\cdot \:3!}=\frac{6720}{12}=560

8 0
3 years ago
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