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tresset_1 [31]
3 years ago
5

A large aquarium contains only two kinds of fish, guppies and swordtails. If 3/4​​ of the number of guppies is equal to 2/3​​ of

the number of swordtails, then what fraction of fish in this aquarium are guppies?
Mathematics
1 answer:
Jlenok [28]3 years ago
6 0

Answer:

\frac{8}{17} of fish in this aquarium are guppies.

Step-by-step explanation:

Let x be the number of guppies and y be the number of swordtails in the aquarium,

According to the question,

\frac{3}{4}\text{ of } x=\frac{2}{3}\text{ of }y

\frac{3x}{4}=\frac{2y}{3}

By cross multiplication,

9x=8y

\implies \frac{x}{y}=\frac{8}{9}

Thus, the ratio of guppies and swordtail fishes is 8 : 9

Let guppies = 8x, swordtail = 9x

Where, x is any number,

Since, the aquarium contains only two kinds of fish, guppies and swordtails,

So, the total fishes = 8x + 9x = 17x

Hence, the fraction of fish in the aquarium are guppies = \frac{\text{Guppies}}{\text{Total fishes}}

=\frac{8x}{17x}

=\frac{8}{17}

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Answer:

200

Step-by-step explanation:

We have:

4(-1+2-3+4-5+6-7...+100)

We can rearrange the numbers to obtain:

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From the left, we can factor out a negative. So:

4(-(1+3+5+7+...+99)+(2+4+6...+100))

In other words, we want to find the sum of all the odd numbers from 1 to 99.

And the sum of all the even numbers from 2 to 100.

Let's do each one individually:

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We can use the arithmetic series formula, where:

S=\frac{k}{2}(a+x_k)

Where k is the number of terms, a is the first term, and x_k is the last term.

Since it's all the odd numbers between 1 and 99, there are 50 terms.

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We originally had:

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And we're done!

Note: I just found a <em>way</em> easier way to do this. We have:

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Let's group every two terms together. So:

=4((-1+2)+(-3+4)+(-5+6)...+(-99+100))

We can see that they each sum to 1:

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Pick which one you want to use! I will suggest this one though...

Edit: Typo

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