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3 years ago

8

A jar contains 960 beans. Of them 600 are Azuki beans and the rest are cranberry beans. what is the ratio of the Azuki beans to

all the beans.

2 answers:

ira [324]3 years ago

7 0

600:360 simplify
10:6 simplify
5:3
Hope this helps.

victus00 [196]3 years ago

5 0

600:960 or, you can reduce it to 5:8

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When 2(3/5 x + 2/3 y - 1/4 x -1 1/2 y + 3 )is simplified, what is the resulting expression?

vovangra [49]

Answer:

Simplifying the expression: $2(\frac{3}{5}x+\frac{2}{3}y-\frac{1}{4}x-1\frac{1}{2}y +3)$ we get $\mathbf{42x-100y-360}$

Step-by-step explanation:

We need to simplify the expression: $2(\frac{3}{5}x+\frac{2}{3}y-\frac{1}{4}x-1\frac{1}{2}y +3)$

First we will solve terms inside the bracket

$2(\frac{3}{5}x+\frac{2}{3}y-\frac{1}{4}x-1\frac{1}{2}y +3)$

Converting mixed fraction $1\frac{1}{2} y$ into improper fraction, we get: $\frac{3}{2}y$

Replacing the term:

$2(\frac{3}{5}x+\frac{2}{3}y-\frac{1}{4}x-\frac{3}{2}y +3)$

Now, taking LCM of: 5,3,4,2 we get 60

Now multiply 60 with each term inside the bracket

$2(\frac{3}{5}x\times60+\frac{2}{3}y\times60-\frac{1}{4}x\times60-\frac{3}{2}y\times60 +3\times60)\\2(3x\times12+2y\times20-1x\times15-3x\times30-3\times60)\\2(36x+40y-15x-90y-180)$

Now, combine like terms

$2(36x-15x-90y+40y-180)\\2(21x-50y-180)$

Now, multiply all terms with 2

$42x-100y-360$

So, Simplifying the expression: $2(\frac{3}{5}x+\frac{2}{3}y-\frac{1}{4}x-1\frac{1}{2}y +3)$ we get $\mathbf{42x-100y-360}$

3 0

3 years ago

Please answer question image below.

anastassius [24]

Yes it has a constant rate of change which is 3/2

3 0

3 years ago

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Alinara [238K]

Answer:

The answer is 3 hours I think.

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Which one is correct

bekas [8.4K]

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Step-by-step explanation:

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3 years ago

What is the sum of the first 51 consecutive odd positive integers?

Angelina_Jolie [31]

We call:

$a_{n}$ as the set of <span>the first 51 consecutive odd positive integers, so:

</span> $a_{n} = \{1, 3, 5, 7, 9...\}$

Where:
$a_{1} = 1$
$a_{2} = 3$
$a_{3} = 5$
$a_{4} = 7$
$a_{5} = 9$
<span>and so on.

In mathematics, a sequence of numbers, such that the difference between two consecutive terms is constant, is called Arithmetic Progression, so:

3-1 = 2
5-3 = 2
7-5 = 2
9-7 = 2 and so on.

Then, the common difference is 2, thus:

</span> $a_{n} = \{ a_{1} , a_{1} + d, a_{1} + d + d,..., a_{1} + (n-2)d+d\}$
<span>
Then:

</span> $a_{n} = a_{1} + (n-1)d$
<span>
So, we need to find the sum of the members of the finite series, which is called arithmetic series:

There is a formula for arithmetic series, namely:

</span> $S_{k} = ( \frac{a_{1} + a_{k}}{2} ).k$
<span>
Therefore, we need to find:
</span> $a_{k} = a_{51}$

Given that $a_{1} = 1$ , then:

$a_{n} = a_{1} + (n-1)d = 1 + (n-1)(2) = 2n-1$

Thus:
$a_{k} = a_{51} = 2(51)-1 = 101$

Lastly:

$S_{51} = ( \frac{1 + 101}{2} ).51 = 2601$

4 0

3 years ago